rllOPOLOGY
Lipman Bers
Notes by: and
Jacqueline Lewis Esther Rodlitz
27
New York University 1956-19.57
-1-
TABLE OF CONTENTS Chapt. I
POINT SET TOPOLOGY NO'cation and Preliminary Definitions ........... . Vector Spaces Scalar Products and Hilbert Spaces • .c: _ • • ,. • • • • • • Normed Vector' Spaces . Metric Spaces c • • • • • • • • • • • • • • • • • Continulty in a Metric Space •••••••••••••••••• 0 ••••••••••••••••••••••••••••••••
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Cartesian Product of Two Metric Spaces ••••••• Topological Spaces (Definiti9n) ••••••••• Ha~sdorrf Spaces •••••• o • • • • • • • • • • ~.~ • • • • < • • • • o Closed Sets o • • • = Interior, Exterior and Boundary Points $
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I!lduced Topology ....... eO" . . . . . . . . . . . . . . . . . . . . . . . Continuity in a TopologicB.l S.pace ••••••••••••• Homeomorphism ••••••••••••••• ~.~ ••••••••••••••• Ax1.:;ms of Countjab:tli ty ••••• " ........ ~ •••••••••• Convergence .~ ••••••••••• Completion of Hetr:'..c and Normed Linear Vector Spaess •••••••••• ~.Q.&~oo.~.o • • • • • • • • • Baire Catego~y Theorem ••• ~ ••••• ~~ ••••••••••••• Compactness , •••••••••••••• l« •••· . . . . . . . . . . . . . . . . Fundamental Cube in (Separable) Hilbert Space • Connectedness c •• ...... . Topological Product ••••••••••••••••••••••••••• Metrization Theorems ........................... . 0
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SIHPLEX AND COJVIPLEX Closed Convex Hull of a Set • • • • • • • • • • • • • • • • • • • Points in General Position ..................... The Geometric Rectilinear Simplex ••••••••••••• Ba.rycentric Coordinates ••••••••••••••••••••••• Definitions of Faces of a Simplex and Properly Situated Simplices •••••••••••••••••
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24 25 26 27
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CARNEGlr IRSTITUi~
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OF CONTENT.§. (Oontinued)
The Canonical n-Simplex ••••••••••••• , •••••••••• Geometrio Oomplex e c • • • • • • • • • • • • • Dimension of a Complex ••••••••••••••••••••••••• IKI, The Space of the Complex K ~ •••••••••••••• Polyhedra and Triangulations ••••••••••••••••• Subcomplexes o • • • • s • e _ • • • • • • • • • • r-Dimensionc.l Skeleton •••••••••• "•••••••••••••• 5t&r of a Vertex ••••••••••••••••• ~ •••••••••••••• Abstract Complex ., •••• $, •••••••• " •••• " ........... . Gcometl'ic Re8.1iz6. tion of an n-Dirnens ional AbstJ..'Clct Ccmplt.:;:;r in R.2n+l ,. •• " ••••••• < • • • • • • • • Sil:-1plic. ial JYIe.ppings •••• ~ • 0 " 0"" .......... " • Isomorphism of Two Abstract Complexes ." ........ . •
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Ordered and Oriented Simplices •• ~ ••••• ~ •••••••• Support of a Simplex •••••••••••••••••••• "••• e , • Integral Oriented r-Ohains •••••••••••.••••••••• Elementary Chains •• Boundary Opera.tor CJ ••••••••••••••••••••••• , ... .. Support of a C.l:':.a:i.n ............................. . Poincare Rela·::;iol1, 6 2 x = 0 cc~.v." • • "~e·~e • • • • • ~ Boundaries, Homologous Chains ~ a!:d Cyoles ••••• " r-Dimensional Integral Homology Group of K ••••• Scalar Product of Oriented Simplices 0
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33 34 34
35 35 35 35 35 36 37
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42 Coboundary Operat:)r 0* ........................ ~ 43 Coboundaries, COhoillologous Chains and Oocycles • 43 and Chains • ., •••.
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r-Dimensional Integral Cohomology Group of K e • • Incider.. ce and Coincidence 1'latrices • Chains Over Arbitrary Abeleare Groups •••••••••• Cochains ••••••••••••••••••••••••••••••• d as a Cochain Mapping ••••••••••••••••••••••• Hr.(K,G) and Hr(l'C,G,r) ••••••••••••• Homomorphisms f, of Chains ~ Induced by Simplicial IVlappings •••••••••••••••••••••••••• 0
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-i11TABLE OF CONTENTS (Continued) Chapt. ~f == fd • • • • • • • • • • • • • • • • • • • • • • • • Induced Cochain Homomorphism (f )
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••••••• ~ •••••••••••••••• o • • ~ • • • • • • • • Baryoentrio Subdivision of an Abstract Complex e Alg. Dim. GmK = Alg. Dim. K ••••••• 0." ........ . 53 Baryoentrio Subdivision of an Abstract Complex • 53 Maoh of a Ge:nnetrio Complex ........ c . . . . . . . . . . . . . 53 Cones over Simplioes and Cha.ins ••••• 54 K:;:ooneoker In';:·~!.~c t'D r ••• " •••••• ,. ••••••••••••••••• 54 Ba~ycenters
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Subdivision ~f Ch~ins ••••• e.~ • • • • c • • • • ~ • • • • • • • • Superdivision of Chains •••••••••••••••••••••••• .... -m ==0-m""\. rr'*~ m* = \ J r;-!l1* CJ ...... 00 DIU \:31 ••••••••••••••• Spe:t'ner Mapping, (S5.mplioial ApproxilflS'.tion to the Identity) •• " ••••••••••••••••••• ... m Spernel" 's Lemma, l () x = x • ., ••••••••••••••••••
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DIMENSION THEORY .. Coverings: Ordar of, Finite, Refinement of ••••• ~9 Definition 1 of Topological Dimension •••••••••• 59 Monotonic1ty of Dimension •••••••••••• ~ .......... 60 Lebesque Number of an Open Covering •••••••••••• 61 Diameter of a Covering ••••••••••••••••••••••••• 61 Definition 2 of Topological Dimension •• " ••••••• 61 Nel"ve of a Colleotion of Sets •••••••••••••••••• 6,3 Nerve and Order of a Covering •••••••••••••••••• 63 Stars ........ 64 Top. Dim. lKj Alg. Dim. K •••••• ~ ••••••••••••• 65 e-Mappings "." e. • • • • • • • • • • • • • • • • • • • • • • • • • • •.• • • . . . . 68 ~ Pal"t1tion of Unity Subordinate to a Covering ••• 70 Dimension Theorem for Compact Subsets of 73 Unit Cube I in RZn+l ............ ~ ................. 74 TheSpaoe IX • o •••• " . . . . . . . -• • • • • • • • • • • . • • • -. • •• • •• .14 II • • • • • • I I . ' • • • • • • • 0 " . . . . . . . . . . . . . . . . . .
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CO,NTENTS (Continued)
Chapt.
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Imbedding Theorem for Finite Dimensiona.l Compact Metric Spaces •••••••••••• a • • • • • • • • • • • Lebesgue Definition of Dinension for a Separable Metric Space •••••••••••••••••.••••• Urysohn-Menger Definition of Dimension (Inductive) ••••••••••• Compactification •• c • • • • • • • • • • • • O • • 6~O.~Q • • • C • • 9 Dime~sion of a Separable Metric Space in Terms of a Gompactification •••• ~ ••••• c • • • ~ n-Sphere ~ ••••••••• ~ ••••••••••• &o • • • • • • • • • • e Top. Dim~ R_.... = n ••••••••••••• e • • • • • • D • • ~ • • • • • • • 0 ••••••••••••••••••••••
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78 78 79 79 79 79
OF DOJYIA IN
of Contracting Mappings ••••••• ~ •••••• Principle of Non-Expanding Mappings ~ ••••••••••• Lebesque Number of a Collection of Closed Sets in a Compact Space •••••••••••••• ~ Covering Theorem for an n-Simplex ••• c . e • • ~ • • • • • Brouwer Fixed Pain t JJ.1heore:r.n •••••••••••••••••••• Retraction Mappings O • • • • • • • • • • QO • • • • • • • • , • • • • • • Theorem on the Invariance of Domain •••••••••••• Appltcation of tile Theorem on In'ITariance 0f Domain .... ~ ••.' ••••••• " " ••••••• oe c • ~ . . . . . . ~ • • Approximation of a ConveA Compaut Subset of a Banach Space by Finite Dimensional Spaces .0 Schauder Fixed Point Theorem (Weak Form) ••••••• Mazur 1 s Lemma ..................................... Schauder Fixed Point Theorem (strong Form) ••••• The Spaces Co and 01 ••••••••••••••••••••••••••• Holder Continuity •••••••• G • • • • • • • • • ~ • • O • • • • • • • • Equicontinuous Functions and the Theorem of Arzela (Aseoli) ••• ~ ••••••••••••••• , ••••••• Applications of Fixed Point Theorems ••••••••••• V
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81 83 83
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H01"10LOGY THEORY (Part I)
Groups •••• o • • • • • • • • • • • • ~ • • • • • • • • • • • • • • • • • • • • • • 103 Factor (Quotient, Difference) Group ••••••••••• 104
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TABLE OF CONTENTS (Continued)
0.
Natural Homomorphism of A onto AlB ............. 104 Linear Dependence and Independanoe ••••••••• ,.$ 105 Rank of a Group .............................. ".105 Maximal Linearly Independent Set •••••••••••• ,. 105 Rank A = Rank AlB + Rank B ........... " •••• "...... 107 Generators oo •• • • • • • • • • • • , .. ·• • • o • • • • • • . . . . . o 109 Direct Sum o • • • • • • Q• • e • • • 109 ".~
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Fundamental Theorem of Abelain Groups ••••••••• Betti Numbers and ~~rsion Coefficients •••••••• Nctaticn Q.~ • • • • • • o • • • • • • • ~.o • • • • • • • • • • • • a •• s • • Euler-·P~Jince.rs Formula ••
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HOMOLOGY TI~ORY (Part II) Allowable Homomorphism •• ,ae . . . . . . . . . . . . . !1 • • • • • • Algebraie Homotopy Operator ••••••••••••• ~,~ ••• TheoI'.em on a Homomorphism Between HI'(K) and HI' (L) ............ , , •• , ••••••••• ~ ••••••• , • Homology Groups of a Cone •••••••••••••• ,. ••••• " Hr{K} -;;:; Hr('-i'K) •• "." •• 10 ~
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combinatoria11y Close Simplioial Mappings •• ~ '._ c;K, t1l6 Oylinder Over the complexK •••.H H ... '~ t7$
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113 114 115 116 116 118 121 123 125 126 126
128 129 129 130 132 134135 139 139
TABLE OF CONTENTS (Continued) Chapt.
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Homomorphisms of HI'(L) and Hr(K) Cohomology Groups of a O-Complex and a Cone ••• •
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HOMOLOGY THEORY (Part 1111 Ordered Chains •••••••••••••••••••••••••••••••• Ordered Homology, and Cohomology Groups ••••••• Isomorphism of the Oriented and Ordered Homology Groups ••••••••••••••••••••••••••••• Isomorphism of the Oriented and Ordered Cohomology Groups ••••••••••••••••••••••••••• Rings and Ideals G • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Exterior r,lultiplice.tion (Cup Product), 1\ ••••• Cr as a Ring ••• ~O • • • • • • • • • • • • • c • • • • • • • • • • • • • • • ~uotient Rings .~ ••••• e • • • • • • • • • • • • • • • • • • • • • • • • HI' as a Ring •••• :t., • • • • • • o • • • • • • • • • • • • • • • • • • • • •
VIII HOMOLOqy THEORY (Part IV) Graded Groups and Rings ••••••••••••••••••••••• Differential Forms of Degree r •••••••••••••••• Grassman Algebra over a Vector Space ••••• , •••• Graded Group with a Differential Operator, d •• Graded Ring with a Differential Operator •••••• Exact and Closed Elements of a Graded Group (or Ring) ••••••••••••••••••••••••••••• Derived Group ••••••••••••••••••••••••••••••••• Allowable Homomorphism •••••••••••••••••••••••• IX
l44
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149 150 150 151 152
153 154 154 156 157 158 161
1'1 162
~OMOLOGY
THEORY OF POLYHEDRA Simplicial Approximation •••••••••••••••••••••• Simplicial App~oximation to the Identity (Sperner Mapping) ••••••••••••••••••••••••••• Existence ot a Simplicial Appro~imation to a. Continuous Mapping ••••••••••••••••••••• Uniqueness ot the Homomorphism of the Homology Groups Induoed by Difi'erent Simplicial A.ppro~1mations ...............................
166 168 169 170
-viiTABLE 9F.Q..oJ:ITENTS (Continued) Chapt.
Page
Homotopic Nappings •••••••••••••••••••••••••••• Homotopic J:..la.ppings Induce Same Homomorphism of the Homology (Cohomology) Groups ••••••••. 6 Fundamental Theorem of Combinatorial Topology c Homology and Cohomology Groups of a Polyhedron X •••••••••••••••••• ~ ••••••••••••• X
li.QIJIOTOPY Glli'}1£3 Ct>art I} Curves in a Topological Space Product of Two Curves •••••••• eo • • ~.~ • • • • • • Homotopic CUI'vas ......... a " • • • • eo • • • • • • • • • • • • • • Fundamental G.r01J.p of X at a Point XI F(X,x) or 1f, (X,x) •••••••• Inner Automorphisms c . o • • • • e • • • • ~ • • • • • • • • • • • • • • F(X~x) ~ F(X,y) 1f X and y Can be Connected by a Curve in X " ........ " ... e • • • • • • • o • • ,. • • • • • • • n-Curves ••••••• f,(I~.' c. • • • • • • • Product of n-Curves ••••••••••••••••••••••••••• Homotopic n-Curves e • • • • • • • • • ~ • • • • • • • • • • • • • • • • • n-Dimensional Homotopy Group I In (X,x) ...... ", •• Co~nutativity of l1n for n > 1 •.•••••••• ~ ••••• Product of a l-Curve and an n-Curve e~
171 172 173 173
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HOMOTOPY GROUPS (Part II) Commutator Subgroup ••••••••••••••••••••••••••• Paths ••••••••••••••••••••••••••••••••••••••••• Elementary, Special and Simplicial Curves (Paths) •••••••••••••••••••••••••••••• Eimplicial Equivalent of a Special Curve •••••• Mapping, 4·., af Special Paths into Chains ....... Homotopio Curves are Mapped (by into Homologous Chains •• " ••.••••••••••• "" •••••• ".. For K Connected, H~(KJGo) ~ lIr (IK/) ••••••••••
*)
XII
175 175 178 178 178 180 180 181 181 181 182
185 18& 187 187 187 188 191
CONTINUOUS JYIAPPINGS OF THE n ...SPHERE INTO ITSELF Homology Groups of the n-Sphere ••••••••••••••• 192
;"'viii-
TABL?. OF' CONTENT§. (Continued) Chapt. Fundamental Group of the n-Sphere ••••••••••••• Degree of a Mapping ••••••••••••••••••••••••••• Construction of a Mapping of Arbitrary Degree • Veetor Fields " Singularities • • • • w • • • • • • • • • • Degree of a Vector Field •••••••• , ••••••••••••• Vector Fields of Exterior and Interior Normals to Sn •••••••••••••••••••• ~e" • • • • c • • • Brouwer Theorem for Sn (n Even) Brouwer Theorem or.. Tangentia.l Fields ••• ,. ............ lfopfis TheoI'em: E,(f) = S(g) imp:!..ies l' ':;' g .. cu.
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XIII HOMOLOGY AND COHOMOLOGY THEORY OF COMPACT SPACES Partia.lly Ordered Sets •• o.c • • • • • • • • • ~ • • • • • • • • • 202 Directed Sets .............. ~ ••• v • • • • • • • • • • • • • • • 202 Direot System of Groups ••••••••••••••••••••••• 203 Inverse System of Groups •••••••••••••••••••••• 203 Projection Mappings ••••••••••••••••••••••••••• 204 HI' and Hr as Direct and Inverse Systems •••••• ,. 205 Direct Limi~ of a Direct System of Groups ••••• 209 v r-DimensioneJ., (Oech) Cohomology Group.:! of X, a Compact Space, H1'(X} = lJ~H~(N(a)) •••• ~O~ 209 Strings •••••••• a , • • • • • • • • • ., • • • • • • • • • • • • • • • • • • • 2:09 Inve~se Limit of an Inverse System •••••••••••• 210 Hr(X) = .Jjm Hr(N(a.» .......................... 210 Cafina1 Sets •••••••••••••••••••••••••••••••••• 213 \/ Isomorphism of the eech and the Simplicial ' Groups £01' X = IKt •••••••••••••••••••••••••• 214 Alexander Co~mology Group ••• o • • • • • • • ~ • • • • • • • • 217 v Equivalence of the Cach and Alexander Groups for X Compact •••••••••••••••••••••••• 218
XIV
INDEX
Part I - Finite Dimen-
Urysohnfs Lemma ••••••••••••••••••••••••••••••• 219
-.J.A-
TABLE OF CONTENTS (Continued) Chap~.
Page
Tietze Extension Theorem ••••••••••••••••••••••• Degree of a Mapping (on Sn) ••••••• ~ •••••••••••• Properties of Degree ••••••••••••••••••••••••••• Rouchets Theorem ••••••••••••••• w • • • • • • • • • • • o • • • Admissible Mapping ~ ~ Degree of a Mapping .,o • • • • • •••••• e •• o •• 2,trong Form of Homotopy Invariance and R01:;.che fS Theore:;.n ;: •• eo . . . . . . . . . .
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Jordan Curve Theorem Invariance of Do:':nain I11dex ~*~ • • • ~4I'.".,.C! • • • • • • • • • • • • • Q.,.e • • • • • • Applications to Function 'I'heory ........ '
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LOO]'.L DEGREE J"ND INDEX (Part II - Banach Spaces,
-;[chaud~r-L~ray-~~Ciry)
Lemmas on Mappings in Rn ••••••••••••••••••••••• Admissible Mappil1gs ~ •••••••••• o • • • • • Completely ContiLl.2011S Mappings ........ Extentions of Contlnuous I:1applngs Allowable Ap}J~'oximations . 0 . 3 • • • • :> • • 0 . . . . . . . . . . . . Jordan-Brou't-"er .. Leray Theorem ,. .................. .,. ~ Invs-l'iance of Domain ••••• " .............. ~ 0
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254 254
x ;~rra
ta and Addenda
The page ts given followed by the line. An asterisk means counting from the bottom of the page. ~'lhen the original text Is not given, insertions and corrections are underlined. f some x G X ••• ••• is ~n abelian group with respect to addition for which a product ax ~ V is defined whsnever •.. . .• in a normed l:inear ••. .. 31 should read 3 i whenever d(xo'x) < 6 l.u.b. d(fy,gx) • •• 0
6 - Sol.. 7 - 9-l:-
9 - 14-;.. 10 - 1 12 12 14 14
-
..
XE.X
18 14*
Q._A)o · .. if B = A G where G is an open set •.. ..• any open G eX, with x € G, there .•• Take Gl =
5
Fa~(to)
2
•.. for each to E (0,1).
8"
17
--
. . . :-=
l6~~
15 16
8
~
'I....
17 - 4olt18 - 9-;1-
n
=
co
Then (1\ 11
H'
-i
=x E X --
r/2.
<
18 - 6* 21 -
8~~
21 23 24 24 24
4oli-
.. • space X contained in some topological space X' is ... · •• Bore 1 ...
1
• •• < •••
11
A sequentiall~ compact ••. • .• a sequentially compact ..• d ( x, xN .) < 1 IN
..
12 7ol~
------.-1
24 - 4-l"
· .• form x
25 .. 7 25 .. 8
•.. RN is · .• of
= {t: j} ... sequ~!1ally
{Xn} ~ich we
compact .••
denote simply by {xn1 whose •••
xi
25
-
25
- 14
25
30
- 15 - 7.;;- 12
31
- 1.;:-
25
13
33 ·';'"13 34 - 11 34- ~.,;23 35 36 - 1 3.,;, 36
"(~ 1="2 , C;'S
•••
,~ M,
0,
)11
•••
Replace by d(x(M),x ) M
n
· ..;> . . 1 (s.l - i
-c: n ) 2
-------
11m
xn
n--;::.(X)
--_--1 N Xi - (x i' ... , Xi )
--------€i < B/k
Aj-;:-Oa
--si!nplex · . .point · . • point 51"~~.0 , " ' , ___
ir in
IT( I Iv I "
not K ••• not K •••
"
a r "\j
"'1 a l,a i + l , ... ,a i +k1 aS3ign
the corresponding face
of {e i' •.• , e i + k} for i +k < r.
37 38 40 41 Lt 6 46 59 59 60 60 61 61 67
- 6'" - 8 - 6~~"'ii"
-
12-~-
- 12-><- 4* - 12 - 17 10 - 14 -
-
••• K not k •••
f(a.) ::: b. not bj l J Cr(K) not Cr
laxl~" a:f 0 Let x, not set x ;j .;:. : C r ( K, G, A
/\.
r) --;: .
{H ~ \. P J
C r+ 1 (l{, Gz
r)
should be ~H{3l t
)
add lip. 24" each G.l is open in X " fA
1
n
HI' ••• ,A
-
n
Hg \ ". .\
-< -"i;";{j
9
{j
6 ~~.
dim X < n should read dim x < n Cr not Cr
9~~
()m not sm 67 - 3~~ {)m not :rm 67 - 1~~ o 68 - 10":- X is interior of X
73 - 3
o
0
X ::: 0 not X
:J
0
xii
11 75 75 - 5" 15 79 81 - 11 83 - 2 84 - 4.;<-
7.;"
87 93 94 101 102 105 111
- 5~:- 4~~ - 3 - 6~:. 10
-
12 112 - 9 112 - 14 113 - 3 III
119 120 127 128 132 133 136 137 139
-
139 149
-
153
-
3 4 9';l-
8.,,, 15
••• such that d.,( x, x") < e C~ not C
~-Jhenever
AC B .: : : Ilx' -x" I
K =
1x/llx II
{a i
, ... , i
.
a
2 l~ ~
}
o r unlabelled point in the second diagram should be x' Le!...lA.l(t) ..• Consider the restriction •.. C.l not G.l Lipschitz.. AlB ~ AI (~means is isomorphic to) The principal conjecture of combinatorial topology is that hO:118omorphi.c complexes have isomorphic subdivisions. The It con jecture ll as stated in the notes 1s easily seen to be untrue. The cone and the point have the same homology groups but they are obviously not horneomorphi£. · .. Chapter IX ••• ;t not .;:; not ~ ',l;1
r
r
t· ql , ••• , 't.::qr (a,b,cl · .. i . e. y - y! delete ttv.Jell ll f:
::/ (~x •..
Cr(T\:,G)
if'd='dIJ
9~"
10 1 • •• r zeros ••• 12";- Sl,S2""'~ 7''- "c.x + Gd x = X O -A
7?~
14
d( fxt , fX") < 6
(Cl*t:)X
=
_
xl
c(ax)
- - - . ; . . _ . - - - o ; .. ..:......~
coboundary should read cocycle
xiii
-
153 157 164
-
- 4·
167 169 170 170 171 171 172
15 3 14
-1
11-;"~
-
(d im .; = dim Ol-l'".;) insert "space" to res.d "vector space over F'." "over D. into r-forms over DtI should read "over D into r-forms ove r !.::::..It f(St(a»)C •.. "m 3 fll should read "m such that fll f:
1M I
9
lKI f <;
ILl
->
...
IL I
2
~
-
12
has been used for algebraic homotopy (relative to (jn& ermL)
4-ll-
1M I = Ierml'( I
-
3,,1172 173 - 8 174 - 8 174 - 14 1" 174 1 178
-
"'i\~
-
•• . 1/N + l/rJ
I 1/ N
)t. :=
2!N •..
for ~l and f liN ---. Hr(Y,L) should read Hr{Y,G) Hr(Y,L) should read Hr(Y,G) if f
N
fl.
aa- 1
Lemma:
N
ca
178
-
12 -;}
178
-
5'~-
h(x) .":: 3-1xg
181
-
11';~
a-l(t l ,·· .,t n )
3
.•• p. 105.) • • • t ••. ••• AG lS an ~nvarlan
187
-
12';'"
(all)
187
-
8-J..~
"'="'l-··A s
10
g(sv)
12 12 8';l-
A ;;: ~
185 185
188 188 189 189 189 192 192
- 11';'"
-
_
4.l~ A
-2
-
4"
_ i'('
... i~~rs~-2.~ a] .
~(A)
:=
a(1-t l ,t 2 ,·· .,t n )
2m r - a
= N
f(t v _ I )
q(hl:L
not x \jf not x •.• consisting of S ... \jf
a1so~
xiv
193
-
14-l~
••• H (s ,G ). n n 0
193
-
12~~
knot g
L
/'..
£i s
,
194 - 5
f(v} ::
196 ... 14 196 ... 15 197 4 8 197 197 - 9 209 11* • 213 - Lt." 217 - 15 217 - 7* 220 - 1 221 - 13* 223 - 13-l} 225 - 2* 231 - 1
dilation dilation Consider F(x,t) = F(l,x} Let t :: 0 To get Lemma !t f : 0 should read ~ ~ 0 last term on right should read '$( x o ' ••• , x r +l ) q/2n should re ad q/2 n X should read F Sn is the same as Sn in previous chapter. deg(cP,G,p) = 0 lim y nj
234 ... 6 234 - 13* 237 - 31~ 249 ... 6~~ 249 - 4"OX'
delete <
-
-
~II."
252 - 1 252 - 3
253 - 3
i
Ji
...
deg(~,~,Di,Dj) should read deg(wcl>,Di,D j )
~ is interior of B
add G is bounded Change to "Finally cP = x - l-l<- will be approximated by an admissible map ~l = x where takes B into a finite dimensional subspace A of B. The inequality should be taken over all x £
'*
11
11
8.
o
G should read G. \jI should read cP
1
Chapter I In this chapter we shall present a brief review of general topology. Many proofs will be merely sketched, and still others will be omitted entirely. Furthermore, we offer no pretense of completeness~ but shall state only whatever may be directly or indirectl'] utilized in our more detailed study of algebraic topology; and finally, as '-Jill be seen later, we shall not concern ourselves with the most general topological spaces, but will always assume the Hausdorff separation axiom (unless the contrary is specifically stated).
-Notation 1'.fe shall use the ~;;rnj~2l 0 to denote the number 0, the null vector, the empty set, the neutral element of an (additive) Abelian group, etc., without specifically announcing the fact if the meaning is made clear from the context of the discussion. 2) A ::=:::::::-.:-.p B is to be read itA implies Btl 3) By a E.~J;, we mean a collection of elements a} x e X means x is an element of X b) X = ixlx is means that X consists of the elements x such that x is,P, where P is the defining property of the set x. c) xey or y~,1X means a e x::::=:~:::> a c Y. It is read IIX is included in ytl or Ux is a subsel of ytt. (-.-d) X:jY = L alacX or acYl is called the union of X and Y. XflY = [alacx and a8Y) is called the interl J ~ction of X and Y. e) Let Y eX. The set of element s in X that are not in Y is called the complement of Y with respect to X. It is written X - Y. f) Let Ct be a collection or family of sets. v,Te define the union and intersection of sets A in t<: 1)
p}
,
",
2
i) 11)
~A::::
{alaEA for some AE..Q.
(1
f a Ia G. A for
A ::::
11..'/
g)
all A
~
e.
.3
aj
De .J1orgal1;.. Law~ Let Cl., be a collection of subsets of X. It follo·H~ from e) and i') that: i)
ii) Manp!~eas
X X -
0:r:
4J A
~
n
Q
A
==
n
(X-A)
OJ
= U(X-A) 0./
Fun.ctions: Let X, Y be t"lr!O arbitrary sets. By a mapping f of X into Y, written f: X --> Y, we shall mean a rule that assigns (uniquely) to each x ~X an element y IS Ye We shall denote the element y 15y fx or f (x). Note that we have said that X is mapped in:~£ Y. This means that not necessarily every y~ Y has the property that fx :::: y for some x ex. For the case where ~y y e Y is the image of some x Ii: X, we shall say that f maps X ~ y. For a subset At: X, we define: f (A) :::: {y Iy
~
Y and y :::: fa for some
ThiCl is called the direct
~m~~~
aC~}
of the set A.
Similarly for a subset Be Y, we define the
-inverse .- .....- -image of .....~
B under f b.y:
1 r 1 f- (B) ==tx1x€.X and fx€:BS
Also [-l(y) :::: r- 1 ({Y}) ::::
·{xlx€. X
and fx ::::
Using this language, f maps X ~ Y whenever f(X}
a
=Y.
Let x" Xl ~ X with x :f x' • If fx t:- fxt, then we say that f is a lI one -to-one" or 1-1 mapping of X into Y. In other words, for ye:.Y, f-ly is either a set ronsisting of one element or the empty set.
3. We may now state precisely what we mean by a finite or denumerably infinite set. a) S is a finite set if there is a 1-1 mapping of S onto the set-of natural numbers ~1,2, ••• ,n:rfor some n. b) S is said to be denumerably infinite if there is a 1-1 mapping of S onto the set of the natural numbers. By a countable set, we mean one that is either finite Or denwnberably infinite. Compos~ mappings: Let X, Y, Z be three arbitrary sets. Assume that we have two mappings: g: Y---ll>o Z f: X ~ Y, which we may write as
X~Y.J~Z Now fx6 Y and so gf'x£. Z. ~omposit~ mapping:
Thus we may define the
h=gf:X~Z
Algebraic Concepts: The following definitions and elementary facts will be needed in subsequent discussions. \'I!e assume that the reader is familiar with the concept of a group, in particular, an Abelian group. A ~ctor space V, (or linear space) over the real numbers in an Abelian group with respect to addition for which ax €: V whenever ~€.. V and a. is a real number" Also, for a.,{3 real numbers, and x,y€ V:
i) ii)
'" ,
!(x+y) = ax + sy, (o.(3)x
= a.{ {3x)
(o.+{3)x = a.x + {3x
Ix =x
The elements (vectors) Xl' x 2 , ••• ,xn , in V are said to be linea;r',ly de:l2~n~ent if there exists real numbers Yl' Y2'··.'Yn with at least one Yi
o.
y.x. ::: l
l
- . ..
:f
0 such that
otherldse the vectors are said to be linear'ly
independent~
...-
A vector space has (alge'l?,.J?aic) dime~ion n if there exist n linearly independent vectors and if every n + 1 vec-
tors are linearly dependent. Every fini1J8 dimensional vector space with dimension n is iR0IDoX'phlc to the vector space whose elements are ordered n-tupl.es .. A linear map of an n-dimensional space into another ndimensional space is onto i.f and only if it is
!:!..
Scalar. Pl:2.duct s __8n~ .Hi.lbe~t Spa..£~~: L scalar P.2:9_duct in a vector space over the real numbers is a rule which assie;ns to every pair of elements x,y a r'eal nlli11ber denoted by (x, y) sat isfying the following properties: 1) (0,0)::: 0 (~:be first tl·!O appearances of 110" are as the null
elemen~
(vector), the third is as the real number 0)
f
2)
(x ~ x) > 0 H' x
3)
(yx,y) =,y(x,y) for y any real number.
4)
(x,y)::: {y,x}
5)
(x + y, ~) ::: (x;z) + (y,z)
0
As an immediate consequence of' these properties, we have the Sch1-Jarz inequality: (x,y)2
~ (x,x) (y,y)
A complete linear space in which a scalar prodl..lct is de.fined is called a Hilber~ Spa~. (A space is .<:.omplete if every Cauchy sequence in the space converges to some, element of the space)
Some Examples of Scalar Products: 1) The elements of the space are ordered n-tuples .of real numbers. We define a scalar product of x = ~1,x2, ••• xnJ and y (Yl' Y2' ••• Yn) as
=
It is quite easy to verify the five scalar product axioms. This space is called the n-dimensional Euclidean space aud will be denoted by Rn. The .elements of the space are inf~ite sequences x ~ (xl ,x 2 , ••• ) of real numbers such that ~ xi < ~ 2)
1
The scalar product is defined as
Convergence of thie sum is assured by the SchHarz inequality. This space will be called :1w (The notation is suggested by the common usage of w to represent the first transfinite ordinal number., .that is, the order tY"0e of the set of natural numbe!'s '{l, 2,.0 It will also be called '£'2' the sepa!'able Hilbert
e}
spac~.
3)
The elements are x(t), real valued continuous funct~ ions defined orJ 0 < t < 1. The scala!' pI'oduct -is defined as (x,y)
=S: x(t) y(t) dt
This space is usually denotedly C(O,l}~ Normed Vector~paces In a vector space with a scalar product, we may define the norm of an element x, written II xII , as
/I xII =
(positive square root) In Ri , 1 - 1,2,3, this is merely the :Euclidean length of the vector. v'{x,x)
It follows immediately from the scalar product axioTllf! that the norm has the following properties:
=0
1)
II
2)
II x II > 0 if ~ j:. 0 IIYx Ii = Iyl -11 xII II x + yll = Ify+xl! fix + yll ~ II x If + 1/ y II
3)
4) 5)
0
II
We shall call a linear (vector) space in which we have defin.ed a norm satisfying 1) -5), a !?;orrned linear (v:,ector) ~pa~~. That iS I a ~ in a linear space is an ~gnment of a real non-ner:a!.::t:!e nurabe!:. t.o every element of the space which satisfies properties 1) - 5). We have seen that a norm can always be introduced into a linear space with a scalar product. The converse is not always true. This means that if we are given a normed lir!ear space, we cannot always introduce a scalar product "t-Jhich will give us back the original norm. In fact this can be done if and only if the norm satisfies the condition
I(x ... YII2 +
Ux "'"
Y1l2;='211 x II'~ + 2.
fI
y
If
(This condition is easily seen to be a generalization of the law of cosines for triangles) Metric Spaces In a normal linear space we may introduce the idea of a distance d(x,y) between two points x and y. We define d(x,y} =
fI
x - yll
It follows immediately from the properties of a norm that
1) d(x,x)
=0
d(x,y) > 0 , x ~ y 3) d(x,y) = d(y,x) 4) d(x,y) < d(x,z) + d{z,y) 2)
Any linear space in which we have introduced a distance function d(x#y) satisfying 1) - 4) is called a ~etric space. Although every normed linear space is automatically a metric space, we cannot alwa,)Ts introduce a norm into a metric space which VIill Give us be.ck the original metric. For example, consider a space whose elements are infinite sequences of real numbers. We define
=L co
d(x#y)
1
That will satisfy all the metric space axioms, but an attempt to derive a norm from this will lead to a violation of the third norm axiom. Let S be a metric space. The diameter of S, written diam S is defined as dj.a.m S ::: 1 ",1.l1J b.
d tx>Y')
x,y,€,S
--- -
S is said to be bou.nded if and only if diam S < cD • S will be cD.lled totally bounded, if for every € > 0 # . S can be covered by a. fipite number of sets Si such that n diam Si < E for all L· That is, Set) 3i 'lrJhere dia.m. Si <
e
5.
for £ ::: 1~2, ••• ,n. From repeated application of the triangle inequality (axiom 4) He see that !2~.~1 boundedness' boundednes~. The converse will be true only for finite dimensional spaces. For, consider RO#. Let S={iIX€.R\V and 1/ xII ::: Diam S=2, so that S is bounded. However S is ~ totally bounded. To see this, we need only to consider elements in RWof the form
.:p.
iJ.
,
8. For E = ~" no two such xn can be in the same set 'frJithout violating the condition that the diameter of that set be less than ~. In a metric space X, the ball of radius r with center at Xo is the set
A'set S in a metric space is called 9~ if, whenever xes" there is some number r > o such that B (x)C S. That r is, every point in an open set must be contained in some ball lying entirely in s. Pr0E..er·t~~"~.2.!..0Den
Se~s..in
a Hetric Space~ i) 0 is open. (0 c ontalns no points and c
An B is open (For a point x€:A () B" take t:be ball of smaller radius). As a consequence n
nAi will be open whenever each Ai is open. It
iv) Let a be a family of open sets. Then (j A is open. 0..,. A and a s a l'e suIt 'frJill have a ball (Each xc. U A is in some q,.., around it lying in U A) ~
Let X be a metric space, S a. subset of X, and x an element of Xi) We define tbe distapce. bet'frJee~ the P2.int x and the set S as
We shall say that S is
~~
if
It follows immediately from these definitions that S is closed (or open)~' ~ x-s is open (or closed) and using properties i) - iv) of ,open sets 1<7e arrive at the following PFoperties of closed ,sets.
=
i) 0 is closed (X-O X is open) ii) X is closed iii) Let be a family of closed sets. The
na.. A
a
is olosed (De ~1organ Laws and property iv) of open ;:tets}
n i V)
U A. 1 ).
is closed whenever each Ai is closed.
Let X, Y be metric spaces. 'I'hey are said to be :J"sometric if thel e is a 1-1 onto map f:X -~ Y which pr'eserves distance. As metric spaces, we may consider X and Y as identical. ~ltin~ity
in a Hetric Spac~ L8t X, Y be metric spaces •. We shall say that f:X -+Y is
continvous at a point Xo€ X if given anye > 0, there is a
o
> 0 such that d(fx o ' fx) < C whenever d(xo'x) < S (although X and Y are diffe}~ent metric spac6s, we have used the same
lid" to represent tbeir respective dist ance functions. This I!inconsistencyll in notation will be repeated, and nG doubt similar on~ as well, whenever there is no danger of confusion). f:X ~y is contihuous on X if it is continuous at every point in X. The
netri.~ Space ~ Let X,Y be metric spaces.
Y is bounded.
Consider the
0
space whose elements are the mappings f: X ---i>' Y. (If X has n elements and Y has ill elements, then there are evidantl;v
n?
maps of X into Y.
This suggests calling this space of functions duce a metric into yX by defining
yX).
We intro-
1.(,)
d{f,g) = lub
(fx,gx)
x€.X
Axioms 1) -3) are trivially satisfied. inequality, we note that
To prove the triangle
d(fx,gx) < d(fx, hx) + d(hx,gx) ~
is true for
~
d(f,h)
+
d(h,g)
X€X and consequently for
lub d(fx, g:x.) xE.X
= d(f,g) car~~n
PE.2...duct of Two Hetr ic Spaces:
--
Let; X,Y be metric spaces. The cartesian Product of --".---..X and X XY = Z, is a metric space whose elements are ordered pairs (x,y) such that x G:X and y€ Y. The distance between two points zl = (x11Yl) and z2 = (x2 ' Y2) is -..--,,-
r.,
Examp1!: Rnx Rm
= Rn+m
{up to isometry}
Topological Spaces We have already defined an open set in a metric space. In order to generalize this notion we start with a collection of sets which satisfy the propertie.s that we have found to be true for open sets in metric sps.ces. We define a topological space X to be a non-empty set of elements in which we distinguish certain subsets, called ~~en sets, having the follm..ring properties: i) O,X are open .ii} IfCl is a collection of open sets, then ~A is open.
11
iii)
If A,B'are open sets, then Af),B is open (the same s.t~roent for any finite number of open sets follows ..J.,mm.ediately by induction). Rema~k: A metric space is automatically a topological space~ Arbitrary unions of balls (open spheres) in a metric space are the open sets of the topology$ In general, we shall be interested only in a particular kind of topological space. A space is called Hausdorff if, in addition to satisfying the conditions required for it to be a gereral topological spacej it satisfies the so-called H~us~2Ef! sepaEa~ion ~xio~:
iv) Giv,en tHO distinct points x,y€ X" there exist open sets G, G' such that x€.G, y£G t and G(lG' o. Remark: A metric space is Hausdorff. Indeed,
=
x -:f y G
'::d;
d (x, y)
={zJd(Y,Z)
<
=r
1]
> o.
G
= ~ Jd (x, z)
<
3-} and
are the desired sets.
A set F in X is said to be ~sed if X - F is open. From this definition and tb,e properties of open sets we have the following properties of .closed .sets: ...... i) 0, X are closed ii) IfUis a collection of closed sets thennA is a... closed iii) If Al , A2 , ••• ,An are closed, then Ai is
-
tJ 1
closed. Let A be a subset of a topological space X. the closure of A, written A, as:
A =(\
We define
B
B closed
B::JA
A may
be equivalently defined as the smallest closed set containing A. It is also immediately evident that A is closed
A=A
<___=> A = A
and
12 A point x is called an interior point of A if there is an op6n set GCA such tbat xc. G o The interior of a set A~ is a set A = xIx is an interior-point of An equivalent definition is
f
A}
~
o A = X -
(X - A)
A point x is called an exterior
poin~
of A if
X€X-A'=(x-A)o A point wh.::lch is neither interior or exterior to A is
called a ~9und~~r p~~~f A. The set of all suc~ points is the set theoretical ~~ndaE.l or the fr2ntie~-2LA. It is left to t~e reader to verify that i) The frontier of l~ is a closed set ii) A is closed~ ~> A contains its boundary. Induce~ ~opologi~~
Let A be a non-empty subset of a topological space X. We shall say that a set BCA is open in the topolo{W iI!duced on A or is open relative to A if B = An G where Gis an open set ,in X. It is easily verified ths.t this defines a topology on A. No't'J let A be a non-empty subset of a metric space X. We have two ways of assigning a topology to A i) The metric in X gives X itself the topology in ;,IThich the open sets are (unions of) balls in X. This topology induces a to~)ology on A (as described directly above). 11) A is also a metric space since it is a subspace of a metr1c space. Its metric automatically supplies it with a topology. Fortunately methods i) and ii} yield the same topology. An open set in i) is of: the forrll G where G is open in X. Thus G is the union of balls in X. Clearly An G is the union of' balls in the metric space A, and so A{)G is open in i1). Similarl.y for the converse.
An
13 continuity in Topological Spaces Let X,Y be topo1o~ical spaces. A map f:X~y is continuous at xeX if, whenev~r fxSG, where G is open in Y, there is an open set Hex such that xE Ii and f{H)C"G. The map ;f' :X~ Y is continuo~ .2!! X if it is continuous at every x€ X. !!:~~ark: For a metric space. this coincides wi th the previous definition. !heorel1!:l f :X-. Y is continuous on X<~ For every open set Bey, f (B) is open in X. ~~: Assurae that f is continuous on X and B is any open set in Y. If f-1(B) 0, the theorem is true since is open. Now let us assume f-l{B) 1- O. Let xe f-l(B). From our definition of continuity there is an open set H~X such that I
°
=
x
Therefore every point of f-1(B) is an ino
.-
teriol'" point ,of f- 1 (B), so that fwlCB) ;:: f- 1 (B) =X ... (X.~l(Bl) ~ which is open. Conve1:'sely if Be y ===9 f ..l(B) is open in X, we may choose H
= f .'.... (B).
HomeoFU.9rphisme or T£po1ogical £.lappings .9:efini1.?-£E..l.: A ?0!!l~9!llorphism or to'p?l~8..ic!.l mapping is a 1-1 (onto) bicontinuous map. definition 2: . A homeomorphism is a 1-1 onto map which . preserves open sets. Theo~em: definition 1~~) definition 2 ~oof: The second definition of continuity demands that the inverse image of an open set be open. Hence a 1 ..1 bicontinuous mapping preserves open sets. Conversely, f is an open map ~ f- 1 is continuous. In addition f is 1-1 and onto, so that f is continuous. -,-~
Basis and Sub-basis c:f a TopolofE:9al Spac.e, Let X be a topological space. A collection of open sets B is called a ~asis of X if every open set in X can be express.ed as the union of sets.in....B. -" CARNEGlf '~N'~"lTUTB O~ UCr.IH.QlOGY. ..JBRARY
14 A collection of open sets S is called a sub-basis of.X --..-...if open sets may be? expressed as unions of finite intersections of sets in S. A set ACX is SElid to be dense in X if A = X. X is called separable if it contains a dense sequence ~ampl~: The real numbers are a separable space, with the rational numbers as the required dense sequence) \rJe state the folloTrJing theorems without proof: ~heorem: X has a countable basis ~ X is separable. Theorem: X is a. separable metric space =)- X has a countable basis. Nested Basis at a Point Let X be a topologica~ space. Suppose that ror a point x~. there is a sequen.ce i } of open sets with the proper-
X,
£G
t~es:
i) x€ each Gi ii ) G.:) G-1 -'-1 J. iii) Given any open S6t GCx., there is some Gi such that' Gj, eG. . This sequence {GiJ is called a nested basis at x. !heo~!!.!: X h~s a countable basis ). there is a nested basis at every xc X• S J;'1. ~ is the countable basis. Choo se a'J. y .E!,9of: Suppose point x~'X and let;({~- (iilAi Ii and XEAiJ. Take Gi AI:' G2 Al A 2 , G3 11 A2 nA3 dnd so on. ok
•
p::::
= n
n
=
e
=
Aocording to the usual conventions, a space with a nested basis at eve';t~y point is said to satisfy the fi,~~ axiom of cot.mtabil:i.ty. A space Hith a countable bas,is (in the large) satisfies the ..second of-countability • - axiomWe have just shown_that the se~ond axiom of countability ~ the first axiom of countapility. The converse is not true. Examples 1) Suppose x is a point in a metric space X. The sets ,~ 1 ") 8 i = Lyld(X"y) < l' fomn a nested basis at x. We have just proved that a metric space satisfies the first axiom of count.... abilit:y-.
J
--- ---
15 2) Let us now consider the space of real valued func~~ on the. interval I = (0,1). Take some t o € (0,1) For any pair of real numb8rs a < 13, consider the set of real valued functions
These sets form a sub-basis for a topology in Lvhich convergence of elements is pointwise convergence of functions (See next section). It is easily verified that the topology generated by this sub-basis is Hausdorff. However the first axiom of countabili.ty is not satisfied, from vJhich we conclude that this space cannot be given a metric which 1--Jill give us back OUI' oY'iginal topology. In other words, pointwise ponvergence of functions canr:..ot be described by a metric spacee
...
£.~r Q;~
".
....
Suppose that X is a topological space and 1.. Xn~ a sequence of points in X. written
We say that [xn-~ converges to x, -'
or
lim x n......".
=x
if, given any open set G containing x 1 nearly all Xi (all but a finite number of Xi) are in G. Examples Let us look at the space X defined above of real valued functions defined on the unit interval. In the previous section we stated that conv'ergence was pointwise. Indeed, let f€ X be a funct::'on whose graph is shown in the diagram. An open set containing f consists of all func t ions g E X for
16 which a.
< g
(to)
<13.
Evidently f n-,,> f means that fn(t o )
~
f(t o ) for each to (0,,1). 2) For the space of ~ontinuo~ functions f(t) on the closed unit interval" we may describe convergence with a metric by defining d(f"g) = max te-[O"l]
If(t) - g(t)1
Clearly" conversence here is uniform converp:en~. Let X"Y be topological spaces. The mapping f:X~Y is called sequent!,a].ly ~}tinuou~ at Xo (j X" if whenever x n-> x, then fXn
--'!:>
fx
Obviously" the follo'l-dng is true: ~~~: f is continuous f is sequentially continuous. The converse is not true except in the case described in the following theorem" 'VJhich is submitted wi thouQ proof. Theore~: Suppose X"Y are topological spaces and X satisfies the fix'st axiom of countabili ty. Then, f:X --9l> Y is sequentially continuous =) fis continuous.
--7
Conversence in a. Metrl..c Spac~ Let X be a metric space~ to a point xE X whenever
A sequence
as n
~
~l..Xn?j
in X
converge.~
ClD
A sequence .f.x.n} for which d(x i , x j )
.........g,.
0 as i,j ~
aD
is
called a Cauchy sequence. ·A metric space X in 't'll'hich every Cauchy sequence conver'g3s is said to be ....... complete. In other words, X is complete if v-iiJ.enever d(xn,x.m )-+ 0" there is an element x€X such ths.t
.,...--
d(xn,x.)
-+ O.
Naturally, in a metric space, a convergent se-
quence is a Cauchy sequence.
17 '\AJe state the follot·ling theorems without proof: :rl'~~or~~:
Let X be a metric space.
Then thel'e exists a com-
plet~ metric space Y containing a subspace Xl satisfying the propel~ties
i)
ii)
X:1==y XI
is isometr ic to X
The space Y is unique up to an isometry. The~l1l:
Let X oe a normed linear space.
As a wetric space,
X has a com:pletion Y as described in the imrnediately preceed-
ing theorem ... Then addition, multiplication by scalars and a norm can be introd'lced into Y in such a way that Y is a complete Ihlear space (Ba?J8.ch spa ce) and the addition, multiplic::'1.tion by s.3alars and norm on Xl will agree with the norm on X. Tl~:::. B~.ire Category !..~~.?r~
Iro prove the catep;ory theorem, we shall need a lemma
which extends the theo:eem on ne sted . intervals to cOY!lplete metri c SDaces.
Le~m~':
Let {J?i} be a sequence of closed non-empty sets in a
complete metr'ic space such that
i) ii)
F j JFj+l dia~il
F. J
~
0 as j
~
co
co
Then
DFi ~(:X.
That is, there e:;dsts one and onl;y- one point 1 in X that is in each F,. l
}'roof:
x,Y€
C0 a)
p'niq~~~ - If there are tvlO points x ::J y such that
n Fi'
then d(x;y) = r > O.
But for j > some N, diam
1
Fj
<
6/3,
so that x and y cannot be in all the Fi unless x = y.
Existence - Since we have assumed that the Fi are non-empty, we may construct a sequence {Xn} by choosing b)
o ~ d(Xn,xm ) ~
0 or
r
x n } is
1.8 a
Cau~y
£x·n~
Furthermore X is complete.
sequence.
converges to a point x in X.
all F i •
Therefore
x€
vIe must now show that
To do this v-le as surne the contrary! x (" F i
:;:':=)
xE. X - F i , an open set
-;>
There is a ball ar'ound x that lies entirely outside of F i But this contradict s the fact that x·n vIe shall now state and prove
tV-TO
~ X.
equivalent forms of the
£air~Gategory ~~~oE~~
Theorem:
Let X be a coy'!plete metric space.
~Gn ~i; C
co
Then
n Gis n
for all
dense in X.
Let p be any point of X.
Proof:
Un = X
a sequence of open Gets in X such that
J
n.
Suppose that
irie lUuSt show that
00
d(p,
n
Gi) == 0, or that ;for any IS > 0, there is a point
1
x€
nco
G1 such that d(p,x) < €
~
1
Since G., == Z, d(p,G l ) == 0 for any point p ... for sorle point xlS~ C" Let S ==
~ld(X,P)
d(p,xl )
open, there is a closed ~,
~X.
That is,
Since Gl is spher'e'_~l
about
....
-_. ---.-.........
... " .. -
• whose ra d ~us r l < E..;,.". Ie w~ t'a S '1 C G-1 0
'vJe can rn8.ke r l
8 1 C 3.
small enough so tlla t
Clearly 31 contains an ODen
sphere S~ with center at Xl-
Since
\. ~........
G2 == X, d(x l ,G 2 ) ;;: 0, so that thel'e is an x 2 E. G2 such that x 2 €
s~ • Now
- -,._
.......
../
... ....
-
.'
19 G2 is also open.
This gives us a closed sphere S.2CG2'
whose center is at x 2 and whose radius r2:<
€/22.
we can make r 2 small enough so that S2C Sl.
Sililarly there
Again
is an open sphere S~C 8 2 about x 2 ' and an x3E:. G3 with x3E-
s;.
Continuing in this manner, v-Je construct a sequence of nested closed spher es ~ Si ~ whose radii tend to zero and for which
. c j SkC: Gl , G2 , •• o,Gk " S$
there is a point Hence
From the lemma on nested spheres,
y in each Sk and consequently in each Gk •
~
r}
00
G
n
and d (p .. y) < €
1
n co
so that
Gn is dense in Xe
1 In order to state the second form of Baire!s Theorem, 1'11'e must define some terms. A set P.CX is called ~her:e d~ if A does not contain a ball <=) X - A is dense in X. A set is called thin or of first category , A . if it is the countable union of nowhere dense sets" A s.et is called ~~ or of ~.con~.ca tego.r:v: if it is not thin. Theql'em: A complete metric space is thick ]?I:'oo!,.: Let X be a cOlilplete metric space. Suppose X is thin. co Si where each 3 1 is nowhere dense. vie may reThen X 1 place Si by 8i F i , so that X F i • Now Fi 1s a closed ________
_. T
=u
Q
=
=
nowhere dense subset of X. Consequently X - Fi is open and
n 00
dense in X..
It follows that
(X - F 1) 1s dense in X.
But
1
co
(J)
ex - F i) =U F i
X - () I clusion '0
1
= x.
= X which leads us to the absurd con-
20 Compas!~~
Let X be a Hausdo::ff space. to be
.£~pa.ct
if
ever~r
A subset A of X is said
open covering of A contains a finite
subcovcr ing. ~rk:
Compactness is what we mj. ght call an "absolute ll
propel"ty of a set.. Th~.t is, 1r-Je can shov.J that A is compact with respect to X (the open sets of the covering are open
in X) if and only if it is canpact with respect to itself under the induced topology (here the open sets of the covering are open with respect to A). Being open or closed is not an tiabsolute II property since a set might very v.rell be closed with respect to the induced topology on A and not be closed in x~ Theorem: A compact subset of a Hausdorff space is closed.
. Proof:
Suppose that A is a compact subset of a Hausdorff
space X.
q€' A. Gp '
We shall show that X - A is open.
X is Hausdorff - .
9
Let prE X-A,
thal'e .exist disjoint open sets
Cq such thi'\t p€ Gp and q € Gq •
Keeping p fixed" we find
for each q e. A a pa1I' of dis joint open sets as described above ~
Clearly AC U '~q'
Since A is compact there are a fin-
A
n
ite nuraber of G
q
sucb that ACt) G " 1 q±
Furtherraor e, if Gp ' i
is the corresponding open set cont2.ining p, and for which
GPiIlG% = 0, we have
n PEn i=l n
where
()
is an open set in X-A.
This is true for all
i=l PE X-A.
Therefore X - A is open and A is closed.
Theorem: The m1ion of two (and therefore finitely many) compact sets is compact Proof: Obvious
...
-
21 Theorem:
Let f:X
compact
-~
PI'oof:
Let G =
uou.s
' i'J
~
Y be continuous and onto.
Then X is
Y is compact.
fG cr.} be. an
open cover ing of y,
i' is contin-
~-l( Go)}
is an open covering of X. But X is comn . pa cite Therefore, XC U f- 1 (G ). And s1n'ce the mapping is onto, YC(J G • i=l ~ cr. .. i=l ~ Theorem. A closed subset of a compact space is compact. proof: :=uppose A is a closed subset of a compact space X.
--
Consider any open covering G of A.
G~fX-l1 is an open covering of X. is
X - A is open.
Therefore
Since X is compact, there
finite subcovering of 7, n (X-A)
U (U 1
G
cr. i
),:) X
n
Clea:>:>ly AC U G ,which means A is compact" °i 1
Tb.eorem.:
Let Y,' be compact.
Then f:X
~
Y is 1-1 onto, and
continuous -=) f is a homeowlorphism. Pr~f: vle must show that f- l is continuous, or that f is an open map.
Suppose G is an o}')en set of X.
Then X - G is
a closed subset of a compact space _ ) X - G is compact =)f:(X - G) is cornpact is open.
-..::> f(X
- G) is closed =-=)Y-f{X-G)
A Hausdorff space X is called sequentially co~~act if every sequence of elements in X contains a subsequence Hhich converges to some element in X. A Hausdorff space X is called conditionally sequentially compact if every sequence contains a subsequence which
C01"1-
'verges) although not necessarily to an element in X itself. Remark: In Rn , compe.ctness( ~,sequential compactness ie Heine Barel Theorem<::::::;>Bolzano weierstrass Theorem. x is said to be an accu,,"llulation P2!nt of a set A, if every o)en set cont8.ining x contains a point of A other than x itself.
Incidentally~
x need not
oe
in A at all.
22
Theorem: If X is cO!l1p-:::,cl_aruL S~1ri-sl"ies the first countability axiom, then X is sequentially compact. Proof: Let ~ x n } be a sequence of points in X. i)
If xn
=x
for infinitely many n, (x,x ••• )
is a convergent subsequence. ii) If no element is repeated infinitely many times, then we may consider the subsequence obtained by removing those elements which are identical with elements that have appeared before. Let us denote this subsequence of distinct elements by ..fx ~. Nowfx ~ must have an accumulatl... n j ( nJ ion point. If not, X may be covered by a far-lily of op en sets such that each contains only a finite Dllil1ber of
{Go,-zj
the xn -
Go.
n
But X is compact, and so
xcU
Ga.. which implies
1
that some G
a.i
~
contains infinitely many x n co
Let p be an
L
8.~cumulation
:Joint of..fx.1. Every open set L loj contains P, also contains infinitely many of the
Xn
-
At p, there is a nested basis (first countability axiom),
call it [GiJ.
Letfx i
i be a subsequence of [Xn}chosen in
the obvious way, letting with i
I
j, etc~
~
=
some Xi ~ Gl ,
x2
= some Xj E G2
Clearly
After proving a few lemmas, we shall state and prove a partial conv.srse of the above theoI'em: In a metr'ic space, sequential compe.ctness ;> compactness_ !!emrn~: A sequentially compact metric space is complete. pro£!: Suppose that [xnS is a Cauchy sequence in a sequentially compact metric space X. Th_ some subsequence converges to an element x in X. Clearly, this implies that the original sequence converges to x. Indeed
23 d(x,xn ) The term d (XUi
;,lXn
= d(x,x~ni } + d(x ,x ) n ) < E for n, n i > some N since [Xnl is a
Cauchy sequence, and d(x,x
.
xn'i
~
x as n i
n.
\ <
e for
n i > some M since
~
~ <0••
A sequentially compact metric- space is totally
Theorem: bounded. ~roo[:
11]e want to show that given anye > 0, there are a
finite nUlY.ber of points aI' a 2 , ••• ,a n in X such that for any xE"X, d(xJa j ) <E for at least one a j • Let us assume that for some E > OJ we cannot find a finite number of points with the desired property. Then, choosing any point in X, call it Xl' there is a point x 2 in X such that d(x l ,x 2 ) > € , an x3 such that d(X3 ,xl ) > d(X3 ,X 2 ) >E and so ono fx n
e
and
1rJe have constrLlcted a sequence
3with the property that
d(x.,x.) >G'for i J.
J
f
It is
j.
quite evident that this sequence carmot contain a convergent subsequence, "\rJh.lch contradicts the assumption that X is sequentially compact. Theorem: In a metric space, sequential compactness
Sicom-
pactness.
~oo!:
G [Go.}
Let = be an open covering of X. show that G contains a finite subcovering.
We want to
To each point p E X, we [flay associate a re a1 positive number *r(p} such that the sphere of radius 'Y (p) about the point p is contained in some G. This can be done since a Go. is open and every II
CX
is in some Go.-
It will be shown
that there is a 13 > 0 such that y (p) > 13 for all p € X. Let us assume that no such 13 > 0 exists. This assumption leads to a sequence of points ~xi1 i11. X such that y(X i ) < viously Y(*i)
f.
tends to zero as i tends to infinity.
ObIt fol-
24 lows from the sequential compactness of the space that a subsequence of f~ i1 converges to some point Xc:o in X. But around Xro there is a sphere that lies entirely in some Ga. This sphere contains infinitely many of the xi' vihich contradicts the assumption that Y(x i ) ~ O. Therefore, a i3 > 0 exists such that y(p) > ~ > 0 for all p. X is a sequentially compact metric space :::::9' X is totally bounded ? X may be covered by a finite number of spheres of radius (3/2, Each such sphere lies entirely in some Ga.. The finite union of these Ga will cover X. S~ Th~£~~: Al\compact metric space is separable, SUDpose X is a compact metric space. Then X is totally bounded <=?) there are a finite number of points in X, call them xll,x12, •• t>"X.J.nl,SUCh that if x is any point in X, d(x,xli ) < 1 for at lea~t one xli in the set. Si~ilarly
!!2~f:
there exists a finite set of points x2l,x22,···,x2n
2
such
that if x is an~T polnt of X~ then d(x,x 2 :t ) < ~ for some x 2i in the set, and so on.
In this manner we constru.ct countably many finite set:], Hhose union is also countable. This sequence is dense in X. For, 3:i..ven any x€ X and any E > 0, there is a positive integer N such that 0 < < ~, and con-
t
sequently a point x Ni of the sequence for Hhich d(x,Ni ) <
The fundamental cube in the separable Hilbert space li 2 or Rw) is the set of points of the form x::; ';i with
o~ o<
tl ::: 1 1
S2 ~ ~
• •
'•. e
i
25 Theorem: The fundamental cube is compact Proof: SincejP2 is a metric space it is enough to show sequential compactness. Suppose x.nl is a sequence in
£
xi
= (~i,
fL2
s~, ... ,S;, ••• )
Let us as'iume that we have already shown that a closed bounded set in RN is compact (prove it by induction). F'or any 11, 'lrIe may select a subsequence of { Xn} vJhose first N cordin8.tes converge, j .• e. j';'i "01
~2 -> S2
'·""1'
Then, .given any E that i f
The term
i
1:'
-.;>
i
M
there is a positive inte",.'er II such
> 0,
.L (~ - S~)
2
and so on.
<
0/2
1
first M coordinates converge.
for n large enough since the
co The terra ~
n2
(Si)
<
f.1+l for n large enoll.gh since it is the tail of a. convergent serles. i.e.
Clearly lim xn = lim x (Ir) I'~co
-
Connectedness , Definition 1: if 0 and X are closed in X. Definition 2: if X cannot be
N-~ CD
A topological space X i2 said to be connected the only subsets which are both open and A topological space X is said to be connected covered by two non-empty disjoint open sets.
26 Th~o~:
Proof:
Definition I <~ Definition 2 Suppose XCG1 UG2, where G1 ,G 2 are non-empty, dis-
joint, open sets. Then X - G1 :: G2 is closed, so that G2 is both open and closed. Therefore G2 is either 0, which we-have assurned not to be the case, or X. ;Lf G2 = X, then Gl :: 0 'VJhich is, al.s~ contl'ar;r to assUmption. " Convers'ely,.. if .ACX, with A 1- 0, A f X, and A both open and closed in X, then X - A is both open and closed in X. ~Ve then have XCxV(X - A) X(\(X-A)=O
Whici1. is a cov6Ping of X by two non-em.pty disjoint ... open sets, contradict5.p..g definition 2. Theorem: Let X be connected. Then f:X ~ Y is onto and continuous .~ Y is connected Proo:': Use definition 20 Product Let X,Y be topolo~ical spaces. of X and Y~ denoted by
Topo~_o?_!cal
The topological
produc~
Z :: X X' Y
is the topological space whose elements are ordered pairs (x,y) with x~X and yf::. Y. 'rhe sub-basis consists of the sets
where G is a set of a sub-basis of X, and G is a set of a x y sub-basis of Y. The reader might find it profitable to convince himself that this actually is a sub-basis for a topology. Let X, Y be m,etric spaces. Consider the, topologies iJ.1duced on X and Y by tt.eir respective metrics. It is easily verified that the topological product XX Y is homeomorphic to the Cartes ian Product X X Y. That is, the product of the induced topologies is the same as the induced topology of the (Cartesian) product.
27 To conclude this outline of general topology He shall state a few theorems, but omit the proofs. The reader is strongly urged to fill in the many gaps we have left by consulting the literature for a more detailed account of the subject. 'rheorem: The product of complete metric spaces is complete. Theorem: The product of compact topological spaces is. compact Theorem: .The product of Hausdorff spaces is Hausdorff. We say that a topological space is metrizable, if a metric can be in.troduced into it that will give back the original topology. A topological SPECE.; is called ~eparably metrlzable if it is metrizal:le, 8.nd, as a metric space, it is separable. A topological space X is called normal if, v.rhenever A, B are closed in X and A () B = 0, there exist dis joint open sets GA"::)A and GB::? B. ~E'.:;~~,r~~: This is another example of a separation axiom. In a Hausdorff space v.Te can lIseparate" two distinct points; in a normal space we can "separate" two disjoint closed sets. Th~~: The necessa:;,"y and sufficient conditions for a topological space to be separably metrizable are i)X must satisfy t~e second countability axioIDQ ii) X must be norma1 4 :fl1.etrizat;ion Theo£~: A normal, Hausdorff space satisfying the second countabj.li ty axiom is homeomorphic to a subset of the fundamental cube of the separable Hilbert space.
Urys~hn
28 Chapter II
Simplex and Comple~: Notation: RN Hill denote N-dimensional Euclidean space o Definition: The ~~men~ joining the t-dO points x and y in RN (or any vector space) is the set of all points p which have a representation in the form, p = Ax + (l-A)Y, 0 ~ A ~ 1. Definition: A set is said to be convex if whenever it contains two points it contains the segment between them. If AI$A2.~.are convex sets then
Theorem:
fdAu
is a convex set.
!!:oof: If the int€L'gGction is either empty or consists of one point it is convex. Other~'1ise let x,YG' Au- Then x., y " of
n u
each A , and since each A is convex, the segment joining x u
.
a.
and y lies in each Aa.. Hence the segm~nt lies in the intersection and the intersection is convex. Definition: The closed convex hull of a set S is the intersection of all the closed convex sets containing S. Theorem: If S consists of a finite set of elements xO' ••• ,Xn then the closed convex hull of S is > -
Proof:
O~
n
>
1.=0
The set is obviously closed. n
.
y' = /"' i_
Aix J,..•
~1 J,.- O
To show S is convex let
Any point p on the segment joining y and y'
is of the form: P
= (.I.y
n t + (l-(.I.)Y' = ?_ [IL Ai + (l-lL) Ai] xi 0 ~ iJ.i ~ 1 ~:.:::O
_
I
Since ~ Ai = ~Ai ~tJ.Ai
,
=1
t
and Ai ~ 0, Ai ~ 0 we have
,
+ (1-(.I.)"'i = 1 and (.I.Ai + (l-(.I.)"'i:::' O.
Therefore the points
on the segment lie in the set and the set is convex.
n Now we must show that if y = ~ "'ixi' "'i .::. 0,
29
Z
Ai = 0,
then y is in every convex: set containing xO'.'.'xn • The proof is by induction. Any convex set containing x O,x1 contains all points of the form:
=
Thus the theorem holds for r 1. Asswne that the theorem is true for r the point
=n
~
1.
Gonsider
The point
has the property that
, From the induction hypothesis, y' lies in every convex set containing x 1 , ••• ,xn ' and consequently in ~very convex set containing x O,x1 , ••• ,xn , y = AOxO + (l-AO) yf
It follows immediately that
= AOxO
n A. + (l-AO) ~ l-~- xi
n
=~ Too
"'ixi
is in every convex set containing xO",.,xn ' ~efiqition:
in
~neral
linearly
The points xO'."'xn , n
~
N in RN are said to be
positj.oE if the vectors xl-xO,x2-xO"",xn-xO are
independent~
30 To show that Xo does not occupy a special position in this definition we shall PI' ove that if Xl - x O!.,! .,.J-:xn . . x.o-.al"e-~~··· linea.rly in·dspendeRt. .. t.hen... x·O-x11··,.x2-xn..... ·.r-· 'l"~x are linearly nn independent.
implies a O = ••• =a.n-1=0 dir'ectly from the definition that a necessary and sufficient condition for the points xO,.",xn to be in general position is that the matr:i.x r-i.i
foll~ws
.
I
/
1 1 (x .. - x ) o
. ~(~ J.
{
n <
N.
~'inition: ~al
The points x O" •• ,xs in RN are said to be in ~en ... pos,ition if any N + 1 of them are in general position.
Theorem:
If
XO/ ••• /X S
then there exists an
e
is a set of points in general position > 0 such that d(xpYi)
<€ for
=
i O, ••• ,s implies that the points YO""'Ys are in general position. Proof: Since XO, ••• /X . s is in general position the matrix ~Xi
of every set of N+l points is of rank N and hence each of
these matrices has at least one Nth order determinant. Since the determinant is a continuous function of the components of ea ch point there exists an ~K > 0 such that d(xK'YK) < ~ K
31 implies that each matpix containing y in place of x K is of . K rank N. This can be done for K = O, ••. ,n and therefore by choosing £2. == min tI K the points yo".' 'Yn will be in general K
positi.on. Theorem:
Any N+l points xO"."x N can be brought into general
position by an arbitrarily small displacement. !!£o£:
Consider eO
eN == (0,0, •• 01). Form
=
(0, ••• ,0), e l = (1,0, ••• ,0), ••• ,
These N+l points are in general position.
o -< t For t == 1, __ /\(;'[.(1» ~
-<
10
== _ /\,(e.) and hence w. 1\ (Yi(l» ~
is of rank N.
The determinants !~(Yi(t»)1 are polynomials in t, and since they do not vanish identiaally we have that for arbitrarily small values of t, tK < G:, 16(Yi(t K » I f O. Therefore the points YO(tK)""'YN(t~) are in general position for 1 J... d(y.(tK),x.) < G.. ~ ~ \..
Any finite system of points xo, ••• ,x s can be bro~ght into general position by an arbitr'arily small displacement.
Theo~:
(N!l)
Proof: Given (i > 0 He must bring k == sets of N+l points into general position. Take any set of N+l points. By the previous theorem these can be brought into general position by an arbitrarily small displacement say & I" From an earlier theorem we know that there exists an ~2 > 0, such that if these N+l points are each moved a distance less than ~ 2 they will remain in general position. Now take any other set of N+l pOints and bring them into general position by moving the a distance less than G2 " Hence we have two sets of points in general position. IThis process may be continued until all
(N!~
the sets are b""ught into general position, where in each case i is taken so that ~l < ~'/K.
e
1
32 Defini tiol1: A geometric rectilinear n-dimensional si:"plex, or briefly_an n-sircple~ is ·the closed convex hull of n+l points, ao, ••• ,an , that are in general position. The points in general posit,ion .ar-e--ssid -to--sp.a.n.-..th.e.-simp..~~~.. gnd- AI:e. .Q?ll.ed the verll.~.• Notation: A taol" .,an }
=
Recalling the definition of closed convex hull we have that the simplex consists of all points which have a repre.sentat ion in the form
-
1, A. > 0 ~
n
This representat :10l1 is unique for if y
But
;>
I-Li-Ai
(i-Li-Ai) =
= O.
° and since the a i
= >-1=0
!-Liai then
are in general position,
The Ai are called the barycentric coordinates of
the point a We see that the si;rnplex is a bounded closed set; in RN and is therefore compact. Definitj.on: A p-dimensio:qal face of a simplex is a subset: p p {Y Iy Z AiXi' 2: Ai = 1 Ai =. 0, i ~ P < n; Ai = 0 for i >p}
=o
0
Hence a face is itself a simplex. An o-dimensional face is a vertex, a l-dimensional face is called an edge, Definition: Two simplices are said to be properly situat~~ if their intersection is either empty or a common face. We see i;rnmediately that two faces of a simplex al'e properly situated. Definition: An inner poi~~ of a simplex is one for which Ai > 0 for all i. Every point of a simplex is an inner point of a uniquely determined face.
33 Theorem: Every inner point of a simplex is a midpoint of a segment lying entirel JT vdthin the simplex. r Proof: Let x (!;A ::; l,a o" •• ,ar } Then x = Aia i and since C i=l x is not a vertex at least AiAj f 0, i ~ j. Choose e such that
'2:
Yl
Let
=x
+ f2{a i -a j )
e (a i -a j )
y2 = x -
Yl+Y2 2
Then
Y1 Y2 ~A and x:::
Theore8: a face o
No vertex can be the midpoint of a segment lying in
Proo.£.:
Assume a K =
•
~(YO+Yl)' YO'YI r
Yp
1
since Y1
1
i
~(AO
=~ 1.=0
p
Y2 there exists an i,j
i
1
j
j
6- A
= 0,1 i
+ Al) > 0, 2(a o + a 1 ) > O.
I-
j
i
j
such that AO > 0, AifO
But from the uniqueness
of the barycentric coordinates we must have
a K = Oan+···+laK+···oar which is a contradiction. Theorem: Proof:
The vertices of a simplex are uniquely determined. Il11l11ediate consequence of the two previous theorems.
The canonical n-simplex is the set of all x that admit representation in the form
~hition:
34 Then x = (AO •• ,An ) and we see that the barycentric coordinates AII ••• ,An are equal to the cartesian coordinates while AO
=1
n - ~.
~=r
Ai'
It is easy to see that any suaplex is
homeomorphic to a canonical simplex and hence any two simplioes of the same dimension are homeomorphic. In the special case of an N-simplex in RN the irmer points of a simplex are the interior points. The points which are not inner points form the set theoretical boundary or frontier. Theorem: The diameter of a simplex S is equal to the lenGth of its longest "edge,
.
Proof: Let x and y be Rny two points in the s,mplex. Assume x is not a vertex, and that diam S = d(x,y), Then x is the midpoint of a segment lying entirely within the simplex. Denote the ends of this segment by PI ,P2" Then sinco S is convex the segments yP 1 and yP 2 will be in S and since we are in
RN either d(xy} < d(yPl ) or d(xy) < d(yP 2 ). diam S.
Hence d(xy) <
Therefore x,y must be vertices of S.
Defini t ion: £:. geor'letric re ctilinear complex, or simply a complex K, is a finite set of simplices in RN satisfying (1) If S is a sll1plex of K, then every face of S is also in K (2) If Sand T are faces in K then Sand T are properly situated. )vertex1 . of some SiIllL face,,) plex in the complex, Tbe d.imension of K is the maximum of the dimensions of its simplices. Let K be a complex with vertices aO, ••• ,ar • Then any point in K can be written uniquely in the form A S-vertex}
1.. face"
x =
n
~--
1:7:1
of a complex K is a
A.(x)a i ~
vertices of a simplex. ordinates.
where~A. ~
=1
and Ai = 0 except for the
The Ai are called the barycentric co-
35 Definition:
The . space of a complex _- - K, denoted by
IKI
is the
set theoretic union of all simplices in K.
IK/
is compact since it is a finite union of compact sets.
Defini~ion:
A E...().lyl~ed££!! is a space 1tJhich is homeomorphic to
IKI.
Such a homeomorphism is called a triangulation of the polyhedron. Defini t ion: A sub,complex L is a subcollection of the s implice s of a complex K which itself forms a complex. Since
ILl
Definition:
is compe.c'!; /KI -
ILl
is open in IKI.
The set of all simplices of a complex K whose dim-
ension does not exceed. r is called the r-dimensional skeleton of the complex K. 111hi8 is obviously a subeomplex. Definiti.on: ---,
If a is a vel'tex of the complex K, the set of all points vJhich are inner points of some simplex having a as a vertex is called the stal' of the vertex a in K and is denoted st (a) ~ Definition:
The set of all simplices in K which do not have a
as a vel'tex is called the subcomp1 ex. opposite th.e
v.~:r::.tex
a,,-.and
is denoted aKa It is clear that aK is a complex. As an immediate consequence of the definitions we have
and therefore st(a) is open in lKj. Every point in K belongs to some star. since every point is the inner point of some simplex in K. ~heore~:
The intersection of stars, St(a O)' st(a1 ), ••• St(ar ),
of a com.plex K is non empty if and only if {a O,a1 ,. •• ,ar } spans a simplex.
!!:2~:
Suppose {ao, ... ,arJ is a simplex of K. st(a i ) conta-d.ns this simplex minus all its proper faces not containing
ai' i.e. tao,· •• ,ar~
-fa1' ••• '8.i-l'ai+ll .... ar1;:: st(a i )·
36 Therefore f) st (a i ) contains ao, ••• ,ar minus all its proper faces. This set is non empty since it contains all points of -the form x = ~Ai ai' Ai > 0, 2).i = 1. I'
Suppose iC'o St (at)
'I
O.
Let x denote a point in the in-
tersection. Then x€ st (a i ), 1.=l, ••• r and therefore belongs to a simplex which belongs to K and which has a O'.'" ar among its vert ices. Therefore taO"" jar} '" K. . ... Definition: A finite set A of distinct elements aO, ••• ,an is called an ~bstl"act complex with vertice~ a O"" "an if 1) certain non-empty subsets of the set A are distinguished and are called abst~?e.ct simplices of the complex A, 2) if S is an abstract simplex of A then every non empty subset of S, i.e., every face of S, is also a distinguished subset of A 3) every set consisting of a single element is a distinguished subset The abstract simplex S = tso, ••• ,sr] with r+l vertices is said to have dimensi0n r. The maximum of the dimensions of the abstra.ct simplices of A is called the dimension of A. p~f.in.i tion:
aO, ••• ,ar "
Let K be a geometric complex with r+l vertices Consider the set of elements aO, ••• ,ar " These will
for'm an abstract complex A provided that the distinguished sub~ sets of A consist of those vertices which form a si~plex in K. K is then called the geometric realization of A. Tpeor~:
Every abstract complex A has a geometric realization
K. Proof: We shall construct the natural realization of the abstract complex A with vertices aO, ••• ,ar " Take the r-simp1ex E = eo,···,e r and to each abstract simplex laio" •• ,a ir
1
1
assign the corresponding face of EJ reio, ••• ,ei~.
1
The re-
suIting set of simplices '..v111 be a geometric complex since the
faces of E are properly situated. This realization will be called the natural l'ealization, N, of A. Theorem: Any t"l;W realizations of the same abstract complex are homeomorphic. Proof:
Let K be any realization of A, and -aenotethe vertices
of K by cO' ••• ,cr -
I>1ap
INI
onto
IKI
by assigning to the point l'
: : >-F:i
A. c.
~ ~
elK I.
It is
left to the reader to show that f is a homeo::norphism. Hence every realization of A is homeomorphic to the natural realization. Theorem.: An abstract complex A with vertices a O"" ,a K of dimension n has a geometric realization K in R2n +l , and the vertices may be prescribed arbitrarily close to any given set of k+l points in R 2n +1 • Proof: Given k+l points in R"c_n +1 we can move them into general position by an arbitrarily small disp1acement# Let cO,.~.,cK denote these points in general position. point c j •
2 =
If S ::: tso,ou,srJ is any simplex of A, denote by
f Go, ..
ponds to Sj.
Assign to a j the
o'Or}
the geometric simplex in which
OJ
corres-
It follows immediately that the resulting set,
K, of simplices of R 2n+1 satisfies the first condition in the definition of complex. To show that k satisfies the second condition let S ::: [so, ••• ,sr] S' ::: {s;,'••• ,s;
, T
1'
= rto, •• o,tp~be
simplices of A and
T' ::: {to, ••• , t;J be the corresponding geo-
metric simplices of K.
Let V ::: {v 0'.' .,
"t}
denote the set
of all points vJhich are vertices of either S,t or T'. Since dim A ::: n, r ~ n, p ~ n and therefore t ~ 2n+l. Hence V is a
38 simplex in R 2n+l and since Sf and T' are faces of V they are properly situated. ,l)efini t iOJ:?: .. _Let· '-a.O~· ......... ,a..r'·be··the·-ve!"t ia~g'6f Ha--c-omplex K and bO, ••• ,b s the vertices of a complex L. of K into L f: K
~
The simplical mapping
L
is given by f(a i ) == bj where v,Te require the mapping to be such that vertices which span a simplex in K map into vertices which span a simplex in L. Let x ~ IK! • Then x = 2,:.-A i a i where Ai ::: 0 except for the vertices of a simplex.
Then the 'ipiecewise fl linear mapping
determined by the vertex mapping is given by f(x) == ~Aif(ai). We call f a sinlplica1. lilapping of J.2.xaTI!p.~:
IK I
into
IL I.
Prove f is cant inuous.
Definit:'Lon: Two abstract con~lexes are called isomor9hic if there exists a 1-1 ma::>ping of the vertices Hhlch preserves simplices. Two geometric complexes are said to be isomorphic if their corresponding abstract complexes are isomorphic. Theorem: If 2 abstract complexes ai'e isomorphic, then their geometl'ic realizations are homegmorphic Proof: The proof is left as an exercise. If Land K are any two complexes,L is called a subdivision of K if
Defin~tio~:
1)
ILl
==
IK/
2) to any subcompl ex Kt of K ther e corresponds a subcomplex L' of L, such that /Lt I = !Kf I.
39 Chains and Cochains: Def~n~ti..~: An ordered simplex is a simplex whose vertices are given in a specific order. Notation: A = [aO, ••. ,ar ]. To every l' simplex there are (r+l) 1 ordered r-simplices. De£.19:.?-~i0t.:!:
Two ordered simplices are said to be similarly if one is an even permutation of the other.
~.!:!!~d
Definition: Two ordered simplices with the same vertices which are not Similarly oriented are said to be oppositely orient£~. Notation! If Sand 'r are oppositely oriented simplices we will write S = -T. Using similar orientation as an equivalence relation lrJe define an ?riented si~plex as an equivalence class of similar'ly oriented simplices. An oriented simolex will be d enot ed by giving one of its representatives in round brackets, i.e., (a O,a l ,a 2 ) is a representation of the class consisting of (a O,a l ,a 2 )p (a l ,a 2,a O) and (a 2 , aO,a l ). DefiE~ti~:
DefiI2l:~:
The support of an ordered (or oriented) simplex [aO, ••• ,a] (or (a.O, ••• ,a r )) is the geometric simplex -l
r .-)
(,. a. 0' • ar S Definition: An integral oriented r-chain, or simply an r-chain, over the complex Z is a functLm, x, which assigns an integer to every oriented r-simplex so ti.lat x(-Si) = -x(Si) (Le., chains are odd functions). If x(S.) = a. we denote the chain x as a linear combin~ ~ ation., ,?u.S., with the provision that if S~= ~ ~ ... -SJ"' then 0 •
,
-
a i = -a j
O
We extend the definition 0,1'. an for oriented k-simplices with kfr.
l'
chain by letting x = 0
Definition: Consider the set of all oriented simplices of a complex. Then a chain Xs where
3
=T
3
= -'r
otherwise is called an elementary chain. In standard notation an elementary chain is denoted by the oriented s ir,plex to which it assigns the value one. There is a 1-1 correspondence between oriented simplices and elementary chains. Theorem: Any chain can be written as a linear combination cf elementary chains" Proof: Let 31 " ",SK be oriented r-simplices such that Si
f
~ Sj for i ~ j, iGe., no two simplices have the same
support~
Then if x is any chain k
x
= --;,: :
i~
x(S.) x. l
S1
where tne sum of tlrJO cnains is defined as the sum of their functional valUes. The sum of 2 r-chains is an r-chain and we have that the r-chains over a complex K form an abelian group, denoted by Cr{K). In this group the chain t-Jhich assigns the value 0 to every simplex is the zero elenlent. define multiplic8.tion of a chain x = :>ais. by an -
l
integer ~ as ~x =,2:~aisi
Defini~i.2!!:
The boundary operator () is a mappinG of r-chains into (r-l)-chains C> : Cr (K) -'> Cr _1 (K)
such that () (x+y) = a.x + () (a xi = aOx
for an elementary chain C)( a x) = 0
Oy (a is an integer) for r = 0
41 and for r > 0 '"\
o(aO,a1,···,a r )
=~ i> (-1) i (a O,a1 ,···a1\ i ,···ar ) J.=o
. /'I:'
where a K denot'es the-·~rm--ti')-be·le-f't· out of- th-e-·--¢hain. We see that 0 is a linear opere.tor, i.e. a homomorpb.j.sm and hence comple te1y defined once the boundary of an elementar·y chain is giyen. Por example ()(aO"a l ) = a 1 -a O i.e. it is a chain which takes on the value 1 at a 1 and the value -1 at a Oo
C)
(a O,a 1 ,a 2 ) = (a 1 ,a 2 ) - (a O,a 2 ) + (a O,a 1 ).
Definition:
A
simplex
S
is said to enter in a chain x if
x(S) :J O. Definition: The support of a chain x, denoted by txl is the set theoretic union of all the simplices which enter in the chain. Ixl is compact and 1) Ia,x I = Ix I a + 0 2) !x+YI C Ixl u Iyl
3) loxl C
Ixl
2;d X;: a2x=o, where x is any Theorem: Poincar~.2:el~~ chain =0 Proof: 1) if x is an O-chain then J x = 0, dUx = " = O-chain and therefore 2} if x. is a I-chain then dx adx = o 3) Since ~. is linear it is only necessary to prove that
uo
02
of
In thlssum each term appears' twice, once when the i th term . is dropped first, and again when the jth ter!Q is dropped first. In the former case the coefficient is (_l)i (_l)j-l since after the i th term is dropped the j th term becomes the j_l.st term, and in the latter case the coefficient is (_l)i(_l)j. i . -1 . j Since (-1) (~l)J + (-l)J(-l) 0, the terms cancel each others
=
Definition% A chain x is said to be a boundary, or homologous to....Q, x !"V 0, if x : .: dyo Definition: A C;h~~,x x Ny, if (x-y) tVO .. Definition:
13 ~!i.d
to be homologous to a chain y,
A chs.in x is called a ~yq~ if C)x = o.
°
The sum of tltJO c;{'..;les is a cycle: Q)x = 0, dy = implies that () (x+y) = d x + (J y = O. Therefore the r-cycles form a subgroup Z (K) of the group of chains over K. I' Every boundary is a cycle. The SUIIl of two boundaries is a boundary: x = d y , x, (;)y' implies that x + Xi ;;y +~T= d(Y+Y')4 Hence the bounds.ries form a subgroup Br(K) of the group of chains over Kc Thus:
=
Definition: comp~~
=
The r ... dimensional integral horilology group of a
K is the factor Group
The elements of H {K} are homology classes of cycles, I' where t1/lJocycles are called homologous if their d:i.fference is a boundary. Definition:
Let Sand T be two oriented simplices and define
their scalar product
43 S = T S = -T
otherwise
Definition: The scalar product of tHO r-chains (x,y) is defined by the following properties 1)
(x+x t , y) = (x"y) + (x'"y)
2}
(x,y)
:::
(y,x)
3)
(ax,y)
:::
a(x,y}
4)
(x,x) >
o
if x
f o.
Choose a set of oriented r-simplices LSi ~ such that no two of them have the same support. If x,y are two chains, x
=? ~ J.
aiS., y == J.
>
f)-;S-; then (x,y) =). aJl.; and hence we ...- J. ~ J.
--,-... ... J.
J.
haVe an orthonormal system for oriented simplices.
The .£.C?£2.~ndary oper'ato!:, CI"'\ <)(0, maps r-chains into r+1 chains in such a l:ay that if x is an r chain and y an (r+l)~chain, the scal~r product Defirdt_~:
( a~;-y , y) Theorem:
dol(-
= ( x,
ay ) •
2f~ x=O
Proof: Let y be an arbi trf\ry (r+2) -chain, x an arbitrary r-chain ( ()~!- c)* x,y) ::: (O";('x,dy) ::: (x, ()~y):::(x,O)=O. Hence the scalar product of any chain with J-l!-c)":'x is equal to zero, and from the properties of the scalar product 2/ ()*x=o. Definition: 9PiO~S.•
t.0
A chain is said to be a
0, x~..t 0, if x
:::
~oboundary
or cohomol-
ify.
Definition: An r-chain is called a cocycle if !"'c) ~!-x = O. Every coboundary is a cocy1ce. The sum of two coboundaries is a _coboundary and therefore the coboundaries form a. group Br(K). Similarly the cocycles form a group Zr(K). Thus
BrOnC Zr(K)
44 Definition: The ~.:.~er~s ianal cohomology group of the complex K is the factor group
zr (K ). = If' ( K) Br(K)
The elements of Hr(K) are cohomology classes of cocycles. Let K be a complex whose distinct oriented simplices
(S~ F ~ s~ for i F 0 1
j) are
- simplicest - s I ices: irilP
11
•
•
n
..
simplices:
Every r-chain x may be
v~itten
Hence the boundary of an elementary r-chain" I'
e ij
i
The matrix (G j ) see that EI' has a.I'
o : ;:
c) -lSI' CI
_
= Er
is given
0
•
is called the r-incidence matrix.
I'm,'IS
~
r-l
Sj
I' >
and ar- 1 columns.
r - \ r-l = ~ c.. r ~ ~ iJ· J,K
i - LIS i· (/ S . j J J
Since r. 'It:;;
r-l Sr-2 J·K
K
we have E r E r-l
=0
The coboundary of an elementary r-chain is given by
We
by
45 where (eI j ) is called the r-co~nc~~ence matrix. ( G>~~S:: l'
:r
sr+l);:: ~ e r
K
ij
(Sr+l j
,
Now,
Sr+l)
K
;:: e.r K s 1.' nee (Sr+l " 1. J
1 ) = ><.j SrJ:7+ ,~ v K-
= (S~,us~+l) = (s:, l
:z Z.
r. 'CJ
r+,l gr)
Ki
II
;{
;:: ,; r+l v Ki Hence the (r+l)-incidence matrix is the transpose of the rcoincidence matrixa ~lniti~: Let G be an Abelian group. We ~efine an r-dimensional chain over K 'Hi th coeffj.cients in lG as a function which assigns to every simplex in K a value in G, with x(-S) = -x(S). There is a natu;r'al way of defining mu.ltiplicB.tion of g(;' G by the integers; i~e~ o ' g = 0" 1 c g _. g, (n+1 ) g ;:: ng+g, (-n) g ;:: -ng. We shall now proceed to extend our previous definitions of chains etc. to an arbitrary abelian group G. Therefore any chain with integral coefficients caL be lilultiplied by an element of G. Hence every chain with coefficients in G may be written x = >g.S. ' - J.
:...
As in the case of integral coefficients we have that the rchains over K with coefficients in G form a group, Cr(K,G). Definition: The £9_~ndary operator maps r-chains into (r-l) chains
a
is a homomorphism which
46 in such a way that
d gS = gd s
a-
for ever~T oriented simplex Se,J\.,ga,in·· 2 --=:. 0- and- the· -b-otmdaries_. form a. subgroup Br (K,..G-}o1'\·-Cr (K,G). As before, a cycle is defined as a chain whose boun.dary is zero, and .we can show that the cycles form a subgroup Zr(K,G) of Cr(K,G). Thus we see that Br (KJlG) C Zr (K,G)
r
l Definiti~: Let G, be two abelian groups. An r-cochain, .!; is a homomorphism of the group of r-chains in G into
r ,
~ : Or (K,.G) ~
r
Since'; is a homcmorphism, ';(-x) = -';(x). The sum of two such functions is aga1n a homomorphism and the r-cochains form a group Or (K$G,
r ).
As an example of a cochain consider
1\
G,G,
r
fl
;\
3
abelian groups
II
• To every gG G and every geG we associate a product = ~ g where 't ~ so that
r,
i\
t\
"
1\
(€l + g2)·3 = gIg + g2 g A A A g (gl + g2) = g gl + gg2 e
r. .
~
We S8.y ths.t G and G are paired to
J\
Set x
/I.
= .>gisi and !\
x =,2:.gtSi be two chains vdth coefficients in G and G respectively.
.
Then the mapPJ.ng /)
0
f
. ,\ . x ~nto xx ::; "",,, o:::::;..gigi ~s a co-
r
chain. If in particular G,G and a1'e additive groups of in .. I' tegers, GOI then Xx is the scalar product and every element of Cr(K,Go,G o ) is of this form. Definitiop= The c,?bounqary operator chains into (r+l ) ...·cochains
0*
is a Iilapping of r-co-
defined by
(a -*~ )x -" = .; (a x)
(r+l )-chain.
where Ig is an r-cochain and x is an
47 Theorem:
::: 0
c)'''' () ~~ sx ::: () ~~ d x
Proof:
:::
s<:'
J. () -- sO = o since s ~x
is a
homomorphism D~,finit~~:
A .£.£.c..I.9}e is a cochain wbose coboundary is zero. As in the case of r-chains Hith Integral coefficients we have that the coboundal'ies form a subgroup Br (K,G, of
r)
Cr(K,G,
r
CI'(K,G,
and
l
Me coc~eJ.e8· ·foJ?l11 a ,su.bgroup Zl'O::,G,
),
r l
)
and Bl'(K;G:s
r )QZJ? (KsG;. f.'
).
r ) of
Thus
.
B:1..,(K~G) C Zr (KpG) C Cr (K~G)
Hence we can define the
~!rJolornr
group of the complex K with
!:~pec.:t_~o G
Z (KG) H (Ie G) ::: -.~ ~ ---ry r' Br K,G
As the e:roup whose elements are homology classes of cycles where tl,vO cyc:J.es are homologous if their difference is a boundary. Similarly we can define the cohomology group of the complex
I.'i: w~th~spec~,_!.~._.(!~
• ::: . Zr (KG [' )_ Br ( K, G, 1'-1)
as a group whose elements are cohomology classes of cocycles where two cocycles are cohoillologous if their difference is a cOboundary. To every simplicial mapping f: IKI ~ ILl we associate a /'mapping f of chains over K into chains over L, if fa.]. = fa.J for some i 1 j /,"\J _ Co ) f:
(a O" ' . ' 8. r )
- L(fa. o, • •• ,far) if fa l
1
fa j , i
1
j
48 Thus the mapping is a homomorphism .'\./
f: Cr(K,G) ~Cr(L"G} TheoF~~:
Let K" Land M be complexes" f and g simplicial mappings such that
K.!~ L..lt> M. Then g • f is a simplicial mapping and hence induces a mapping ~ ,....--./ 7'.;.;1 .;;v g • f of chains over K into chains over Mo Then g • f g • f, and hence the curls can be dropped.
=
Proof:
obvious
Theorem: A simplical mapping co;mro.utes with the boundary operator: \,)fx = fc)xc E~of:
1)
Let x be an a-chain. Then fx is an a-chain and hence C)f,X = O. "ve also have Ox = 0 and sin~e f is a homomorphism f<:J x = fO = o. 2) Assume x be an elementary integral I-chain. Let x = (aO;a l ) and assume fa O ~ falc Then f(aO,a l ) = (fa.O,f a 1 )
d f(aO,a l ) = fal -fa 0 0(a Oa l )
=
al - aO
f <)(aO,a l ) = fal - faa Assu.me faa = tal·
3) case i)
since f is linear.
Then
Let x be an elementary integral r-chain, r All ta i are distinct. Then
>
I
49 f (a o ' • • • , a r ) :::: ( fa. 0' ••• , far) __ i
I)
(-1) (faO, •• ~fa., ••• fa ) elf (a 0" • • • ,a r ) => '-.1 r 1
_ ~()i( 1\ ) -..c:.:::... -1 a O"·" .a i ,.· oar i
f
-
i
h.
'> (-1) (faO, ••• fa., ••• fa r ) 0 (a 0' ~ • • "ar ) ::: ........ 1 1
case ii)
Two of the fa i coincide while all the others are distinct. Without 10ss of generality we can assume that faO ::: fal f ( a 0" •• ~ a r ) ::: 0
c)f ::: 0 "'l --i A C/(aO, ••• ,ar ) ::: ~(-~) (aO,···a., ••• a ) 1 1 r
f(;)(a o'· ",sr) = (fal,···fa r ) - (fa o,fe'2,···fa r ) since faO
= fal
definition.
terms i,n which fa O and fal a',?pear are zero by
Hence f
cJ (a O' . ' . ,ar ) ::: O.
Case i3, i) If any 3 or XlJ.ore of the fa i are equal then it is easy to see that everything is zero. ~'Je define a hOlTIol"llor-phism f-l:- of r cochains over L into r-cochains over K f~~r(L) _->Cr(K)
by requiring that if xQCr{F::), t;E:,Cr(L) then
f~tx ::: ~fx where
fx~ C r
Theorem: Proof:
(L)
<:Yl-f"*:::
f-ll-
CJ.r.-.
This means
cf:-f1~
= f~-c)'~~
If two cochains are equal then they each give the same
value to every cbain.
Hence we want to shov.J that if x e Cr
then
c) ~(of -;~x cl~~ * ~x = f .;~ C) x
= f,-llo cY~x.
Now
= t!,f Ux =
CJ"'tfx
= f -:~ (')~l;x.
on
Barycentric Subdivision'J.nd Sperner Mappinp:s: Definition: the point b
bl:u~ycenter
The s
=~ '1=0
a}
of the simplex A = fa j " " lOS
is
i.e., that point of the simplex whose ..S!l a.; J.
barycentric coordinates are all equal. lITe see that the barycenter
1)
of a O-simplex is a vertex (i.e. the O-simplex itself). 2) of a I-simplex is the midpoint of the edge 3) of a 2-simplex is the midpoint of the triangle, etc. Definition: A sequence of sets Si is said to be in ascending ord~ if Si c: Sl+l for all ij it is said to be in descending orde! if Si ::; Sl+l for- all i. Definition: The barycenters of the faces of a simplex are said to be in {aSC~ndiI?-g } order if the corresponding faces of the descendJ.ng simplex are in {dascenda.~~g } order. escen lng Let b i denote the barycenter of a simplex A., i = 0, ••• ,r. ( d" J J. If b , ••. ,b are in {/:scen dl?g ~ order, they are in general posio I' .~e3cen lng) tion. To show this we 'ay assume, without loss of generality, that Ai is an i-simplex, Ai = {ao,.··,a i }. Then b o = a o ' b l is on the line aoa l but b o I a o and hence b I i b o ' etc. The proof, which is done by considering the barycentric coordinates of the points involved, is left to the reader. Theorem: Let A be a simplex with (r+l) vertices. Consider all ( "small \I) simplices spanned by barycen ters of the faces of P. arranged in descending order. Then (i) every small simplex is contained in A, (ii) every point of A is an inner point of exactly one small simplex. Proof:
( i) All b,g,rycenters of faces of A belong to A and A i.s convex. (ii) Let d be a point in A. v·ie renumber the vertices or A in such a way that the barycentric coordinates of d do not decrease. In other words, we have
51 l'
d==L~A.a.
o
and
1
0 < A
1
a
-< Al
... <
<
A r
Note that the point d determines the order of vertices unique 1] except that a. and a. may be interchanged if A. == A ., Set 1 1 J
J
i \Ai .- ~-f-Lj
A == 1-10 0 so that all I-1j > 0 and if j == C) • "\file have
if and only if Aj > A.-l (A > 0 0 J
~j > 0
+ .,. +
=
,
j=O
(~
+ •. ,r + ~r )a o
o (a 0 +a l + ••• +a r ) + ~1(al+a2+···+ar) + .•. + ~rar - (r+l)l-1 o b o + r~lbl + .,. + Il r b r
where
~
1
== r+ 1- j
r
!~ j
8. k
Also
Thus d belongs to the sLnplex spanned by the barycenters b o ' b l , " ' , b r arran9;ed in descending order. Furthermore, d is an inner point of the simplex spatmed by those b j for which ~. > 0 and this sicnplex is uniquely determined by the coordiJ nates Aj" Uniqueness is clear, since the computation may be reversed. To sh011'] that any two small simplices are prope rly s i tua ted we note that every point is determined by a set of equalities and inequalities of the Ai' Hence a point in the intersection of two small simplices satisfies two such sets of inequalities. Since such a set of inequalities determines a small simplex, the intersection of two small simplices is a small simplex.
52 Definition:
Let b O' ..• ,b s denote the barycenters of the simplices of a geometric complex K. The barycentric subdivision of K is the complex (J- K defined as follo1'J's: 1) The vertices of (;1\ are the barycenters, b i , of all simplices of K.
2) The vertices b. , ••• ,b. J. O
J.r
span a simplex in ,rK if and
only if, in some permutation, the b i . are in descending order. J Note that if K is a O-complex then K
= IJK.
\.'Je must
verify that () Ii:: is a complex &nc. that it is a subdivision of 1\. ~;L i~
a cO'1l.pJexc Obviously J};: is a collect ion of simplices.. Tb,e fact that every face of a simplex in UK is a simplex in v~ K follows immed:Lately from the definition. Now i'lle must show that any tv'JO faces are properly situated. If the tv-JO faces~elong to the same simplex then by the prev ious situated~
theorem they are properly of a simplex Ai E. I:
i
= 1,2,
Suppose that Ti is a face Al 1- A2 - Then
Tl contains all, po :Lnt,s of IE! whose barycentl'ic coordinates satisftr Ai <~~c< l . • Thus to determine Af} Tl drop out " 0 ..., ,
-
J.I'
of this set of inequalities all A'S which do not correspond to vertices of A.
Therefore AnTI is a simplex in K and so is
A () T 2' and we are back to the first case.
(l K is a subdivision of K:
From our pr'evious theorem we
have that if d,e { a O" . ' ,art <2' K then d<2{ b O" . ' ,b r } E: U K, and conversely. Hence IK I I uKI. The fact that for every sub complex K' of K there is a suocomplex uKi of U'E: with,
=
1Kt I = I q- K' I follows from the fact that uK is a complex. We shall denote by umK, the roth subdivision of K.
53 'rheorem: The algebraic dimension of (I'mK sion of K Proof: The proof is left as an exercise.
= algebraic
dimen-
Definition: Let A be the abstract complex 'f,.lith vertices aO" ••• "am and distinguished subsets (simplices) so, ••• ,slJ.. subdivision of A is that abstract complex, (J'A, whose vertices are sO' •• "SI1- and whose simplices are given by the condition that s. '~P.,s. span a simplex if and only
TlJ:~b6.rycentric
1.r
1.0
if there exists a permutation of the si' call them Sj' such tha t s.~ .:J s ~ :) " •• :;; 8 ~ • ~o
~l
Jr
It is clear that vA is an abstract complex and fined as above is a geometric realization of OAt
vK
de-
The mesh of a geometric complex K is the largest diameter of any simplex in K, ice •. the length of the longest I-simplex belonging to the cOi:rJ.f\lex. n .1\_ < Theorem: If dim K = n, then mesh -'" ...... n+l mesh K Proof: If n = 0 then K consists of a finite no. of points and so does sK. Hence mesh u~K = mesh K = o. We need only consider the edges of a. simplex ::-jiG-u K ~'lhich has the barycenter of A as one endpoint. Among these we may restrict ourselves to edges ba i where 8' i is a vertex Of A. Choose one of the a 1.' as oriain, say. a O • Then b = -1-._ '-' ;r+l "";' i:.,..,.,-:", a.1. and ~~..nition:
(/.~
Ib-aol
= !bl .
where
- -1-1 r+
t
1.=1
I :$~
a.1. .....-1.=1
is the diameter of A.
since r < n
Ib
- ao
I :::
and since mesh K mesh
(j
we have
K :: sup lao'" b I < n~l mesh K
54 By repeating this process we have that mesh Defini~:
(j'-'
mh ~ (n~l)lU mesh K.
=
Let~)
(a O'" .,ar ) be an oriented simplex in a complex K, and let a be a vertex of K such that (a,aO, .... ,ar ) is also an oriented simplex. This simplex is called the cone over S with vertex a and is denoted by as. If as is not an oriented simplex in r;: then it is undefined. For an elementary chain gS, where S is an oriented sim~ plex, we define [gas, if as is defined.
agG
== LotherWise un
Definltion:
The
ax, with vertex a O"l7er a chain
~o~
x = ;; giSi is given by ax = .::~ag S = i i
~ ~
g as if all the aSl." are defined, and i '
otherwise it is undefined. Ifaxl and aX 2 exist then a(xl +x 2 ) exists and we have
The !\renee·Itor index
Definition:
~ of
x
is defined by
0'x
... t
Note that d : CO(K,G) Theorem:
If x ==
~
= ~gi
~
G is a homomorphism
giSi then the boundary of ax is given by
a'x
1)
oax = x-a
2)
Oax = x-a dx
if x is a O-chain if x is an r-chain
r > 0
Proof! Since .0 and d t are linear operators it is sufficient to give a proof for the case of elementary integral chains.
55 1) If x is a O-chain, x = (a O )' then ax == aaO _"\ t X :"\ t Qax = aO-a, U = 1. Hence Oax = a O - (cl x)a '\
2) if x is an
ax
=
chain, x == (a O' . ' .,ar ), then
I'
(a,ao, ••• a r )
Uax = (aO,alP ' .ar}-(a,al , ..... ".~)+( a.,.aO.'~..JJ ... -....ar-J+ ..... +(_l}r (a,ao, ••• ,a r _ l ) Definition:
=x
- aax
Let K be a complex and q'K the barycentric sub-
division of K. Then the ~ubdj.vision!§x, of chains over K with coefficlents in G is given by
1) C"'x 2)
=x
i.:'
X
1>:3 a O-chain
(jgS == gb vc) S
if g3 is an elementary r-chain, r>O, and b is th.e bary~enter of S .--~ 8 3) If x = 2:' gi 8 i ti.len ux =~ ' - v gi i if.!e must show that b (f"'C)x is defined. For an elementary O-chain, x = gao' V'x = gao is defined since every vertex in K is a vertex In (j K. Asst;.me that () x is 0efined for an r-l chain. Then if x I'
is an r-chain, x == '-:->.
'1=1
cl 8 i
g i 8., v· x = Y ~
'--
is an 1'-1 chain and hence
g. b i 0- i) 8 .• ~
(j US i
~
j.s defined and
b ClG0 8 i would, if defined be an r-chain as desired. But this follows from the definition of barycentric subdivision. Note that u(x+y) = (Ix + (ly \\le give as an example the barycentric subdivision of an
elementary I-chain, x == (aOa l ) b<JOx
= b (jC> aOa l = bCT(a1
- ao)
Since a 1 -a O is a O-chain b
Cd x = b(al
- a O)
= (aOb)
+ (ba l )
56 By induction on m we define the mapping
by 1)
C;'nlx
:::; x if x is a O... chain
tl~ansposed
\ille shall call the (!m·:~
er' «(j InK)G, I')
cr m
~x
homomorphism of cochains
~
er
::
S (,J'mx
(k, G)
• • . a super d ~Y..l. ..~.?:.£E.. Theo~:
(1) The subdivision operator commutes t-vith the boundary operator .. c;:) (T fiX = () me) x. (2) The superdivlsion operator commutes with the cobounda.ry operator, d-l~ Om':4 = (j'm* (')~s. ~ro0f.:
We shall give a proof of 1) and leave it to the reader to
fill in the details for 2). Let m = 1 If x is a O-chain d (jx = c)x = 0
u Dx = vo
: ;: 0
If x is a I-chain x = ()x = (aob) + (ba l )
c)
Ux
The proof will be done by induction.
(ao'~)
= b - a o + a 1 - b = a 1 - a o = () x ::; (/c)x
Assume the theorem true for r-l chains r > 1. r-chain then o-'x ;::: b (lVc,lX Using tbe forrrllla for the cone we have that
Let x be an
UU.x = 6" Ox - b f) O....... t0 x where now c)x is an r-l chrl.in and hence by our induction hypothesis
c)vx = v-ox and since C)2;x. :: 0, ClO
=0
b
(JUcJx
and bO :::; 0
du;x. =u<:~.
57 Therefore ~Gx = G ~ X
fOl'"
all r.
j(:;111X
Induc tion on
111
gj. vas us
={)m OX
D§finitiog.: Let KO be the O-dimensional skeleton of KJ i.eo, the set of all vertices of K. A Sperner ~apping !, is a mapping (not necessa:t'lly continuous or uni.que)
........
such the.t L maps the inner points of a simplex onto one of the vertices of that simplex. It is easy to see that t is a Sperner mapping if and only if x E !KI impliec x € 5t( x): If x. E IKI then x is an interior point of' some simplex A ={a.o"~o, a r} in K .. Hence t'x = a i and x f=.: S·C~. s-i}' The converse is trivial. lJ~w cons:'der (J"rllK; and the restriction of 't- to the O-dimensional skeleton of GmKe 't now defines a vertex mapping of (5mK into K~ 1:Je may extend (the vertex mapping) 't linearly so that it becomes a simpl12ial mapping
7: !GlnKI
-->
!K!
Clearly, this simplicial mapping is characterized by the fact that x E. IKI impl:!.es that x E St(t'x) (the star is in K of course). From now on, l..Jhenever vJe speak of a Sperner mapping, we shall mean this :;articular simplicial mapping. It is also known as a simplicial aY;)D.£.?l:imatigp. to the identi ty (see Chapter IX) c Tile indu.c ed chain homomorphism of Or (G mK1 G) into C (K,G) will also be denoted by 1'. I' Defini tion: A transposed Sperner mapping ' ' 'c' '" is a homomorphism of oochains over K into cochains over 5 l11K 4
such that ('t'-i}s)x
= s('l'x)
Theorem: (Spe:r:ner f s Lemm,a) If x E. Cr (6 mK,G) ; then 'tGmx = x Proof: The proof is. by indu.ction. Let aO' ~'" a r be the vertices of a complex K. 1) If x is a 0 - chain, x L giai' Then, since Gmx x, we have
=
=
So '- "'" =
'i,..,In..,. ::: "i'v
,. ~ "'"
~
~
gi 7' \.< 9. i ::
X
2) Assume the theorem holds for (r-l)-chains o tve mall prove that it holds for elementary integral r-chains. ~ takes each ilimen~a£y r-chain of G mx into ei ther :!: x or o. Thus "'c' (r;lUx ) ::: ex. tiT emus t show that e:: 1;. cdx :::LGm(OX) :::
ox
so thatc :: 10 This ppoves the theorem for elementary integral chains c For arbitrary integral chains, '\'rie use the fact that 1;' and GIn f'.re hornomol~ph:L::ms. ,., Ii'l T.heor~: If s € c- (6 I K,G, I~) then (6 m~l-t~(.s :::
Proo!.:
S
J··..:tst use the above proof and t~~sx :::
stx.
59 Chapter III - Dimension Definition:
Let X be any set.
J = {G~,G;, ... ,Gr J is
Theory
The collection of sets
called a covering of X if
XCG l UG2U~ ... UG r • If each Gi is open, cover_~E8. of X. it.Ie sa~l that the £,r:der 0'£ a covering
f;. is
~ i~l if iJ~ in 1+2
1+1
of the(""G i , but no xSX i$e. for some subset .,> G. ,G. ,9'.' ,G. . l ~l ~2 ~+l
xQX is in
o
called an ~
I
but for !J;J..l subsets fG. ,G i ,0 L ~1 2
some point of the Gi
of G,
,I +2 •• ,G .. JI. G.
r)
~i:.+2 ' 1
~j
= o.
f1em~:r'k:
Al though only ~.!.~ cuverings vlere defi::.1ed, we might just as well have considered infinite coverings, since all spaces to be considered (for the time being) wi11 be compact.
ii
Defi£itio£.: SU1?Pose that = ~ Ga-~ and )J :;::tHp?; aye both cov~I t J ~ - Q er ings of X. .n is s8.id to be a Eefir!eme'p~ of 11 if every H{3 is contained in BCmE. Go!
For the major part of tb.is cUscuss ion of Dimension 'l'heory, we shall assume that the space X is COD~act and metrizable (and consequently se~arably metrizable o Chapter I, p. ). Two defin.itions of top r.Jlogica1 dlmenslon VJ]:d.ch Hill be shown to be equivalent for crn!~act metric spaces, will be presented, At the end of the discussion, a few i,-rords vJi11 be said about separable (non-compact) metric spaces, and the dimension of such a space will be given in terms of the di~ension of its "compactificationl!. Definition: The (topological) di;.Wens:b0n of X, written dim Y..., is not greater than n if every open covering has a refinement of order not greater than n. Dim X = n if dim X < n, but dim X is not::::. n-l; that is, there is an open covering of X wh:i.ch has no refinement whose order is less than n.
60 If no such positive integer n exists" then t-Je write dim X co •
=
-
It is easily seen that dimension is a topological invariant" Indeed, the definition involves only £PE!~ ooverings, and open sets are preserved under homeomorphisms. ~~:
Topological dimension is an absolute property of a set. 'rhat is, for ACX, dim A n regardless of our choice of coverings by sets open in X, or sets open in the induced topology on A.
=
~f.:
Suppose that ACGI UG 2 U
X then[A()G1 ,Af\G 2 , ••• ,A()Gk,5 ing of A by sets GA
"". UGk ,
where each Gi is open
= [GA1,GA2, ••• ,GAk1 is
a cover-
Let dim A < n mean
which are open in A.
i
that there is a refinement [HI, ••• ,H.t} of [Gl, ••• ,Gk3whOSe order is not greater thBn n. a refinement of [ GAl""'. n.
Clearly {Ar)Ul, ••• ,A()H e ] is
,G~} whose
order is not greater than
Conver'sely it is easy to see that if din! A
~
n in the in-
duced topology, then dim A ~ n using sets open in X. As an immediate consequence of the definition: i) dim 0 = -1 (Given any open cover.ing, ta.ke the covering consisting pf no sets as a refinement. The order of this covering is -I). ii) dim fx}= 0 if x is a point (from a given covering, choose a refinement consistin~ of any single set of the given covering which contains x). Theorem:
Let X,Y be co.mpact.
Then XCY
~
dim X .:: dim Y.
Proof: Suppose that XCG1 U G2 U ••• UGr where each Gi is open in Y. X is a compact subset of a compact space =) X is
Y-Xl
closed:::::;> Y-X is open ::::::;:>{G1 "" .,Gr , is an open ooveriDg of Y. It dIm Y;=::2l,then{G1P ..... ,Gl',Y-X}has a refinement of order ~ n. Let [HI, ••• ,Hs} denote S~Ch a refinement. Consider the suboollection il , ••• lk ) of this refinement,
[H
,H
61 for which we have thrown away all Hi which are contained in Y-X.
Clearly
i Hil , •• • ,Hik Sis
a ref'inement of
f Gl , • • • ,Gr ~
whose order is < n. In order to present the second deflnition of topological. dimension, it is necessary to first prove a lemma, ~:
Suppose that X (compact metric space) has an open covering. Then there exists a 0 > 0 such that if i)ACX, ii) dlam A < 0, then AeG a for some Ga.in the coveringo (5 = lub 0 is the Lebesque numb'er of the cavering.. p.rq 0 a will be called a. Lebesque l1iitrlber) 0 Proof: vJe must show ths.t there exists an r > 0 such that if xEX then Sr{x) is contained in some Ga. We may then take 0=2r.
:s
Assume that no such I' > 0 exists. Then for r I -- 1/2, there is an Xl€- X such that Sl/2(xl ) is not contained in any Ga - For r2
= 1/22 ,
there is an x 2 €-X such that S,,\_2(X 2 ) is not cant::
tained in any Go.- Similarly for any positive integer n there is an xn such that S n(x ) is not in a.ny Gr , . This defines a 2....... n "" sequence [xi} in y.~ In a metric space, compactness implies sequential compactness. 'rherefore a subsequence {Xi .-~ converges to a point Xb X.
Now
fGa.~ covers
J
Xi so x is in some Ga.
Fur-
thermore this Go. is open; hence there exists an r>O such that S (x) CO , contrad.icting our assumption. I' a 4 Eefinition: Suppose that )1 = a is an open covering of X. We shall say that the £2~ ~ove~g ha~ diameter d, written
fG 1
diam b~_~" if lub(diam Go.} a
= d.
Definitio~1:
Let X be a compact metric space. Then dim X < n, if for e';;ry ~ > 0 there is a i'lnite open covering of XI such that diam =:. ~ , and the order of is ~ n. As before, dim X n if i) dim X ;: n, ii) ther, is some (;Z > 0 for which there is no finit.e open covering!f, where
It
diam
!J.:::. G
ft
=
and order of
/J :l.s
~
nr.I.
fo
62 Theorem: The two definitions of (topoloGical) dimension are equivalent. Pr~f:
Suppose that dim X < Consider the open covering
h
n~
using the first definition. = tS--E/2{X)J where xEX. Since
X is compact, there exists a finite subcovering
It'
=
f s ~/2{xl)'·"'S e/2(Xr )J. it
Using the triangle in-
equality, diam t ::: diam j} ~ E. Since dim X ::: n, there is a refinement)) = of //, v1hose order is < n. Finally each Hi C some S ~ /2 (X j ) --? diam Hi ::: diam S E./2 (x j ) .::: C:,
tHI" .. ,HsJ
satisfying the conditions of the second definition. ii) Now sup~ose that dim X ::: n, according to the second definition. Let jJ be any open covering of X. It has a Lebesql.le number <5 (see previous Lemma). From the second definition, there is a finite open covering such that
0',
b
f a) diam < <5 Ij t b) order of!J is < n
!l
t
Clearly f:j is s- refinement of of definition 1.
f; ,
satisfying the conditions
are aL--r..cst in a position to p:eove some really significant results in Dimension Theory. But first we must define the nerve of a collection of sets and review some of the properties of the ~!ar of a vertex of a complex" The first really important theorem we shall pr'ove will utilize all these not.ions in addition to Sperner f s Le~lllna (See end of Chapter I I) co '\rJe
DefinitiQ]1! Let AI, ••• ,Ar be a collection of non-empty sets. The nerve of the collection, is the abstract complex L, whose vertices are the symbols al; ••• ,ar " The ~ertices aI' • • ",aiJ, j:-
will span a simplex in L if and only if
n 1
:f
Ai j
o.
\-Je must show that the nerve actually is an abstract complex" This becomes evident when we observe that if the intersection of a collection is non-empty, the intersection of any sub-collection is non-empty. Thus the faces of a simplex in L are thems.elves L. Similarly for the intersection of two simplices of L.
63 It is easily seen that the order of a covering is equal to the Algebraic dimen~ion of th~ realization of the nerve of the c~vering. Indeed, if the order of is r, then some
£1
n
r+l 1
G. :; 0 but all 1
n+2 () Gi
= O.
But this means that L contains
1
r-simplex, na.:m.ely the one ,spanned by gl' ••• ,gr+l' but does not contain an n+l simplex. Perhaps an illustration would be worth more than additional verbal explanation. The figures below are, from left to right, a collection of sets Al, ••• ,A S' and the 3-comp1ex L, which is a geometr:i.c realization in R3 of the nerve of this col1ectiol:lc L conslets of the (-simplex {a2,a3,a4,as1, its proper faces, the l-s,imp lices ,ta2&11 andla3&11 and the O-simplices
f&1 ~
,
i
~ a 2 ,.,.!! [as
I·
Note that
& 2-sirr.plex of L, since Al nA 2 nA3 = 0
f a l &2&3 \ is ll2!
Let us review the definition of the star of a vertex of a complex, and some of its properties. Definition: Let K be a con~lexJ and let a be one of its vertices. The st~of._~, written st (a), is the set of al.l points in IK/ which are inner points of some simplex which has a as a. vertex. It may be defined equivalently as the oomplement of the union of those simplices in K which do not contain a as a vertex. st (a.) he.s the f'ollowing properties: Property~:
Proof:
st(a) is open
st(a) is the complement of a finite union of cl.osed sete.
Property 2: Let a l , •• e ,a..... be the vertices of a complex K. Then [st (~i) form an ope~ covering of IK I.
3
Proof: st(a i ) is open: property 1. Suppose b G-IKI. lJ.'hen b is an inner point of a uniquely determ:i.ned simplex whose vertices are a i ,a •• ,ai.d In this 1
case best(a i .), j=l,o .. ,t.
'if' ,/"\
J
,froperty.1.: A geo:metric realization of the nerve of [st (a i is isomorphic to K~
!
Proof: Let L be a geometric realization of the nerve of [st{ao),st(al),.""st(an}J, witb vertices a~:e:a;, ... ,a;. ~!o ?EIJ Now a~o, ••• ,aiLQ span an £-simplex in L ~/ ('I st ( a. )../. T O. ... A j=O ~j
If we can show that
l no St(a~"'j )
-:f 0 ~)aio'··· ,ail span /'-"
a simplex in K, then we are through, for we will have established an isomorphism between K and L. That is, we will have a vertex correspondence a 1 <~ a:, in addition to which we will have shown that a i , ••• ,a. span a simplex in o lp /V
K
f
~) .. o"" , .. ~ sp ..n .. simplex
in L.
65
L :Now we will prove that span a
s.~·mplex
in K.
~
in K.
~
rj St (a. o
~j
)
f 0 /:\ ~/
Suppose that a i , ••• ,a iO span a simplex
~
0
st (a i ) consists of the simplex fa o '
"j?l
minus its
proper faces. This set is not empty, for it contains those points all of whose barycentric coordinates are positive, Conversely, suppose b G () St (a i .).
£. no St(a i j ) f
O.
Then there is a point
Tbls means that b is in a simplex which has
J
iS7
a io "" ,a;! among its vertices, inrplying that 5 L a io " .• ,a is itself a simplex in L.
Theorem: Let K be a complex. The topological dimension of /KI is equal to the algebraic dimension of Ko Proof' The proof will 'be divided into two parts, In part i) we shall show that if alga dim K n, then dim JKI ~ n. In part ii) we shall derive the opposite inequality. That is, if alga dim K = n, then dim /K/ ~ n. i) Suppose that l"I'lesh K = q (Pecall that mesh K = length of the longest edge). Consider the roth barycentric subdivision. We know that: a) alg.dim ~K = alga dim K = n b) I O'~J = IE I c) mesh CmK ~ (n~I)mq
=
Since lim ( n+1 Xl )m m-> co
u~ ~ (~r)mq
= 0,
for large enough m, we can make mesh
=: e/2
Thus for any vertex am of <)'~,
66 Here we make use of the fact that the order of the covering [St(a~}J is equal to the algebraic dimension of its nerve, which is n. Therefore, given e > 0, we havEl found a fini te open covering /; :; tst (a~) such that diam ~ E- and the order of
J
~iS
n.
But this means that dim X
~
§
n.
ii) lATe shall show that all open coverings of IK I by sets of arbitrarily small diameter, have order not less than n. Suppose that Gl , ••• ,Gl is an open covering of II~I such that diam Gi ~ € where (/t,' is less than the Lebesque number of the ' covering fst(ai>j of ,IKI. The covering!t has a nerve L, and a Lebesque number 5. A positive integer m may be chosen so that
i
"l
We shall construct two simplicial maps, tl: that
rY
~
G'~ ~L
t2: L -..,. K in such a w~y
,
= t2tl is a Sperner mapping.
First let us define the vertex mapping tl: a-mK---!>L in the following way: Suppose that c is a vertex of vIDK. Then c(2st(c}CG i for some i, since m was chosen so that diam (st (c» < 5. live define This defines a vertex mapping of (/MK into L. Does tl map a simplex of O'lilK into a simplex of' L? Suppose that c i , ••• ,c i 0
1p
. 1 ex span a S J.mp follows that
J.' n
vr.-mK •
Then f) st (c i )
o
:f
p
0, from which it
j
p
() where st(c J,) C Gi • j
°But then gi
#gl , •• "Si
o
1
span a simplex
p
in L, proving that tl is a simplicial mapping of
u-mx
into L.
67 Now let us define the vertex mapping
in such a way that
= ai
tZg i
where a i is a vertex of K and Gi cst(t 2 gi ) = st(ai)(Recall thgt diam Gi < E which is a Lebesque number of [st(a i )} .) To show that this is a simplicial mapping, we must look at the image of a set of vertices of a simplex in L. gi , ••• ,gispan a simplex in L~ ~
a i I.'. ,a i
o
p
p
no
p
G. J. j
f
0
0
-7 . n0 St{t 2g i j ) Y. 0
~
span a simplex. in Ko
r.--1= t2tl is the product of two simplicial mappings, and
therefore is itself a simplicial mappinge It is easily seen that 'i': 0"I11:K .-,. K is a Sperner map:
{C)€? Gi
V n1K -=) cG-st(c)CG i where tlc = gi"
CSt (t 2 g i )
= st (t 2 t l c) = st( 'L c).
Furthermore
Therefore
c ~ st ('t' c) ..."
,
showing that i really is a Sperner map, In Chapter II we discussed the manping 'i":Cr( umK)
--!>
Cr(K)
(where Cr (O""mK ), Cr (K) are int.egral chains over (/~ and K ' respectively.) for r=O,l, ••• ,n. Since alga dim K = n, K has at least one n-simplex and consequently at least one elementary n-chain, which we shall denote by x (aO, ••• ,an ). By Sp erner s Lemm.a,
=
'LO"'mx = x This says that there is a chain
68 C"
,....."
such that /..y = x. there is a chain
But x = l Y = t 2 (t l y)
tly such that t2z
= x,
Hhich means
= zG-Cn(L)
= x 1 o.
Since
is a homomor'phisrrl, z f: O. Thus there is an n-chain over. L, and consequently an n-simplex in L, proving that the alg. dim L > n. But L is a geometric realization of the nerve of the co;ering 1.. i ~ , so that the order of i ~ is not less than n, which coru.ple te-s the theorem. To satisfy the m.Ore impatient reader p VIe might pause here and pr evie'~l some of -ell.e coming events of the chapter. First we shall define the concept of an f -mapping, and then prove a theorem (Theorem II) which char'acterizes the dimension of a compact metric space in terms of such a mapping.. If the notion
G
?G
of an ~-mapping, or even Theorem II itself, does not appear to be intrinsically interesting, it is at least necessary to introduce both in order to prove the main results of this chapter, which will be; Let X be a cOIDDact subset of R. Then . n o 0 X i) dim X n if X 1 0 (X is the interior of 0) o i1) dim X < n if X 0
III: ..Theorem .
=
=
Theorem IV: A compact metric sr.)ace of dimension not greater than n is homeomorphic to a subset of the unit cube in R2n +1 • Definition: i)
¢:X ~ Y is cS.lled an E -mapping, if
¢
is continuous on X
i1) d iam
¢-1 Y <
f; •
That is, an G -mapping is a continuous mapping that is "almost l~l ", in the sense that if two points of X are sep-. arated by more than e 1 then they will have distinct images.
69 Let X be a compact metric space. Then dim X < n (-) for every ~ > 0, there is an E -mapping f, which maps X into the space of a complex Hhose dimension is not gre ater than ne Theo~em
II:
,~
..
'
-pp.o.oi' ~ -fruppos-e that dim X < n.
ite
~pen
covering
,f;
Then-:ror-&>-O, X has a f'in=fG1,: •• ,Gr ] such that
i) diam
$:: e
ii) order of iii) Gin X
:f
f
X
is not greater than n
for any i
Let L be a geometric realization of the nerve of We are going to construct a particular f:X ~
.fJ,
with
e -mapping
ILl
which is known as the cF.~nonical mappin~ of X into We first define the r real-valued functions
IL I
The,Xi have the following properties: i) ,/1,.,1 . ./ is continuous on X ii)
Since each x
ex
=
° for
xQX-G~.....
is in at least one Gi , the function
° and continuous for Consider the function
all x EX.
70 The ~ have the following properties: i) ~i is continuous on X ii) ~i = 0
for xl2 X-G i
iii) 4i > 0 for xQ Gi r
iv)
L1 4.J..
= 1
(The r' funct ions 4icons t i tute a part it ion of unity subordinate to the covering
1;.)
We shall nov.T produce a candidate for the desired G -mapping of X into IL I" To this end we define the function r
fx ==
L1 4i (x)gi
f is certainly continuous on Xo We know that the order of /; is < n and so the alg~ dim L :::! n. What we want to show is that f'x € the sum"
IL I.
For a par·ticular x~ X, omit all 4i (x) This gives us: k
fx =
2. k
f'x G:
IL I
so if
J
each
4i
J
. (x) == 1.
It is now clear that
J
if g. ,gi , ••• ,g. span a simplex in L. II 2 l.k
X G. 1
lj
q.lj (x)
-1
0.
> 0.
from
~i. (x)gi.
j==l
where each ~i. (x) > 0, ~ J J=l
=0
But we 8.lready know that x
e
This will be
0 1
G.
lj
since
Therefore f is a continuous mapping of X
where the alg. dim L < n. Is f an ~ m8.pping? Suppose that for x,x 1 X, fx=fx l =y ~ IL I. y is an inner point of a uniquely determined simplex in L, into
ILl
e
say [gl, ••• ,gm~. Then x,xt~Gi for i that d(x,x') :::! diam Gi :::
e .
= l, ••• ,m
which implies
71 'l'1ms we have shown that if' dim X < n: then for every
e
> 0, there is an G -mapping
f:X -~
ILl
where L is a geometric realization of the nerve of a covering of X stl.ch that. diam ~ & and order of is < n. Naturally dim L < n too. Conversely, we now assume that for every G > 0 there is .,;-. an ~ -m8.pp lng
/J
f;
1;
where dim IKI :; n. We vJant to shovJ that dim X < let us prove the following
n.
First
Let ~ be an Q -lilapping~ Then there exists a 6 > 0 ;~that diam q,-l(A) < G whenever diam A < 6. Lemma:
Proof: If we assume that no such 6 > 0 exists, then there is a sequence of sets siC'IKI such that lim (diam Si) = 0 i->co while
for all i. 'rhis means that there is a double sequence of points in X, x~ ,x~ and a corresponding double sequence in IKI,
i 4Q, j
~xi}
3
t
yi]SUCh
=tYl,
that
lim d(Yl' Y~) = 0 i->co while
E.1. where lim co
i~
~
i
=0
Since X is compact, lim i~ co
xl
=
x'
lim i~ co
=
II
X
72 lilrom the continuity of ~,
exist.
lim i..,..
00
But we also know that .,
til)
lim d'Yi lY i
= 0
1->00
=
Therefore ~xt 4x", which means there is a point y ~.xl 'x" in IK I for which
=
=
diam ~-ly > ~
contradicting our assumption that ~ is an ~-mapping This proves the le~na. Given any E=> 0, we shall construct a finite open covering )Iof X, such that diam 'J/ e a:J. d the order of is not greater than n (second definition of dimension). Let ~ be an e-mapping of X into IKI where dim IKI;£ n.
tI
=:
From the lemma, there is a 5 > 0 such that whenever l (A) < ~ • diam A < 5 for A elKl, then d1am Let 11 •• D,Grl be an open ,cQ"t.;rering of IK I such that diam /;::: 5 and the order of is ~ no Set
4-
/j = fG
/J
The following properties are satisfied i} H. is open (~ is continuous) ~
ii} xC
r
0
Hi
1
( I KIe
r.
IJ Gi) 1
iii) diam Hi :: E. (Lemma) iv) order of
fli ::;: [HI' ..·.,~3
order of )J were grea tel" than n tl·~en some
.
n+2
n 1
Gi
~ O. j
But the order
is ::. n
~2
(If the
t · 'J1 Hi j ~Q.. impJ.ying that of )j is =: n by cons£ruction)
This completes the proof of Irheorern II
73 Theorem III:
-
---
Let X be a compact subset of R. n
0
i) dim X
=:
if X
n
0
1i) dim X < n
0
if X =I 0
0
Proof: i) Suppose that X :f O. -
o
"I
Then
Then there 1s a f> >
o
and a
0
point as X such that So (a)c..XCX.
In other words, about the
o
point a in X, there is a sphere of radius 0 lying entirely in
~.
Consider a set of points £a,bl, ••• ,bn~ with the property d(a,b i ) < 6/2~ By an arbitrarily small displacement, these
points may be bro 11ght into general position while still remaining in 8 6 (a). The displaced points 'E,blP .. ,bn span an n-simplex 6'. Therefore
G'c s ~ (a) C Xo C XC Rn and alg. dim u- = dim 10 i < dim X But a compact subset of Rn is bounded. Therefore X is contained in some complex K. Since K is a comp~ex in Rn , we have dim X < dim IKI =: alg. dim K < n n
=:
Therefore dim X =: n~ o ii} To ShDW that dim X < n if X =: 0, we shall construct an ~ -mapping and use Tbeorem II. v'Je already know that since X is compact, and hence bounded, it must lie in same n-simplex in Rn' call it K. For f > 0, consider the roth barycentric subdivision of K, G/mK, where mesh
C-lnI(
< ~
Letting L denote the n-l skeleton of crIDK (L consists of all simplices in o-mK whose dimension is not greater then n-l) we define f: X ~
as follows
ILl
71-!-
a} fx = x if x ~ IL/ b) If x $ 111, then x is an inner point of a unique n-simplex in V-mK. There is a point y, in this sir..iplex o -\ such that yq X (if not X -:J 0). Project the / " , point x through y to the bowldary of the simplex which is contained in IL I. Denoting ! >( / \ this projection on the boundary by z, we take // \.~ fx z. (See diagram for n 2). It is not I \ / \, \ difficult to see that f is continuous. And 1- _ _ .. ~~., since mesh (i'mK < G/2, f is an E: -mapping of X into the spa ce of an n-l COlilplex L. From theorem II, dim X < n. The definitions and lemmas that follow are needed to prove that every compact metric space of dimension < n is homeomorphic to a subset of the unit cube in R2n +l • Actually the theorem holds for non-compact separable metric spaces, but this more general result will not be proved here. The reader miCht take note of the fact that lemmas 2 and 4 will set the stage for the application of the Baire Category Theorem (See Chapter I)n Definition: The unit cu0e I in R2n +l is the set of points "
\
/
,
'3
=
=
_._._.------..~i.
{yJ = {(Yl ,··· 'Y2n+l)] s::ch that
0 < Yi < 1.
Definition: Let IX denote th.e metric space of continuous mappings of X into I. For f, ge IX we define d (f , g) = 111b d ( f
x(CX
x J gx )
(See Ghapter I) Lemma l~ IX is complete. Proof! Suppose that [fn is a Ce.uchy sequence in IX. any point x E X d(.fn(x), fm(x) ~ d(fn,i'm)
!
F'or
so that for a fixed x, ['f n (x).s is a Cauchy sequence in R2n +l , which is complete. Therefore lim l1-!!>
co
75 In addition, convergence is uniform, (see first inequality) so that the function
°
is continuous. Finally, ~ f(x)~ 1, which results from the fact that 0 < f (x) < 1 for all n. Therefore f E IX and n lim d(fn,f) 0, proving that, every Cauchy sequence in IX converges to an element in IX.
=
°
Lemma 2: For a fixed ~ > let Ce denote the set of all f -r!lappings of X into I. Then C<2 is open in IX. Proo~:
We have already shown that if f is an
~
-mapping there
is a 5 > 0 such tL.tI.t d(fx f ,fX"} < E whenever d(x l ,x"} < 5. No·w let f E Cf: e Choose any g e IX for which d{f"g) < 0/2. Suppose g(x1 ) = g(x 2 ) = y G I. Then d(fxl'Y)~ d(f,g) < 5/2 d(fx2'Y)~ d(f,g) < 6/2
Usir.:.g the triang1e lnequality: d(fx l ,fx 2 ) < 6, from which it follo't·!s that d (Xl ,x 2 ) < e • Therefore g is an f: -mappj.ng. This means that for every f Q Co , there is a 5'(6' = 6/2) such tha t g E C~ whenever "-" d(f,g) < o~
-
Lemma..1: A continuous rna~)ping of e. compact metric space into a metric space is unifo!'mly conthluouS. Let X, Y be metric spaces, X being compact.
tro~;::
Suppose
f :X~ Y is continuous. Take a point Xo G X. Consider S5(x o }' where 0 > 0 is chosen so that d(fx,fx o ) < E whenever d(x,x o )<5. J;'ut "such a ball about every point in X. Since X is compact, a finite number of them cover X, say So {xI), ••• ,So (xn ). I n Choose 0 = min (° 1 " , . , on) • d(XI,X") < 0. Lemma
4:
Then d (fx' ,Xli) < <=- 1I-Jhenever
For any fixed Q > 0, C
is dense in IX.
Proof: W.e must show t:tat for f f2 IX and any ~ t > 0, there is an -mapping g such that d(f,g) < ~ t. To simplify the not-
e
ation let u,s use ~ to denote a pO,sitive number not greater than either the original (2 or fr I .
76 Then our task is to find an -mapping g such t;f1.at d(f ,g) < r;' • Since dim X -!: n there is an open covering l:f = [ Gl , ••• ,Gr whose order is < n and whose d5.a'Tleter can be made arbitrarily small. v-Ie shall requh"e that diam /:; < .; where s is chosen in the following way:
3
i)
s<E
ii) d(fxl ,fx 2 ) < e/2 whenever d(xl ,x 2 } < .; (uniform continuity). For each i, let xi be a fixed point in Gi • Then fX i = Yi GI. The set of points tYl"'.'Yr3 mayor may not be in general position. By an arbitrarily small displacement, we may bring them into general position in such a way that the "displaced!! points, which we denote by tYl' ... • ,yrJ, remain in I. vJe require
To be sure that the Yi remain in I, we move the points Yl'." ,yr , into general position by displacements not greater than ~/4. If some of the points lie outside, we may perform a contraction mapping on these points if.Jhich will "pull them into I" in such a way that no point is moved by more than a distance e /4 by this contraction. At worst a coordinate of 12 E one of these points may be either 1 + ~ or - 1r. Then we must find the number
t,
such that
V(l
(]
+
e -"21) = -4
(1
1 "2 + E) Ij: ="'2
If Yl' •• "Yr are in general position, but lie outside of I, then "'Je per .form the contraction mapp ing given by '(/ (Yi -~)
= ~i
-
~
The poi-n,ts Yl' •• ' 'Yr are still in general position and lie inside I. Let L denote the realization in ~2n+l o~ the nerve of The vertices of L are Yl ,··· ,Yr , (or Yl' •• ' iYr.a if we had to perform the contraction mapping), and Yi ' ••• 'Yi will span a s I s simplex in L if Gi 1: o. j=l s
/J •
n
77 ~Je
shall construct the canonical map g: X ~
ILl
in the same wa.y ae··-was. do-ne-in--rrheorem II, taking gx ;;::
L
~i (x)Yi
ELis an G :~:!.8~: For, assume that gX1 = gx 2 " Then x l ,x 2 are both in some Gi , since 41 (X j ) ~ 0 <;:::;>X j cG i for j ;;:: 1,2. To show that d(f ,g) < E: , take any x EX.
II
d(gx,f'x) ;;::
e:x-fJ:i!;;::
if.::
L:
~i(X)~\-f(x)
~(t'i (x) II Yi~'fxll.:: .2- 4i (x) [11 Y-Yill toJhere x
II
Qi(x}"~
+
II
+
~
Yi-fXIl} =
£. Gi
2:¢i (x) since
Then
f/I
Yi-Yi
ll
<
Yi-Yi 1/ +
II fXi-fXII]
<
f
~ by construction, and
II
fX i - fxll < ~
s.
for
d(x,x i ) < But d(x,x i ) < S holds, since we have only summed over indices i for which xG Gi , and we have chosen so that diam Gi < ~he~m !Y: Let X be a compact metric space whose dimension is not greater than n. Then X is homeomorphic to a subset of
,& ,
s.
the uni~ cub~ in R2,L1+l' Pr.££!: cl / nJ is a sequence of open dense sets in IX,
I
Since
IX is complete, we may use the Baire category 'rheorem to conclude that
ri~l Gl/n
where F is an
is dense in
i!-.
'H:ms there is a map
(; -mapping for all € > O.
This means that F is
a 1-1 continuous mapping defined on a compact space. Consequently, F is a homeomorphism o Remar!: It is also true that for each positive integer n, there is a compact metric space Xn which cannot be mapped homeomorphically onto a subset of R2n , Although we do not prove this here, we may give a simple example for n=l to make the theorem seem at least plausible.
78 Consider two sets of points in R2 , each containing three points (as shown in diagram). Connect the pOint a with each of pOints a, (3, 6'. 'I'hen do the same for band a b c c. The set of points on these nine • • curves may be made into a metric space by using ordinary arc length as the metric_ It is quite clear that this • a. ¥ metric space, whose dimension is 1, cannot be embedded in R2 -
-
-
Many of the results we have obtained hold for non-compact separable metric spaces too. For these more general spaces, we shall present tllree equivalent aefinitions f)r dimension. One is identical to the Lebesque definition for compact metric spaces The secona, knovln as the Urysohn-Menger definition is inductive, and defines dimension at a poin!. POl' example, look at the plane set drawn on the left. Although we haven't defined dirrlension yet, it is intuitively clear that we shall want a definition that allows us to say that the dimension a:t A is. 1, at B is 2, and at C is 0. Ii
-
Lebesque ..definition: .- A seps.rable metric space X had dimension 1:. n, if for every e. > 0, there is an open covering [; of X,' such that diam < G and order of is ~ n. ---~
A
!J
The dimension of the empty set is .:1.. dim X ::. n at a point xe X if there are arbitrarily small neighborhoods of x "Jhose boundaries have dimension at most n-l. A space X has at most dimension n if dim X ~ n at all xt'.. X. dim X=n if dim X ~ n, but dim X ~ n-l is false. That is, there is a point x~ X at which the dimension is not < n-l. 11Ie shall use a third definition which involves the notion of compacti~ication. prysor~~ger deZin~tion:
79 Definition: A is called a cOIDPactification of X if . .i) A is compact ii) AOY where Y is homeomorphic to X. F'rom the Ur~Tsoh1Lk!etrization Theo~ (Chapter I, p. 27) every separable metric space is homeomorphic to a subset of the fundamentql cube in the separable Hilbert space. It has been shown that the fundamental cube is compact. Thus every ~ep ~able metric sp~e_~a~ a compactification. Definition: The dime~sion of a separable metric space is the smallest possible dimension of a compactification. It can be shown that these three definitions of dimension are equivalent in separable metric spaces. Using the third definition;) we shall prove the monotonicity of dimension (already shown for compact metric $paces) and that dim Rn=n. Let A, B be separable metric spaces, A B.
Theorem:
Then
dim A < di111 B. Proof: Let C be a compactification of B. Then since B is a closed subset of a compact space, it too is compact. Thus
-
ACBCBCC so that B is a compactification of A. dim
A <
dim B < dim
Therefore C
holds for all compactifications C of B, and hence for the compactification with smallest dimension, proving that dim A < dim B Defini tion:
The
n-sphe:.~.
in Rn+l is given by
In othe;r words the n-sEhere is the surface of the unit sphere in Rn+l~ Theore~:
Proof:
Dim Rn = n
i)
R contains at least one set whose n
80 dimension is n, namely the canonical n-simplex, whose vertices are a O = (0,0,.,.,0), a l = (l,O, ••• ,O), ••• ,an = (0,0, ••• ,1). i1) Dinl Rn .:: n: We shall define a homeomorphism of Rn and the tlpunctured" n-sphere, i.e. the n-sphere with one point removed. Consider the map: y
- ~ i - Y":x O
,
This is a stereographic projection of Sn onto Rn with x = (1,0, ••• ,0) as the singular point or "north pCDle H of the mapping. 'rhus Rn is bomeornor?hic to a subset of a compact space Sn' v-Ie must now show that Sn is homeomorphic to the boundary of an n+l-simplex. The dimension of the boundary of an n+l-simplex is n, from which it will follow that dim Sn = n, implying that dim Rn .:: n. To see that Sn and the boundary of an n+l-simplex are homeo:illorphic, merely enclose an n+l simplex in Sn' so that the origin lies in the simplex (see diagram for n = 1). Take ~Y' point; x on the boundary of the simplex. Draw the line directed from the origin through the point. It will inter~ sect Sn in a unique point y. Clearly the mapping f, where fx =, y, is a homeomorphism of Sn and the boundary of the simplex. As a result of this last theorem and the fact that dimension is a topological invariant, we may conclude that Rn and ~ and B£! homeomorphic for n f m.
81 Chapter
4
Fixed Point Theorems and Invariance of Domain Let F be a mapping of a space K into itself,
Definition:
F: K
~
K.
Then x € K is called a fixed point of the mappir.g F if F{x)
= x.
Definition: Let K be a subset of a normed vector space. A mapping, F, of K into itself is called a contracting mapping if for all x', X!I~ K
II
F(x f
)
-F(x")
II.:::
6
/I
where 0 < 9 < 1.
XI_X""
Definition: A mapping, F, of a subset, K, of a normed vector space int,o itself is called a non-expanding mapping if :for all Xl, x"€ K
!heorem:
Pr mciple of contractiLg
mappin~.
Let K
=
[ xl/l x II ::.
l} be a subset of a Banach Space and let F be a continuous contracting mappi;ng of K into itself. Then F has one and only one fixed pOint. Proof: l) lI£!<.!ueness; Assume th~t x', x" are two fixed points of F, i.e. I F(x') = x' and F(x") = x". Then since F is a contracting mapping we have
-
II x' - x" II < e II Xl which implies: 2) Xo 4
X'
-
xt!" .
= xu.
Existence:
'\PIe will construct the fixed point.
K and consider the sequence
1. x i 3
Let
82 ·'Ot..]
Similarly 1/ xn-xn_lll ~
911 x n _ l -xn _ 2 ",
and hence
"xn+l-xnll.:: 9 2 11 x n _ 2 -xn _ 2 11 • Therefore, b;y- continuing this process we have
II
xn+l,.. xnll.:: gnll xl-xoll:::
nen
where a.::;:
II
xl-xoll
Using this inequality lJe will show that the sequence {Xi} is a Cauchy sequence.
• Since
e
< I}
¥-Qn tends
to 0 as n tends to (X)~ and hence [Xi
1
is a Cauchy sequence. Eence xn converges to some value Xro and since K is closed, Xoof. K. 1rJe also have xn+l ~ Xoo wrJ.Bre xn+l ::;: F (xn ). Taking the limit as n~oo and using the fact that F is continuous we have
and hence
~
is
a fixed point.
83 Theorem: Principle of non-expanding mapping~. Let K = x/ 1/ xl/ G ~J be a subset of a firdte diruensiona1 nor-med vector space, and let F be a continuous non-expanding mapping ! of K into itself. Then there exists at least one fixed point.
r
=
Proof: Define Fn(x) (1 - ~) F(x) for all integers n. Then Fn(x) is a contracting mapping satisfying the conditions of, _ the previous theorem and hence has a fixed point, say xn~ i.e., xn = Fn (xn ) = (1 - 1) F(xn ) n Since K is comp act, the sequenc.e {xn~ has a subsequence J1... xn } which converges to a point x,-n. Now i V..I
Taking the limit as n continuous, we have
~
(Ii),
and using the fact that F is
Xoo = F(~co ) ..
and hence Xoo is a f:i.x,ed point. Lemma 1.: space X.
(Lebesque) Let F l , ~ •• Ft. be closed sets in a compact Then there exists a Lebesque number 5 > 0 such that
whenever Aex, diarn. A < 0 f and A (\ F i.~ 0 for j
then
J
r
n F 1. f; o.
j=l
= l, ••• ,r
J
°
Proof: Assume that no such 6 exists. Then for 1 :::; 1 t~ere exists a set Al X such that dianl A1<1, Fi fli\t 0 far j=l, ... r and j
l-
OF i. = O. 1
Similarly for
°2 :::; 1/2, °3 = 1/3, ••• ,on = lin, ..•.
J
Since there are only £, sets there are only 2 L combinations of the F i • Therefore one of thege combinations must appear infinitely many times as the particular subaollect1on of the Pi corresponding to 0i :::; l/i.
84 Consequently there is a subsequence [6i~ , which for convenience we aha11 denote by corresponding sequence of sets properties: i) ii) iii)
f 5'\1 1, such that 2 Jsatisfying A'll
there is a the following
lim diam A = 0
v~ <:0
A'll
nr
)1
nF i .. t- O;
j
= l, ••• r,
" = 1,2, •••
oJ
F i . = 0 (note that fF., ••• ,Fi
1
:Ll
J
I'
~iS
a fixed
i.e. it is the same for all v). Choose a point P)1 in each A'll. Since X is compact lim pv. = p exists for some subsequence. We also have colle~t~~,
J
<:0
so that :'..im v~
the Fi
j
r
n
Fi
1
j
CD
e = 0 B.nd d(pco' F i .) v J
=0
for j
= l, ••• ,r.
are closed l:.nd hence Pco is in each F i .•
But
Therefore
J
t-
0 contrary to our assumption.
Lemma 2: (Covering theorem of Xnaster, KuratowSki, and Mazurkievicz). Let S =. lao, ••• ,an be a geometric n-simplex and let be a COllection of n+l closed o ' Fl, .... ,F n sets such that any face raio, ••• ,ao~ , I' ~ 0, of S is covered
y.:;::
LF
!
1
I'
by
U j=O
Proof':
F i . (in particular
Then
o
J
Assume that
no
Fi
nn Fi :f O.
= O. g.
has a Lebesque number 6.
Let K be the complex whose elements are faces of S.
Consider (lmK, where m is chosen so that mesh ()mK < 5 Let ({:': u..... m.,K ~ K be a Sperner map. Since 'L is not uniquely defined let us also require the following to hold
To see that this can be done we consider the simplex . fa .... a.] of which i-is an inner point. By our assumption ~O
t...
~r
r
U j=O
Hence b must be in at least one of the F i.' say Fi k • J
Z-
(b) = a.
~k
vJe
define
(again this mapping is not uniquely determined
since it is clear that b can be in more than one of the F ) ij
.
Let (aO ..... ,an ) denote the integral oriented chain on K which corresponds to the simplex S itself. From Spernerts le:mma
If Sm is an n-sin~lex in ~K, then n
diam s~ ~ mesh O-mK <
Since we assu.med
nn F. o
~
o.
= 0, we have, from the previous lemma,
=
that Fin S~ 0 for SOIne i. But this means that no point of Sm can be mapryed into a .• Therefore n
-
~
86
which contradicts Spernerfs lemraa.
Thus
As a consequence of this lerr~a we may prove the Brouwer Fixed Point Theorem and the theorem of Invariance of Domain. Theorem:
(Brouwer) Let S be a geometric n-simplex, and let F: S -+ S be continuous. Then there is a point xE S such that Fx = x. ProQf: xC2. S ~ x = L Ai a i i sets F O' F I , .,Fn so that
= 2:!-Li a i •
Fx
Define n+l closed
H
if
xG:. F.
~
The collection [Fil satisfies the hypothesis of lemma 2. r
?
let x belong to a face laio, ••• ,a{
Then x
... ~
~ A.. l.
= 1.
j
· ~ S ~nce L!-L.
,
~j
Consider Fx
= ~!-Li.
J
For,
=~.A.~ a. , j=O
"'j ~j
a i ,J
= 1 for some ij we must have Ai. J
>
tJ.i •
But this
j
means that xQ F i .' and so J
[a i
a
, •.• , i
o
1.
rJ
c
r
U 0
Fi . • J
Hence, from lemma 2 n n
Fi ,. 0
b
which implies that there is some point x
~
S for which A. i .:: !-Li
for all i. Since ~Ai = 2:tJ. 1 = 1, Ai' !-Li ~ 0, this can only happen if Ai = ~i for all i. Therefore
87
There is ar..other theorem w~ich is equivalent to the Brouwer fixed point theorem, but easier to grasp intuitively. In or·der to state this theorem and. show its equivalence we ml.lst first give a definition. Definition: A ~ctioQ is a continu.ous mapping of a space onto a subspace in such a way that the subspace r~mains fixed. Theorem: There exists no retra.ction of a closed ball onto its boundary. Wo shall sketch the equivalence proof of the two theorems. SI.!.~,p02e a retraet5.on F' exists. FoL. ('Jt~ tbi r: ms.pping by a mapping which ta~ce& F~x) :l:ltO the point diagonally op~OSit8 it; call this point G(x). Then t:18 pI'crillot mapping is a continuous mapping of the closed ball into itself w1thD~t a fixed point. Since the closed bs.1J. is homeomorphic to a simplex, thls v,;o'.lld contradict the Brouwer theorem. n0nveracly s~ppose there exists //"----...-~:;::~ r a cont~luOUS m£p, F, of the closed ball into itself with no fixed point. We \ .I X could then draw a line from F(x) thru \ I F (.X) x, which intersects the bOllildar-y at \, ,AFru) / some poin~, say ii and map F( x) on , / \J'/ x'. This product mapping, whlcb is tj~ continuous J leaves boundary po:tnts fixed, and hence is a retraction.
/ /x/ \
I
.........._../
Theorem 1: (Inver-iance of an inter-ior point) Let A C Rn' Sur... pose that f:A ~ Rn is continuous and 1-1. Then an interior po:l..nt of A (inter~. or rela ti ve to the topology of Rn) is mapped into an interior point of f(A). Tbeorem 2: (Inveriance of Domain) Let A be an open set in Rn and let f:A ~ Rn be continuous and 1-1. Then f(A) is open.
88 It is quite clear that theorem 2 ·~theorem 1: For suppose that x is an interior point of A. Then x is in some open ball that is contained entirely in A. From theorem 2 this open ball is mapped into an open set lying in f(A), which proves theorem 1. We al so note that if f is an open map (Hhich 1-1e shall prove) then r- 1 is continuous~ and hence theorem 2 may be restated in the following way: -i~
Theorem 2: If A is open in Rn and f ~ A ~ Rn is 1-1 and continuous then f(A} is open and f is a homeomorphism. !:.!'C:2t~
r
If
E.
e A then for some
> OJ S~(a) C Ao
Clearly 3 r (a) contairls an n-simplex S "t,rho se barycents:r is a. We already know that a 1-1 continuous mapping of an n-simplex into R is a homcomorphlsm. n Let 11 cenote the boundary of S. Diagram 1 Now .i).. and (2.} are both compact, disjoi~t subsets of R , and hence f(a) and f(Sl) are both n compact (and therefore closed) and di s joint" Thus ther" is an n-simp:,.·:';x So; with f(a) as its barycen!:;er and .J..
If all the points of Sl are images cf points in S we are through. ]1,::-: Sl1:..1'1e the::6 is a point
C
Diagr'am 2 G 8 1 which is not the image of
any POl:::.lt in S. StJ.b.di vide S = [ao$ ••• ,an1 once. Let St(a i ) ~enote the star of the vertex ai in a'F;,. vIe see easily that
n n
st(a;)
=
Dj,8gram
~
Let Fi = f(St(ai»1) [llJ_-w..~re· fl1 is the boundary of Sl" Clearly [Fi} is a colle-etlon of n+1 closed sets, and since the . map- is l-~-.and a 1s not mapped into ill
n
Ilo Fi = 0
•
Let 5 be a Lebesgue num"Ger. Qf the collection fFt}- We may construct a covering !j{ = {Ho, .... ,HLJ of fil where diam Hi <5/2 and the intersection of any n of the Hi is empty_ Indeed consider o-m.rll' where m is chosen so that me sh O'mSll
~
<
•
bY a m m He may d e. fi ne the O'" •• ,a.f.. H4 to he tIl':;) closure of the stars in .,..mJL, of the a~. ... -v ~ D3fiJ:H3 a collection of sets lFi 5 in the following rnanner._.
. th e ver t ~(jes . D eno t ~ng
0_f'
m (\ (J'1L
We divide lHi~ into mutually exclusive sets ~i such that: i )
~{O
= 7 Hk lei t her ~
~ () F;u
= 0, j = 0, • • • ,
J!..,
or Hk (\ F 0
-f
Then we define
r
~,
l
The collection LFi.l has the follo-;,oJing properties .....,
i) ii)
The F. are closed
UN o
F. =
n
iii)
~
~
.0. 1
( since U Hi =
il. l )
r.,'
rlFi = 0
o
To prove iii) we assume first that x ~ ~'Ll and x t Fi for all io Then s is in at most n-1 H. and hence in at most ~ J n-1 of the Fke Now suppose that x ~ ....'1..1 and x E F s' Assume furthermore that x € Uj and therefore Hj
n FS :I O.
'"-'
If x
f-
F2
o}
90 it is because either x E Fr or ,,, x ~ H.J intersectino F. Since t::) r diam H. < 5/2 we have that x E F -} d(x,F ) < 5/2. Thus if J "" r r x Gall Fr , S5!2{x) n Fr =I 0 for all r. But 5 is the Lebesgue number of the {Fi}' This implies that
n
n
Fi
'I
, 0, which we know is false.
Ii,"\
/""
/ ,-' ' ...... . \ \ We now construct a collection of sets ~~.1 as !,,' '\ <.. 1) '\ follows: \ Let c be the point in Sl which is not the image of .'.X F;, \ any point in S. Construct the !' / \ ~\ /----~~---~l-,-)~---.~\ Hconel! whose vertex is c and .""'--_.. .... ------. whose base is F i ; i.e. the ~~ union of closed segments with c as one endpoint and a point of A A Fi as the other. Call this cone F i • Clearly {Fi~ has the following properties: I, i) the Fi are closed n 1\ ii) nFi == {c~
o
,
.Y
,
IA\ ,,\ I
,
/
_-\\,...-
N
~
iii)
°Un " o
F. == 3 1 1
We now construct the sets V
A
'~-r-
___
F.1 == F.l . U.[f \St(a1.') n(Rn .., Sl)\] .. .l
Recalling that ~ f(S't(a i
o
»
== fa, and observing that
fa ~ Rn - Sl we see that
nno Fi = y
~c " ':.s
Finally, we define the sets
F~~ == f- 1 (F i ) Since f is 1-1, f- 1 is 1-1. the
n
Therefore (\ F~' should be c=O
91 pre-t-'n8.p;'8 of c; but it Has assumed that c has no pre-image and h(.';l1ce
This contradicts Lemma 2.
Then since F~:'
denote any face of S. r
..
0
1.
To show this let Sr ~
r ••• ,ar } = Lao,
fIJL= St(a.) 1.
nJL
we have
that SI'C U F~', and so -
Hence every point in Sl has a pre-image. Applicati~:
The?~~~ 1:
Let qi (~:)~2"'.'Sn) i = 1,2,~ •• ,n be n real valued fUnct10ns of n real variables that are continuous whenever all lSi! < 6. whGre 5 is a given positive nlllnber, and let each ~i(O$~"'O) ::: O. Suppose also, that if a solution of
should exist, it is unique. 1/Jhenevel" _all
111 i I
Then there is an N > 0 such that
< ~Jl, a solution does exist.
!E..£of: ~ == (~1;1 ~2' • .. , ~n) is a continuous 1-1 mappinG of n-tuples = (~lS~2""'~n) into)/ =_(/1 l ,77 2 , ••• ,7"/n). Fl"orn the theor'em on invarlance of domain ~ is an open map, 1. e.,
s
there is an M > 0 and an N >0 (choose N < 5) such that ''1henever each
I '1 i I
< M
we have a solution for all
lSi I
< N.
Theorem ~: Let ~ be a 1-1 linear mapping of Rn into RnThen X maps Rn' onto R • ~
Proof:
--
¢i(~i,···'Sn)
n
n
==
~~ aij~j = rl i
i
= 1,2, ••• ,n.
Clearl;y- the cPi are continuous in all variables Si. From the theorem on invari,ance of domain ~ is a homeomorphism and so fmaps Rn onto Rn'
92 ,!heorem:1:
Let
41
(Sl"" 'Sn.) i :: 1, ••• ,n, have continuous
first derivatives at ~ = (~l' ••• ~n) and in some neighborhood of
s.
Suppose also that det
C)~. ) (~bs ~
_ f o. S
J
Then if ~i (Sl'." IS n ) all
I ';i 1-
11 =
(~)Jl~.··"'7n)~
I( i
I
<8
=rti ,
there is an e > 0 such that if
then ~-l exists and is continuous for
Proof: We shall show that ~ :: (41~e •• ,qn) is 1-1 and then we can apply T~~~~ 1 to obtain the existence of solutions of 41 (Sl'···'~n):: l'li for 71 i "close enough!! to 7'(io Suppose that
~i (~ls~.o;Sn) - ~i {Sl + hl'···'Sn + h n }
= 00
Applying the mean value theorem
=0 This holds fol" all i. Thus we have a homogeneous system of n .. equations in n-unknovms, 11.1 ,h 2 , •• • ,b n " A non-trivial solution exists only if det (
)_
~ !~ J
I
= O.
(';+Qb.)
But we know that this determinant evaluated at s from zerc. Hence from the continuity of the first derivatives of the 4i det
()~ ) ( -.!. c)'; . J
(s+ 9h)
= 0
93 if the hi are small enough. Therefore, in some neighborhood of ~ the solution of 4i (Sl' ••• '~n) = ~i is unique (or is 1-1) • We may then a:.:>ply theorem 1 to
I
1\
~i
= 4i
-
"'2.
i-
That is
and the hypothesis of theorem 1 is satisfied. ~!xed Poi~t
Theorem for Banach Spaces vJe sh~ll extend the Brouwer fixed point theorem to Banach spaces, i.e., complete normed vector spaces. The theorem one gets is the Birkhoff-Kellog-Schauder theorem. This appears in a weak. form due to Schauder$ and a strong from due to Tv'Iazur. ir,Je shall derive both forms.
Lenllna l: Let B be a Banach space, S a convex, compact subset. Then for all e > 0 there exists a continuous map)ing, h, of S into a finite dimensional subset, T J of S, such that Ilh(x) - xII < e. Proof: Since S is compact we can find a finite number of points xl, ••• ,xr in S such that every point in S will have distance < e from at least one of the XiLet~(t) be any continuous function such that
vd (t)
r
, = to
t
~i
'> ="f(:i~). \,.::
Form \j!i(x) Let ~i(x)
=-
*i(x)
.
4:W j (x) J
.
0
->1
0 < t < 1
•
-
Then
>"
T
'¥ i (x)
>
0 on S.
Then ~i{x) is continuous on S,
94 ~i(x)
=0
if d(x,x i ) ~ e and ~ ~i(x)
=1
~
Now let 'r be the convex hull of [xl',.,. .•• ~~l. since S is convex, TC S. Define r
,hex) =
fg'=1
Then
4· (x)x i • J,
Then h(x)eT, and
~
>-
~i (x) Ilx ...x.1J<e ~
II xi-xli
< e
hex) is continuous since it is the sum. of continuous functions, and hence our construction is complete. Theorem: .§chauderts Therorm (weak form). Let B'be a Ban&ch Space~ S a. convex compact subset of B, and F a continuous mapping of S into itself. Then F has a fixed point. Proof: By the previous lemma, there exists finite dimensional subspaces Tn of B and continuous mappings h n such that n==l, •.• ,k and
1/ h n (x)-xl/ -< 1 n Consider Fn (x) ping of S into Tn.
= hn [F'(x)].
Then Fn is. a continuous mapConsider the reC&RstrJction of Fn to TnnS
By Brouwer',s fixed point theorem each Fn has a fixed point, say xn; i.e., Fn(xn ) ::: xn" Now
II F(x)-Fn (x) II < 1.n and as n
~
a:>
This holds in particular for xn' and hence
Since S is compact the sequence (xn ) has a convergent subsequence (xn ). Set lim x = x m • Now i ni
/!F(Xn
i
) -xn
i
/1-->0,
theI'efore
and hence xCI) = F( x oo ) is a fixed point. Le~~~: (Mazur)~ If T is compact, the closed convex hull, H, of T is compact o We will prove that H is totally bounded, which implies that H is compact~ Let H~r be the union of all convex hulls of finite subsets of T, 1 .• e., P~oof:
'r.
H"" =
[:
x/x =
r =>""
i=I
Aix i where r is any finite integer, 2)'i = 1,
Trivially H~l-CH and hence H-;"C H = H. H* is convex for if x, yeH'~ then, by the definition of H*, the closed convex hull of . x, y lies in H~, and hence the segment between x and y lies * ~je will show that -* -* There in H. H is convex. Let x, yeH. exists xv' y" e H* such that x = lim xv' Y = lim Yy. ~
y .,..co
v ..
CD
.)}
Since H
is convex we have tx y + (1 - t }yy
y
~oo
= z.,eH-lI-.
and taking the limit as
we have tx + (1 - t)y = Z e
--* H • -it--
-'~
=
convex and H~3H. Thus H H. Since T is compact, T is bounded. For simplicity assume that if xG T then Ixl < 1. There exists a finite number or points al, ••• ,ar in T such that every point in T has distance < € from at least one of the ai' 1. e., min II x-a i II < ~ for x~ To Now xe H~4 and hence x = 2: AiX i • i Consider y = LAia ji Then
" x - y
II <
;>Ai II x., - a J, .I. i
-
II
< E..
Divide the interval rOtl] into n equal intervals in such a
1 r: way that -n < .s:..r·.
Let z =
Pi is an integer
~
M.
Pi > T n
a. where Ji
Pi
11n
Then
Ila·lI
r
terms in the sum and thus we have
I/y Hence
II
x- z" ~
zll
II
x-y II +
II
y-z II < 2 ~ •
There are only a finite number of the points z.
H*C U S(z,2~), thus ...,.
ii*c U z
S(z,2~)
c...
US(z,3E.)
z
Now
Hence -* H = H
97 is totally bounded.
Theorem: Schauder Theorem (strong form). If S is a closed convex subset of a. Ba.nach space, and TeS is compact, then if F is a continuous mapping of S into T, F has a fixed point. Proof: If T is convex this is precisely the weak form of the theorem. Otherwise consider the convex hull Tl of T. Then F: S ~ '11 ==> F: S ~ T', but by lemma 2, T' is compact. Hence F has a fixed point. App1i9ations vIe shall give some ap~lications of the fixed point theorems to the theory of differential equations. First, however, it will be neces sary· to discuss sorne function spaces and to prove Arzel&ts theorem (also known as Ascoli's theorem). Let Co denote the space of continuous functions of the variable t; unless otherwise stated we shall consider 0 ~ t ~ ~ For x(t)€ Co we define the Co norm of x, denDted by II xII 0 as
II x II 0
= max t
Ix ( t ) I
Hence Co is a normed vector space. By use of the \veierstrass M-test we cap show that Co is complete and is therefore a Banach Sp ace. Let C1 denote the space of continuously differentiable functions and define the Cl norm of x as
II xIII = /I xllo + II x! 16
•
It can be shown that Cl is a Banach space_ We see that this definition can be generalized to th.e space of k-times continuously differentiable functions, Ck Definitio~:
A function x(t) is said to satisfy a Holder condition with exponent a if
where A is some constant.
98 We note that functions which satisfy a Holder condition for a. > 1 have derivatives equal to zero, and hence are constant. Ca.' 0 < a. < 1, is defined as the space of functions which are continuous and satisfy a Holder condition with exponent a.. The norm of x( t) ~ Cu is defined as IIx(t) 1Ia.
Exercise:
•
Prove that Cu is a Banach space.
Defi!!i.t~.9E.: A set of functions [fi(t)} is said to be equico~~~~ if for every C > 0 there exists a 0 = o(~ )
> 0
such that
,
Theorem: (Arzela) Let sC CO. This S is compact if and only if S is closed, bounded and equicontinuous. Proof: We use the' fa.ct that a compact subset of' a metrio space is closed (cllapo 1, p. 20) and that a clpsed set is compact if and onl~~ if it is totally bounded (p.23). We shall show that total boundedness --:.) equicont:i.nuity. Let f' > 0 be given. Then there exists a fipite number of sets Al,.I"Ar such that diam Al < E/3 and SC~ Ai' Clearly, any finite. set of uniformly continuous functions are equicontinuous. Let Yi(t)G:.Ai" Then there is s. 0 > 0 such that i = l, ... ,r.
Now let x(t)es. and we have
Then x(t) lies in one of the Ai' say Al Ix(t) -li(t)/<
t ·
Using the triangle inequality we have
99
Thus S is equicontinuous. it!e mu.st now show that if S is bounded closed. a:ntr'-equicontinuous then it is totally bounded. Given E > 0 there is a 0 > 0 such that if x(t) is any function in S
Consider the set A of all piecewise linear functions, Yi defined by i) Yi(O) = m~whe~e
ill
is any integer, and
ii) Yi (11 0 ) - Yi «(1l-1)0)
=~
for all integers
iii) in any interval no ~ t !: (n-l) 0
71
>
o.
Yi is a linear functiop...
We note that since, Y is defined for 0 .=: t ::. L, A has a finite number of elements. Given any function x(t) ~ S vJe can find a Yi such that /x(t) - Yi(t) I < 3~ and hence
SC V S{Yi 3(; ). Theorem: A set S of continuously dif:terentj.able functions whose derivatives are uniformly bounded, lx' {t} I < a., is equicontinuous.
r
tl Proof:
Ix (t 1 )
- x (t 2)
I.=:
Ix i (t) Idt
.=: a Itl - t 21.
t2
Theorem: (Peano) If f(x, t) is continuous in (x, t) in the who,le (x,t) plane, and if f is uniformly bounded everY1r1here, i.e., Ir(t,x(t»1 ~ M for all t, x, then the initial value problem x{O)
=0
100
has a continuously differentiable solution x = x(t). Proof: l1Te prove the existence in an interval, 0 If x(t) is a solution of (I) then t dt"..". x(t) = j f{~, x(~»
~
t < L.
r'"
'0
For x(t)t2 Co define y = F(x) by
=
yet)
it
f(~-, x(~»
d~ ..
0
"For XE. Co de.f.in.e 1'"
= F(x)
so that
t
j
=
Y ( t)
f ( "L. x ( r-)) d 'L.
o
Hence F is a continuous mapping ot' -Co intoits.Edf; .-in- ..f-act F maps Co into Cl • \tva see that it' vIe can find a fixed point of F it will be a solution of (1). Now
II
'I F(x) Itl =
F(x) 110 ~ LN
II F(x) II
0 +
II
F' (x) lIo~ LM + M = (L + l)M.
We shall prove that F(x) is continuous at a point xl' f(t1x(t» is continuous and therefore for any e > 0 there exists a o > 0 such that Ilxl(t) - x 2 (t)lI o < 0 -:) If(t 1x l (t)-f(t J X2(t)l<e. So for
II
xl (t) - x 2 (t}
IIF ~ )-F (x2 ) III = /I
110
< 0 we have
t
S o
f
(1'"' ,Xl (7.)
- f (
c--',x 2 (~)d 1:' 11
~
L
€
101
Let S = II x 110 ~ LM]. Since IIF (x) 110 ::: LM we have that F maps S into itself. Let L be a subset of S such t:1'':.t
{xI
Henc~~ is closed in Ole
2:"
We claim that the closure of
~" in Co is compact.
This is clear s'ince the derivatives of all functions in ~ have the same bound which implies that they are equicontinuous, This fact together with closure gives compactness e Now S is closed hence L C Sand F: s~ 2:.
.-
By Shauder's theorem] a fixed point. exists/)
Therefore a solution
Theorem: If f(t,x(t), x,(t» is a continuous function of its three arguments defined for all t, x" and x'" and f is uniformly bounded, If/ < M then the second order" non-linear boundary value problem (2) x"(t)
= f(t,x(t),
xf(t})
x(O)::; x(L) ::; 0
has a .solution x = x (t) which is twice continuously differentiable. Proof:
Let y"(t)
= ~(t)
I
,
yeo) = y(L) ::; 0
then (3) yet)
=
°
Thus Y = G(4) defines a mapping of Co into 2 , Define a mapping H of 01 into Co in such a way that if xE:. Cl then H(x) = 4, where 4(t) ::; f(t,x(t)"xt(t». Let F ::; GH. Then F: Cl ~ C2 - If ~e can find a fixed point of F it l-Till be a solution of (2).
102
f is, continuou,s in its 3 arguments H(x) is continuous ~ and ',IIH (x) 110 ~ M. We observe that G is a linear mapping: Sinc~
G{A~ }--=""AGttf> j-'-'
G(~l +
42 )
We shall. ,~stimate IlG II 0 and
= G(~l) + G(~2)
HGlI l •
~ G Co and .thus
I~ I
<
II ~ II o.
(3),~
From
Hence and also
'
IIG(~ijlt = IIG(~flo+
I/Gt(q
)11 0 S
(L 2 + 3L/2)
114110 •
=
Since IG(~l - Q2 j f IG(~l} - G(~2)1 we see immediately that G 1s continuous. Therefore F HG 1s continuous. N~
=
f/F (x) III = llGtH(x)} III
:t
(L 2 + 3L/2)
II H(x) I' 0 ~
(L2+ .3L/2)M =KO
and similarly IIF(x) 112:! (L 2 4- 3L/2 -I- lHIl
K=
s<:..91,
Consider max (Ko,K1>j.
= Kl
°
s
= [x(t)/xE' 1 , "xII < K where Then S is a clGsed~ cOhve~ ball in Cl •
Define :Z:C0 2 such that x
E.Z
(==) If x 112 '<
l1. ~
-
Then
L
is
closed in 02 (not in Cl ). ~CS and hence ~ formed.in 01 will be ~subset of S. Furthermore F: S ~ L ~ Now ~ is compact in C1 (not in C2 ) and since the first and second derivatives are bo~ded~ the first derivatives are bounded and equicontinuous. Therefore F has a fixed pOint. We note that if in addition we assume that f satisfies a Lepschitz condition with respect to its 2nd and Jrd arguments this boundary v~lue problem has a unique solution for L sufficiently small. This can be derived from the fact that F will have a Lipschitz constant involving (L 2+.3L/2), and so for L sufficiently small F will be a contract~ng mapping, whiCh implies uniqueness.
103
Chapter V
Definition: A set of elements A in which-·,a b1r...a~y· operat'1on 'has' been defined is said to be a group if i) A is closed under this binary operation. That is a,b E A ) ab, ba E A ii) a,b,c in A >a(bc) = (ab)c iii) A contains an identity element 1 such that f.oral.l-.a· in A, a'l = loa = a tv) 11'0 every ele:nent a in A, there corresponds an element a -1 st1.ch that Remark: The binary operation, as represented above, is multiplication. However, we shall usually be concerned with additive groups. The change in interpretation is obvious. We shall however, in talking about groups, use the notation of multiplication, as this will facilitate any necessary reference made to texts on group theory. Definition: A group A is called Pbelian (commutative) if for all pairs of eleMents, a,b in A ab
= ba
Definition: Let B be a subgroup of A. Choose any element a in A. By the product as 1f.le mes.n the set of elements in 1-\ which are of the form ab, where b is in B. That is
"-
aB = tab
I
beB}
It can be shown (see any text on group theory) that a group A caD be considered as a ?isjoint union of sets aB, where B is any subgroup. Th5s is called a decomposition of A into cosets of B. Since we shall only be interested in Abelian groups, it 1s true that for any subgroup B,
104
aB
= Ba
for all a in A. This leads us to the notion of the factor (SE~ tien t) group (or differe,nce group if A is additive). Suppose that
A
= alB
+ a 2B + ••• aLB
where aiB () ajB = 0 if i 'I j. Then the 1. sets, alB, a2B, .•. ,a1.B themselves form a group, known as the factor gro2}E of A with respect to B, and denoted by A/B. Denoting the elements of AlB by Bl , B2 , "', B~ we may define multiplication of two elements in AlB as follows: Choose an element b i e Bi for i=1,2, ••• A. Define
where the element bjb k is in the coset Bm' It is quite clear that the elements of AlB are equivalence classes of elements in A in the follol-Jing sense: Two elements of A are called equiva- . .!ent !!'!9d'Ll:~.! B if they lie in the s a.me coset of B in A. liJe shall forego a.dditional explanation, assuming that the reader has been throu~';h 9.11 this before. If not he should consult the literature. Remark: The factor group of A with respect to B exists when A is not Abel:tan, provided that B is an l.nvad.ant subaroup. That is, if for all a in A aB Ba
=
irie define the "natural U homomorphism of A onto AlB in the follow1.ng way: Let f:A .-... AlB be such that for a in A,
if' a is in the coset aiB. (The reader should check that this is indeed a homomorphism.) Recall that f:A ->- C, where A,C are ~roups, is a hO~9.!:phism if' f is a single valued mapping of A into C that preserves addition. That is, for all pairs a l ,a2 in A
105 If f is 1-1 and onto, it is called an iso~!phism. ---:A:1:thoU'gh ~'Je shaTl iibt"pr-ove'lt:"( see Ledennann;. The Theory of Fjnite G,r?.:lps; for example) it turns out that all homomorphisms are "na.tural" homomorphlsms, in the sense of the following: Theorem:
If B is an invariant subgroup of A (If A is Abelian,
all subgroups are invar-iant.) then A can be mapped homomorphically onto A/B. Conversely, if f:A .-;::. AI is a homomorphism then the elements of A which are mapped into the unit element of A I , form an invariant subgroup B of A such that A/B ~ At (j!lIt means
is homomorphic to"). homomoI'phislll. II
B is called the kernel of the
Definition:
A set of elements x l ,x 2 '" . xi.. , of a group A are said to be linearly ind~pendent if Alx l + A2 x 2 + ••• + A.ix.t :::: 0, all Ai be 1ng integers, implies that Al :::: A2 :::: ... :::: AI. :::: o. 'rhe set is called .9:~end~.~ if for at least one Ai =I 0 the integral combination Alx l + •.• + A,.(x( :::: 0.
The reader should convince himself that the following statements are true:
i) If A is a group of finite order (finite number of elements), then E.:..1.l sets of elements ara linearly dependent (hint: take A.1. :::: order of A). . i1) In any group (:i.nfi.nite order), every collection of elements, each of finite order, is linear'ly dependent. iii) Any subset of a collection of independent elements is itself independent. iv) A set of elements containing a subset of dependent elemen.ts is itself dependent. pefiD:i t.~Q.£: The x.~nl!: of dependent elements in A.
!
is the maximum number of linearly inIf, for all pos i ti ve integers 11, tf.:2I'C
exist n linearly independent elements in A, then the group is said to have infinite
~k.
'rhe set of elements x l ,x2 , •.• ,xk in A, is said to be a maximal 1inear:ly indep9.Lldent set, if for any x e A, the
Definitio..!!:
set [x,x l ,x 2 , .•. ,xk} is linearly dependent.
106
Theorem: Suppose {X1 ,X2 , ••• ,xk) .is a maximal linearly independent set in A. Then I'ank A = k. Proof: i).A contains at least k linearly independent vectors, namely x 1 ,x 2 ' ... ,xk " Therefore rank A > k
•
ii) We shall show that any k+l elements Yl'Y2' ••• 'Yk+l in A are linearly dependent. Since the xi form a maximal linearly independent set, for all x in A, x,x l ,x2 ,. ",xk are linearly dependent. And so
where the A's are integers and not all of them are zero. In fact AO .; o. For if 11.0 ::: 0 then either the xi are dependent or x ::: 0, both of which 'V:e have assumed to be false. Thus, if x e A there is a non-zero integer AO' such that k
AOX
: : L1
Aix i
•
Therefore, there exist k+l non zero integers 1-I-j such that Zj
= ~jYj
k
::: f~l ajix i
,
j
= 1,2, •.. k+l
This represents a system of k+l e'quations in k unknowns. Since it has a solution, we know that some linear combination (rational coeff:i.cients) of the z. must vanish. That is, J
Y1 z 1 + •••• + Yk+1zk+1::: 0 where the "'(1 are rational and not all zero. Letting Y be the least common denominator of the "'(i' we see that yy1zl + •••• + Y"'(k+lzk+l where the YYi are integers.
=0
Since Zj ::: ~jYj' we have
107
proving that Yi are linearly dependent. Theorem: Let A be an additive Abelian group with finite rank n and let B be a subgroup_ Form the factor group A/B (actually A-B would be more appropriate since elements of the factor grotp are sets Bl c A which have the property that if x,y are in B1 , then x-y is in B. That is, for any two elements x,y in Bi , x = y mod B). Then rank A = rank
A B +
rank B
Proof: Since A has finite rank, A/B and B must each be of finite rank. Indeed B C A; hence B cannot have more than n li~ nearly independent elements. As for A/B, the fact that A/B is a homomorphic image of A guarantees that rank ~ :: :rank A i) rank A > r~nk B + rank AlB: Let Yl'Y2'" 'Yr. be Ie independer.1t elements of B~ And let Zl,Z2'" .Zp be p linearly independent elements (cosets of B) in A/B. From each Zi' select a repr·esentative element Zi' '\tie shall show that the set
is a linearly independent set in A, proving that rank A > r·ank B + rank A/B. Assume that the set is dependent. Then
where
Ai'~i
are integers, not all zero.
we may conclude that
From
108
which means that in A/B Sinc.e "the Z1 are inde'penden:t '" tJr~
=
1J.2
=
= ~p = O.
Therefore
with some Ai "I 0 which means that tYI J ' " ,Ytt:;1 is a linearly dependent set. i i) rank B = k ==> the re is a maximal linearly independ.ent set Yl'Y2""Yk in B. rank A/B = r;~ there is a maximal linearly independent set Z~.1. ,Z2""'Z r in A/B • Letting zi be an element of Zi' we shall show that {Yl""'Yk' zl""'z;} is a _maxima~ linearly independent set in A. We must show that for any x in A, the set
is linearly dependent. We have two cases: a) !;~: Tnen x = L A.y., and the set is dependent. ]. l. b) x i B: '.rhen x is in some other coset X :/ 0 (remember B is the 0 element of A/B). But since the Zi are a maximalli .. nearly independent set, tx,Zl"",Zr-} is a dependent set. 'l'hus
---
A(I-I.OX -
L
k
I-I.i Z i) =
~ Ai Yi
proving that {Yl""'Yk' zl""'Zr} is a maximal linearly independent set, and that rank A
=n = k
+ r
=
rank B + rank AlB
•
Definition: Let A be an Abelian group. If there is a collection of elements a 1 ,a2 , .•• ,~t in A such that for all x £ A,
109 j,
X ::::
L: n ,a, 1 1
1
where the n i are integers, then A is said to be finitely gene~ted. 'l'he set {aI' a 2 , ••. , aL are called ~~~~ of A. v-ie s tate without proof the following:
1
Theorem: If A is a finitely generated Abelian group and B is any subgroup, then i) B is finitely generated
ii) AlB is finitely generated. Definj.tion: Let A l ,A 2 , ••• ,A1 be ~, denoted by
1
additive groups.
The d:l.rect
is the collection of i.-tuples (a l ,a 2 , •.• ,a-t) wher-e a i e Ai. The rea-ier should verify the following: Theorem: A:: Al (t) A2 CD ••• 8.) AJ!, is an additive group with addition defined as component-wise. Theorem: Let Al,A2, •.. '~t be additive Abelian gr·oups. A :: Al l:t} A2 (£) •.• (~) A/... is Abelian. Definition: cyclic .B.£0uE.'
Then
A group which has only one generator is called a
If' the order of the generator, a, is infinite,
i. e. na =I 0 for all n, then the grou.i.J is called a fr~ cyclic group. Clearly such a gro'J.p is isorn0rphic to the additive group of integers, whi.ch 'we deflote by GO' If a group has one generator· a, whose order is n (order=n mes.ns that ne. :: 0 and ma -I 0 if m < n) then the group is isomorphic to G , the residues modulo n. n
lrTe shall state, without proof, the Fundamental Theorem on -;'1belian Groups. For further discussion see Leder-mann; ~ ~he2roy of Finite Grou.p1., (Chapter VI) ; University ~.t[athema tical 'Texts; or Bers; ~ntrodu£~io~~o Topology, (Chapter XVIII), Lecture notes of 1954-19~;5 term, N.Y. U. ) Theorem:
Every fini tely gener'ated Abelian gr·oup A, can be writ-
ten as the direct sum of r free cyclic groups and q finite cyclic groups,
110 A = Al
(f)
(f')
A
r
®
G1:'
1
®
,'+1 ",,-,,
.
'tl > 1
Gy
q
where 7:'1 is the order of G.1:: ' and i
I
t'l
I .. . I
1";....
1'2
Lq
f
(a Ib means a divides b, or am = b for s orne integer m). The number r is called the £etti number of A, and the ' i are called the torsion coefficients. These numbers are uniquely determined by A. Thus the elements of A may be written as (r+q}-tuples (a l ,u2 , ••. ,ar'~1'~2'···'~q) where the a i e Ai and ~i e G~i' Two elements x,x', in A are equal if i)
ai
:::
ai
fo r i ::: 1,2, ... , r
ii) 'til{I\-~i)
for i
= 1,2, •.• ,q
Theorem: Let A be a finitely generated Abelian group. rank A = r, where r is the Betti number. Proof:
Then
The r elements Xl = (1,0,0, ... ,0)
x2 = (0,1,0, ... ,0) r-th place xr
:::
(0,0, ... 1,0, ... 0)
form a linearly independent set. If we consider the collection X ~ where y. = {al, •. ea 'O'''''~j'.''O) we see that t(y j' x l ' e •• '"r j J r tjYj::: {t' j a l ,···1'j a r ,O,O, ••• O}::: '7.:'ja l x 1 + C j a 2 x 2 + ••• + 'tjarXr" From this consideration it is clear that the set
{y ,Xl' ••• Xr 1 where yeA, is linearly dependent.
Therefore rank A ::: r.
III
Definition:
G:::: G
'"(1
@ G
't. 2
(~
. •• (f) Gl
is called the q
torsion group of A. Clearly, G is the subgroup of .A which contains only elements of finite order. Since A is finitely generated, the order of G itself is finite. Before going on, let us recall some of the ideas introduced in Chapter II. It will be the purpose of the next few chapters to study the topological invariants of polyhedra. That is, whenever a space a.dmits'a triangulation (is-homeomorphic to a finite complex,) its topological properties can be investigated by studyin~ its horeology groups. The principal theorem to be proved in Cha.pter' VIr, states that homeomorphic spaces have isomorphic homology gro:.lps. That is, the Betti numbers and torslon coefficients aI-e topological invariants. The converse of this theorem, known as the principal conjecture of combinatorial topology, has never been proved. Hence the theorem that we shall prove will only allow us to S9.y that two spaces are not homeomorpn5.c if they do not have the same homology groups. 1I'Tben a finite t:'iangulation is not possible we may use the co-homologz groups. But more of co-homOlogy later. For the present we restrict ourselves to finite dimensio~al geometric rectilinear compleyes. Let us present a brief reS'LUlle of notation, definitions, etc J which have al:!:-eady been introduced in Chapter II and earlier in the present chapter. K will denote an n-dLlle!iSional geometric complex. G is a group. (i,~Je assume G is an additive Abelian group unless otherwise stated.) C (K,G) denotes tbe group of r-chains over K, with coeffir cients in G. To simplify notation we shall let r = O,+I,~2, .•. , defining Cr(K,G) to be empty for r < 0, r > n. 'tn1hen there is no danger of confusion we shall write Cr inste~d of Cr(K,G). An element x e Cr may be wr:!.ttea as
112 r
where the Si are elementar:cr-chains~over K, and the gi e G. The bo1indary operator and Cr_l(K,G)
d defines a homomorphism .of C (K G)
-
r
'
;i-x= 0 for all x e C (K,G)
r Zr(K,~ denotes the subgroup of cycles in Cr(K,G), where Z is a
cycle means dZ = 0
Cle arly Zr is the kernel of the hemomerphism ,~. C /Z
r
r
q:: ,)(
That is
C )
r
Br(K,G) denetes the subErcup of Cr{K,G} w'hich ccnsists .of bcundaries. That is, if b e Br , then there is an x e Cr + l such that ~x = b.
Cr
Iz r
~ d (C . r ) = Br- 1
rrhe factcr p.:roup Zr ~ hemclcgy group .of K.
IB r = Hr
is called the r-dimensicnal
Let us .once more remind the reader of what is tc come. The principal task of combinatorial topology is tc determine the tcpclogical invariants of pclyhedre.
The prirlcipal invariants
are the H.omclcgy (Betti) grcups; i.e. the B8tti numbers and the t.orsicn c.oefficients.
We begin by intrcducine the so-called
Euler-Pcincar~. ?ho.r~!!£Eistic, "j...(K) , .of a ccmplex K.
Let us ccns1der Cr(K,G.o ) = Gr , where G.o is the (addi.tive) group of integers. EVidently Cr is finitely generated. Fcr, if
sr
we let denote an or-iented r-simplex in Ie (and hence an ~l~ mentary r-chain; see Chapter II, page 40), then every r-chain can be written as
As a re'sult, Zr' Br , and Hr are alse fini tely generated. As a result .of the Fundamental Theorem on Abelian greups, Hr may be described by its Betti numbers and its tersien c.oefficients.
113 Thus the complex K, assuming dim K == n, may be characterized by the n+l Betti numbers p , and the n+l sets of torsion coeffi. r r r c J. en t s ?; l' ••• ?: , I ' = 0,1, ••• , n. qr Let us denote by aI" the number of r simplices in K. The Euler-Poincare Formula (which shall be proved later) states that
It is quite c1ee.r that once we have shown that the Bettt numbers are topological invariants, we may conclude the j.(K} is also a topological invariant. wbat this means is that ~(K) (like the Pr,~r, ..• tr ) depends only on the space, IKI, of the qr'
complex ){, and. not on the complex i ts.elf • In this way, we see that the complex plays only an auxiliary role. Its principal function is to determine the topological invariants of the polyhedron. Before proving tbe Euler-Poincare Fo~!ula, let us see what it means in the following very simple case. Only the shaded areas anu the edges de:1.ote the space to be cons idered. The ~wo figures dr~wn below illustrate two triangulations of the sa.me polyhedron.
:t Let us calculate !riangulation...!:
r: (_l)r ar
1k for triangulations I and II:
0,0 (nu..'11ber of vertices) == 10 ~
(number of edges)
18
==
0,2 (number of triangles) ==
= 10
~
18 + 8
=0
8 CARNEGIf. lNiTlTOTa
O~ J:.fCrlNOlOGY. J..JSRARY
114
j(K)
Proof:
HI'
= Z!B I' I' rank HI' Br _l
=5 -
7 + 2
=0
•
1
= Pr
= Cr/Z r
~?:;;r =
~> a r
=
PI' + I3 r ?:;;r + I3 r - 1
since u r is the n(.lmber of r simplices, and consequently the rank of C •
I'
13_ 1 =
For r = 0, a O ; ?:;;O' since every O-chain is a cycle. Thus O. For I' < 0, r > n, we may define a = p = O. Then
I'
Ur
= ~r
+ I3 r - l
= PI'
r
+ I3 r + I3 r - l
Multiply5ng both sides by (_l)r and sumrrdng from -co to get
+00
we
The Betti number Po of Ho C9.n be given a very simple geometric interpretation. To motivate the more general discussion, let us look at a few pictures. This in no way is a deviation from our g e11e1"a1 policy of IIdoing topology without picture s II , since this will not affect the general discussion. 1~!e shall find that Po is the nu::nber of "connected compo-
nents ll (definition later) of the complex IC As one might easily guess, looking at the t!rJO complexes drawn below, K has two cO;'.Lnected components 8.nd L has one connected component. ll1Te recall that the elenents of II are equivalence classes of o-cycles. o Clearly, every simplex (a i ) or (b i ) is a cycle. Looking at the picture, we see that (13.1 )
r-.l'
(a 2 ) ' ...... (a 3 ).
That is
·
\'") A ,,:> .ll
115
I
,j/
" ,I }
!
/
!
,
b.
,
"--~----!-.
\-., .. Jv
L
D'i
--.
( L... _ " ) ',- --' ,
;
(a l ) - (a 2 ) =: deal a 2 ), etc. (twe cycles are.hemologous if their diffen~nce is a beundary). But it is net true that (a4) 81nce (a l ,a4) is net a simplex 'Of K. Therefere the rank 'Of He(K) = PeCK) =: 2 since (a1 ),(a4) are two linearly indenendent generators. By the same r'easening He(L) = po(IJ) = 1. That is, (b i ) ",-,(b j ), 1,5 = 1,2,3,4,5. (a l )
·V
Definition:
A cemplex K is said te be
al~ebraically
cennected
if, fer any twe 'Of its vertices a,b, there is a sequence of vertices, 8 1 ,a 2 , ... ,a 8 such that (a a l ), (a l a 2 ), ••• (a s b) are I-simplices in~, i.e., yeu can get frem a te b by travelling along the edges 'Of K. Theerem:
K is al[:ebraically cenn.ected
f
;> K is (tepelegically)
cennected. Preef:
i) Suppese K is''net (tepelegically) cennected.
That is,
n
K ::: L U M, L M = 0, L i 0, :tV! i 0, where L, N, are subcemplexes. Let a be a 7ertex 'Of L, b a ver,tex of ]\'1. AssUJ:ning that K is algebraically cennected, there is a sequence 'Of vertices as described abeve.
If a i is the last vertex in L, i.e. a i + l is in
M, then ~.a1+1) is net in either L 'Or M. But (a 1_ a'+l) is a sim1. _ 1. plex in K (defin:I.tien 'Of algebraic cennectedness). Thus K must be cennected. K.
1i) New supP'Ose K is c'Onnected. Choose any vertex b in If a vertex 'Of a simplex in K can be joined te b by the se-
quence described above, then all vertices of that simplex can be jeined te b.
The set 'Of all suOh simplices in K which can
be jeined to b ferm a subcemplex L.
The simplices 'Of K that are
116
not in L form a subcomplex M. Since K is cormected, IVi = O. Therefore K is algebraically connec-ted.-Definition: If L is a (topologically) connectea subcomplex of K such that K = L (j IY1, 1-1 (\ L = 0, M a subcomplex, then L is called a connected component of K. It is not difficult to see that if Kl , .. oKs are the connected components of K, then K = Kl U • ••
Ki
n
Kj
=0
U Ks
,
i
t
j
Since the continuous image of a connected set is connecte~ the number of connected components of a polyhedron is a topological invariant. Let us now recall the definition of arc-wise.connectivity'. The topological space X is said to be arc-wise connected if, for any two points a,b in X, there exists a continuous function ~:[O,l] -::.. X such that 40 = a,
K is algebraically connected ~
~
K 1s arcwise con-
Proof: i) Let a,b be two points in K. a is an inner point of some simplex S in I": one of whose vertices is, say, a o ' Since a simplex is a convex set, all points y = Aa + (l-l)a o ' 0 .:: A .:: 1, lie in S. Similarly b is an inner point of a simplex..n in K with a vertex boo If K is algebraically connected there is a sequence of vertices al, •.• ,a s such that
are simplices in K. This defines a continuous piecewise linear curve between a and b. Therefore K 1s arcwise connected. ii) Suppose K is arcwise connected. If K is not topologically connected, then it has at least two connected components Land M. Si.nce L,M are disjoint closed sets in a metric space,
117
d(L,M) = 0 > O. Thus if we take points a e L, b e; H, there is no continuous curve lying entirely in 1\ that connects them, meaning that IC is not arcwis e connected. Therefore !rcwise connectedne~-~ . :) topological connect.edness(.....-:?- algebraic c.onnectedness. We recall that if x e C (K,G), then o
x
dX d'X
where
= 2:
g1 (a i )
=0 = L
gi
,
~tx
is called the Kronecker index of the chain x. From a' (x+y) = ~'x + Jl y , we conclude that "(3' defines a homomorphism of Co into G. But Co = Zo since ox = 0 tor all x in Co. Therefore 0' :Zo ~ G is a homomorphism. It follows from the homomorphism theorem that Zo/( kernel of CI') is isomorphic to
~I ZO
=G
But what is the kernel of the homomorphism? '\'lie shall show that for a connected complex K, ~'x 0 ( X N O. i) Suppose x~ O. If (ab) is an elementary I-chain over K, then C)[g(ab) J = g(a) - g(b). And d' [g(a) - g(b)] = O. Since x is a linear combinatIon of terms o[g(ab)], VIe, see that
=
x . . J 0 -, 9 d' x
=0
=
>
ii) Suppose d'X O. Let a,b be any two vertices in K. Since K is connected there is a sequence of vertices al"'.Jaa in K such that (aa l ), (a l a 2 ), .•• (a s b) are simplices in K. Let y g(aa l ) + g(a1 a 2 } + ••• + g(ssb). Then ~y = g(a) - g(b). Thus (a) rv (b). And so every O-chain x over G is homologous to g(a). Since. x ~ g(a), dlX ~ g and so x ~ ~'x(a). Therefore .:3'x = 0 = } X O. Before we continue with the general discussion, let us compute the Homology groups of some polyhedra.
=
I"V
118
......,
,
1\ - ,-- - - - - - - -..-, -,'-
~) ' ~
1
.
~
_ -
1,
f
....~-.
',1
~
~-
C. ~.-- --+-
) --',)
<;::1
""\
f
bi.-, -
r/'
,
:-V.-
Jl
C~\--~~--~-""+t:-
'
i
a·---;L·
I") L"
figure 1
d
'1,,\-.-, .
Cutting the torus along a circu.lar cr·oss .. section (figure la}, opening it up into a cylinder (figure lb) anj finally slitting it lengthwise, gives the rectangle (figur-e lc), with opposite ~'!e
points identified.
must first be sure that the torus actual-
ly j.s a polyhedron, i.e. we must be able to triangulate it.
As
a first l2'uess we might try the following triangulations in the order shown: '--j
I;
I
I
i i
figure 2
119
The trouble with 2a is that each ofthe2.-sim;plices has 3 id"en.--· tical vertices. In "figure Zb ~;1e observe that there are two 2simplices (a,b,d). They should be the same simplex. However it is quite clear that they do not ac tually match. '1'he simplest adequate triangulation is shown below. o
figure 3 Since the torus is COTl!1ected, Po ::: 1 and Ho ::: Go' where we have assumed Go to be the addi ti ve Abeli.an group of integers. (Any other Abelian group will do too.) l,7e recall that the elements of H2 are equivalence classes of 2-cycles where ~ z~ means that z~ = oy. J"ust looking at figure la, we see that the torus itself is a 2-cycle and certainly coes not bound anything since there are no 3-ch8.ins (3-simplices) for the torus. This means that the sum of all the elementary 2-simplices of figure 3 gives us this 2-cycle. To see that this 2-cbain is actually a cycle, we may observe that if we apply the bounda!7 operator to the sum of tbe triangles, we will get a sum of I-simplices (edpes) in which each I-simplex appears twice with opposite orientation, thus making tbe sum zero. Tberefor~ P2 ::: 1 and
zf -
H2 ::: Go • "r-'~~
-I.) ~<;".
zi
120
To comp11te Pl
an~:..J{l-l---l:~··'Us ..
look-·at f'igure la again and make a_g);les.a._~.whict:r""'wlll turn out to be correct). It is not ___ d:tf":fi.cult to see that y and "(' are both cycles that do not bour.d. In addition y + y', i.e. y - yl i dX for any chain x. Furthermore it is intuitively clear that these are the only "independent ll cycles. In order to snow all this, we must use figure 3. Corresponding to y and yl are the cycles z = (ab) + (bc) + (ca) and Zl = (ad) + (de) + (ea) respectively. lrJe must show th.at i) z'f.-'
Zl
ii) If ziti sal-cycle, there exi st integers g and g' such that (z" - gz - glZI) ~ 0 Since ~(f,h,:i.) = (hi) - (1'i) + (fh), we have (hi) + (fh),....., (fi). Therefore we may forget about (fi). Similarly ~(f,d,h) = (fd) - (fh) + (dh) _~ (fd) + (dh) -v (fh) so that we may eliminate (ih), and in a like manner we may for·get about (hi), (ig) and (gf) .so that we now have the pictureshown at right. 11'Je next apply the same argument to ( dh), (hi), ( b g) and (ge) ill that order leaving us with the situation shown in figure 5. It is now quite clear figure 4 that we may eliminate the edges (df), (gd), (eh), (ie), (af') , (cd), (eb) and (ia) respectively leaving us with figure 6. Thus all l-cycles are homologous to tbose cycles 1rJhich remain in the picture, namely z,z'. To see that z,z' are independent we must show that mz + nz I 0 only if' m = n = O. To save figure 5 us the trouble of' writing out /'V
121 a ._len.gthy"--botmd:a..r:y ..e.xpress ion we-·may look at the picture. I Suppose z appears with a mUltiplicity ill, where we consider z as it appears at the \ top of the diagram. Then all triangles which have part 1, of z as 8.n edge must appear figure 6 with the same multipllcity. This continues to all adjacent triangles, so that we finally arrive at the 10'wer z 1-11'1ere it must appear with multiplicity -m due to the opposing orientation. (See picture.) Th us p 1 :::: 2 and Hl :::: Go ® Go • i,'lje may check the results obtained by using the EulerPoincare formula
I
I I I
~-
>
-1) r
(
"--
U
, - ( -1) r P
::::)
I'
---
-
r
Substi tuting ao :::: 9, ul :::: 27, a2 :::: 18, Po :::: 1, Pl :::: 2, P2 :::: 1 into the above formula, we get ul + u2
::::
9
27 + 18
Po - Pl + P2
::::
1
2 + 1 :::: 0
a0
0
:;:
which checks. ii) The HHbius Stri;e! •.'. .-~~, !,,' ...,...,....... _ - - ....... ' ..~ .... -
i "~i.:~·
.i''-
I. II \\\ ;,: !I \. '\. 1 \\')1 \ Ii
it::··
1
~.. ~
I
\'
\.•. I.
,
\\
,'\
\.~
\',
: \ \'"
\
\
; ___ M _ _ _ _ _••
t-,
,j
.. _...... - -" _.... ;-_. -.~----~-P -'-"'-r
\\\
\,
'.\, !
_~
\
!
\.i
'" } __
.~_ • • _ _ • •
\~-.~---.--.--
_. _ _
~
figure 7 '
•
-• • •
. \
122
The NBbius strip is the "l-sided" surface obtained by joining two opposite sides of a rectangle so that diagonally opposed points are identified (figure 7). It is quite clear that figure 7 is an ade4uate triangulation of' the M8bius strtp. The reader should check that the configuration is a complex and that the identifications are maintained under this triangulation. Since the complex is connected, Po = 1 and Ho and Go. Clearly there are no 2-cycles. For if there were a 2-cycle z, every triangle would have to enter in z with the same multipli..city, to insure the cancellation of the tlinterior" edges (e.g. (ae), (cf) etc.) when ~z was formed. But then it is quite clear tha t we c01.l.ld not get rid of the ter-ms (ac), (cd) etc. Thus P2 = 0, and H2 is empty. The easi est way to get Pl is to use the Euler-l'oincare formu.la:
Therefore PI == 1. z
=
\rJe
shall show that
(a b) + (bd) + (d c) + (c a )
is a. I-cycle that does not bound and that all l . . cycles are homo ... logous to z. By the usual ar·gument, z can be a boUt."ld ary of a 2-chain x only if ~ the elementary 2-chains enter into x with the same multiplicity. But again, ~x will con ... ta.in (ac), (cd) etc. That is ~x I o. To see that all other I-cycles aI'e homologous to z, we eliminate, in order, the edges (be), (ae), (ef), (tc), (fa), and (ad) leaving us with figure 8. 'Ihe only other cycle appearing in the picture is
~ -.:-.---.-~ : I
i
:I j
i I
'
L----.-.. ?
...- . -.('i i.
It
I
I !
figure 8
l
123 Zl
but clearly
:::
Zl
(ba) + (ac) + (cd) + (db) + (ba) :::
-z, so that Hl has no torsion coefficients.
iii) The Pro.iectiv~_flane The real projective plane consists of sets of real triples (x l ,x2 ,x3 ), not all zero, where (x l ,x2 ,x3 ) is said to be equivalent to (Yl'Y2'Y3) if there is a real number a I 0 such that X.
1
:::
ay.
1
,.
i = 1,2,3
1~je
may think of the pro jecti ve plane as the set of all straig,ht lines (in R3 ) throue:h the orivtn, Hhere, to each line we may associate an equivalence class of triples {(x l ,x 2 ,x3 the xl' x 2 ,x3 being direction numhers of the line. By requiring that
1,
x~
+
x~
+
x~ ::: 1, we choose a particular element of each equi-
valence class and thus represent the projective plane by the surface of the unit sphere in R3 , where diametrically opposed points are to be identifi,ed. Finally, by projecting the hemispher'e onto the unit circle we get a representation of the projective plane by points inside and on the unit circle, diametrically opposed points being identified. A suit9ble triangu w la t1tion is shown below:
124
figure 9 Since the complex is connected, p o ~ 1, H0 ~ G. Again it 0 is clear that ther'e are no 2-cycles. Therefore P2 ~ 0 and H2 =0. To find PI we mightiJ.se the same method as was employed in the case of the torus. But the Euler-Poincare formula will also a 16, we get give the same result. From a o ~ 9 ' 1
Thus PI == O. HO"t\lever He do have a torsion coefficient 1:"1 = 2. hi'e shall see that i) z = (ad) + (dc) + (cb) + (ba) is a cycle that does not bound ii) 2z = ~x where x is the sum of all the triangles in the complex. iii) All other I-cycles are homologous to z.
12;;
Using the s arne argument as for the torus 'VJe may foy-get about (eo), (o,f), (o,g), (g,f), (fe), (eh), (hg); (ea), Cah), (he), (cg), (ga), (af), (fc), (ee) in this order, leaving us with figure 10. From the b picture we see that ii) and ",...,~---...-. . ,\ /'" '. , .... c>· ~// i '-><:, \~/ iii) hold and dZ ::: 0 gives / I \ us i). / Ii{) \\ ! \ We mention in passing ( "'~ ; -1 _~ that similar results hold if i l~ \ i we choose G as some other Abelian group. But we shall not discuss this matter here. befo~e we get down to proving the fundamental theofigure 10 rem of Combinatorial Topology, which Hill be the business of tho next chapter, we should like to say a few Hords a.bout pseudo-manifolds, and orientabillty. ;
I___
__
•
••. -_ _
I
Definition: An n-complex is called a pseud0.:Ela.ni~ if' the follow lng hold: i) Every s i.mplex of the pseudo-manifold is the face of an n-simplex. ii) Every (n-l) simnlex is the face of' exactly two n-simplices. iii) If Sand T arB any two n-simplices in this complex, there exist a finite number of n-simplices Sl,S2, ••• ,Sk in the complex such that (8 0 :::) 8 and 8 1 , 8 1 and S2'" ,Sk and T'(=Sk+l) ea.ch int.ersect in an (n-l) face; Le. Sj () Sj+l is an (n-l)simplex. Now let us constder an oriented n-simplex S of the pseudomanif'Jld and the associated elementary n-chain which we shall also denote by S. We form
~S
= ~(a
n
o
.
al ••• a n ) ::: ')-- (~l)i i=O
126 where /::-. a i denotes the vertex to be omitted. For each (n-l)-simplex that appears preceeded by a minus-...sisni--we-- -1nt"erchange the position of two 'V'eTt:!"c-e-sin__tlie . · c11.ain. so that this simplex will appear with a plus sign. For G C"·.' example, let S = (a,b,c). /\ Then /" ~(a,b,c)
=
=
( bc ) ( ac) + (ab) (be) ;. (ca) + (ab)
In this manner the bounda!l:. oEer~tor induces !B orienta!12£ R~ the (n-l)-boundary ;' of the n-simplex. In the above example, the induced orienta.t:!.on on the boundary is from a to b, b to c, and c to a.
/
I
\"
'.
figure 11
An n-dimensional pseudo-manifold is said to be coherently oriented j.f the n-s implic es have been oriented in such a manner that the induced orientations on the co~~on faces are opposite in sign. This is so constructed that the bound~ry of the whole complex itself is O. For certain purposes it is convenient to define a pseudomanifold with -a --.-......... boundary. We retain properties i) and ill} of - - ----.the pseudo-manifold, but substitute the following for ii). ii)t Every (n-l)-simplex is the face of either one or two n-simplices. ~nition:
Definition: A pseudo-m8nifold is called ~ientable if it can be coherently oriented. j.~s examples of pseudo-manifolds (the latter with a boundary) that are ~ orientable we have the projective plane and the MHbius strip. Indeed, looking at figure 8, we see that the oppositely identified e~ges, e.g. (ad), both appear with the same orientation, so that 2[(ad) + (dc) + (cb) + (ba)] actually bounds the complex. In the case of the M8bius strip, the edge (ab) appears twice with the same orientation.
127
Definition: The n-d.imens ionalBetti number, p , of an n-dimenn sional pseudo-manifold is called the n-dimensiona! integral of the pseudo-manifold. Theorem: Given an n-dimensional p s eud 0 -mani fold M. 7.'hen i) Pn :::: 1 ~ M is coherently oriented. ii) Pn :::: 0 N is not coherently oriented.
> < )
Proof: Let 3 1 ,3 2 , ... ,Sr be the coherently oriented simplices of M. ffhe chain 3 = 3 1 + 3 2 + ••• I- 31' is a cycle, ainee every (n-l) elementary chain in as appears twice with opposite orientation. 11his follows fro;;l the definition of a coherently oriented pseudo-manifold. Clearly all other n-cycles are multiples of 3. For, suppose z is an n-cycle. Then if 8 1 appears in z wi th a mul tj plici t;,r k, then the other n-simplices which intersect 3 1 must also appear with the same multi~licity, so that when .jz is formed, the (n-l) simplex, which is the intersection of 3 1 and a neighbor, will well appear ti-Jiee with opposite or'tentatJon so that it will cancel out in ~z. Therefore g :::: k3 and p n :::: 1. Moreover in any pseudo-manifold, k[3 l + ... + 3n ], if it exists, is the only n-cycle possible, for no other n-chain will have a vanishing boundary. rrhus either Pn = 0 or Pn :::: 1 holds. If Pn :::: 1 ruld S :::: 3 1 + 3 2 + ... + 31' is a cycle, the pseudo-manifold is coherently ,. oriented. Otherwise pn :::: 0 and the orientation is not coherent.
128 Chapter VI Homology Theory - Part II Definitl.!2E:: Let f: Cr(K,G) ~ Cr(L,G) be a homomorphism for r:: 0,1,2, •.• . f is said to be allowable~ ~f:: fd. Example ~.._of A.llowable Homomorphisms: i) Sj.ml21:ic i81. Mar1pings :
'Ilo every s implic ial mapping f,
we have an associated homomorphism of chains (see pages 47-48) wi th 011' :: fd. ii) m-th B':l!'Ycentric Subdi vis ion:
(par"e 56)
Theo_~:
If f:Cr(K,G} ~ Cr(L,G) is an allowable homomorphism, then f induces a homomor'phism f.,: HI' (K,G) -> Hr (L,G) • ..~
Proof: --._X
i) x -is .a cycle ---_.- -in Gr (K,G) ===';?-fx -is -a cycle -in CI' (L,G): is a cycle ~.-:-·-).h :: 0 =~ ofx :: f2lx :: 0 ( ~fx is a cycle.
ii) x ~ ~ b~~?ar'y in Cr(K,G).w ~fx is §:. boundary in Cr ( L, G): x is a boundary chain <- ) there is a chain y e Cr+l(K,G) such that x ::dy ~-~fx:: fdy:: dfy 0=)fx is a boundary chain in Cr (L, G). Prom this ~. t foLLows that x'" y w >fx'-..J fy, which means tha.t the homology clss3 of cycles containing x is mapped into the homology -c1'ass of c,Icles contalning fx, proving that f induces a homomorphjsm of the
homolo~y
groups.
Theorem:
g: C (L,G) .-,;.. C (M,G) I' r then the cOlnposi te mapping: g f: C -
I'
C{, G)
-> C ( ~l , G) r
is allowable and
(The proof is trivial and left to the reader.) \'Je shall nm'll look for sufficient conditions for two allow-
able homomorphisms to induce the same homomorphism of the
129 homology group.
The question may be posed as follows: uFoI' any
cycle z in Cr(K), when will fz
IV
gz hold?" or, "When will"
fz - gz :;: Jy for some y in Cr+l(L)?1t Defini.tion:
~rJe
shall say that f and g are algebr·aically homo-
topic if there exists a homomorphism Q: Cr(K) ~ Cr+l(L) such that
fx - gx :;: dQx + QdX for all x in Cr(K). 0 is called an algebraic homotopy operaEor. If z is a cycle in Cr(K)j i.e. dZ :;: 0, then
fz - gz :;: dQz + Qdz = dQz This allows us to state the following: Theorem:
Let f,g: Cr(K) ~ Cr(L) be algebra.ically homotopic.
Then f and g induce the same homomorphism of Hr(K} and Hr(L). In other words, whenever z is a cycle in Cr(K), then fz -g(z)-'O. Suppose that 1': Cr(K,G) -> Cr(L,G) and g: Cr(L,G) -> Cr(K,G) are allowable homomorphisms.
Then we
know that fg: Cr(L,G) ~ Cr(L,G) and gf: Cr(K,G) ~ Cr(K,G) are allowable homomorphIsms. Suppose also that fg ~ 1 and gf -' 1 (::, is to be read 1\ is algebraically :homotopic to U ) . This means that there exist homomorphisms 0, Of such that fg ~ 1
Qa
= dg
+ gf - 1 :;: dQ' + 0 'd
Thu.s fg lnduces the same homomorphism of Hr(L) into itself as does the indentity mapping. g~(-. f?l-
::: 1.;(-.
That is, f"og'L::: 1.1'. -1
But this means that l' ~Af
* :;:
g ;1-1 : HI' ( K)
~
'X'").-
:;:
-.>
g* and f
* :;:
Similarly, -1
goil-.
Thus
HI' (L )
1':1 ::: g*: Hr(L) ~ Hr(K) proving that Hr(L) '~ Hr{K). to II • )
(;:' is to be read
It
is isomorphic
130
Theorem: Let K contain a single O-simplex (a). Hr :;: 0 for I' I C.
f!22f:
Then HO
~- G:.~ __
(a) is a cycle that obviously does not bound anything. Therefore Po 1, HO ~ G. Since there are no r-si~plices for rIO, PI' = 0 and Hr = 0 for rIO.
=
Definition: Let (aO, .•. ,a r ) be a simplex of a complex K. Then b(a O" .• ,a r ) will be called the ~~~! (aO, ... ,a r ) with~ .!~ b, provided that (b,a.ol ... ,a r ) is a simplex of K. We shall say that K 1s a. cone if it contains a single simplex and its proper faces. The.'2~:
If Ie is a cone, then HO "" = G and all other Hr :;: O.
Pr:)of: Suppose th'3.t K consists of the n-simplex b( a O" .• , a n _ l ) and all its proper faces. K is connected. Therefore Po :;: 1 -v and HO = G. Now let z be an r-cycle, rIO. Then
where the Si are the e leraentary r-chains and v is that part of the sum which consists of s impU ces in which b does net appear. First let us consider the case where r > 1. Here
o=
dZ :;: d( bll) + r3v - u - bou + dV
Since u + CJv contains no simplex with b as a vertex, it follows that u + dV = 0 and b~u = O. Thus z bu + v = b(-dV) + v = a(bv) prov:lng that z is also a boundary. If r = 1, we have (page 54)
=
o = dz :;: cl(bu) +
~v
As befOl-e, bd'U :;: 0 and u + ~v
=u
- bo'u +
= O.
Hence
z :;: bu + v = -bdv + v
cv
= a(bv)
shall now give another pr'oof of the same theorem. The purpose of the second proof is to acquaint the reader with a ~fe
131 typical construction of an algebraic homotopy operator. In the course of the discussion, it will become quite clear that the construction is really superfluous in this case. However', the reader should pay close attention to the method involved in the constructlon, for later, the algebraic homotopy operator will playa crucial role in the discussion. 2: Let K == b(aO,al, ••• ,an _l ) and L = (c) be two complexes of dimens'i.on n'and 0 respectively. lATe shall construct two allowable homomorphisms Pro~f
.
g: Cr (L)
I
where gf . mappings
N
1, fg
~
~
° (K) r
For this purpose we define the vertex
1.
=c ,
fb = fa i gc = b
i = 0,1, ... ,n-l
Now fg: Cr(L) -;:.. 0r(L) and fg(c) == c. Thus fg = 1. To show that gf' ~ 1, we must construct a homomorphism g: 0ron -;.. Cr+l(K} such that
gf(x) - x
==
2Qx +
O~x
for all x in Cr(K) • The construction will be by induction. For any vertex a of K, gf(a) == b. Thus gf(a)
-
( a)
and we define
::;
Q
(b) for r
...
( a)
= {;( ab)
if a
-I
b
if a ::; b
=0
as follows: (
Qa
=~
(ab) 0
if a
-I
if a
=b
b
We now assume that Q has been defined for chains of dimension ~ r-l. Since g is to be a homomorphism, it is only necessary to carry out the construction for elementary chains. Let S be an oriented r-simplex (or elementary r-chain). Consider
132 w ::; gfS - S - 9ds •
This exp r-e ssion is well defined since di,oJ
::;
Ob fS - dS
- dQdS
as
e Cr-l·
Now
- OlGdS
::;
gfdS
dS
::;
dQdS +
Q~dS
- dGCiS ° ::;
The third equality comes from the induction hypothesis and the fact that dS e Cr _l , Thus w is a cycle. To shaH it is also a boundary, we proceed exactly as in the previous proof. That is, we wri te etc, w = bu + v The chain w is seen to be a boundary of something which we call OS. That is w
Theorem:
=
gfS
-
Hr(K)
S
-
':::1
Hr(crK).
gdS
= ~GS
+ OoS
-
GdS
= dGS
Proof: Since 1:, a Sper-ner mapping, is a. simplicial mappi:1g, and o"t::; aif, bo th(; aua [j' are allowable homomorphisms. From Sperner's lemrr.a 1:0-' = ,J... If we can show that 0"1' ::: 1, then we are through. 'I'rnls our tas}{ is to find a homomorphism G for l...]hich c{'ex - x ::; 9dx + dGx will be cons tructed so that the f'ollovl ing is true: Let S e a'K. Find the smallest simplex S 1 e K, such that S C S 1 • Then I GS I c Is' I (Ix I is to be read lithe support of XII; Chapter II, page 41). Since Q is to be a homomorphism we need only define it for elementary chains. Again the construction is inductive. i) r ::; 0: Let ( a) e O''K • vJe know that a is an inner point of a unique simplex 3' in K. Then 7:(a) ::; at e 3 1 a.r~j o'1:(a) = 'C'(a) ::; a I . To define 0 for r ::; 0, we note that Q
O't(a) - (a)
= 7:(a)
- (a) ==
fod (a ;7:: ( a) )
l
if a :/ a f if a ::; a'
133 Thus we may define
GQ
f:·t(Q»)
=
if a if a
i a' = at
Since a,a' are both in SI, and since SI is a conveX"set,
IGa I
= (a-, a.l-. )-·C S I
•
ii) i~ow suppose that such that if S C Sf, then expression
w
= ifrs
Q
has been defined for dim S < r-l,
IQSI
Let us consider the
Cst,
- s - QdS
where dim S = r. This is certainly a well defined expression since di':'!1 Js = r-l, Bn',l 0 is assu'TIed to have been defined for dim
n.:::
r-l. Furthern:ore w satisfies the following properties: a) Iwl C Sf: (S' is defined as before.) To see this, we invest}gate w term by term. We have chosen S' so that S C S'. From the definttion of 7:, Irs I c S' and
Itf rs I : :
To see that IQ631 C S I , let T be the smallest simplex of K containing as. Then d seT C S I. From the induction hypothesis 8....'1d since dim dS;: r-l, IQ3S1 e T c Sf. ItS
Ic
S.
b) ow :::: 0: (f d :::: dO" and t d :::: 6 T...
Slnce if,"[, are both allowable homomorphisms, Henc e :
Now dS is an r-l cha.in.
Using the induction hypothesis, the
last expression is equal to
c) w
is~ bou~y:
Let b be the barycenter of SI.
Therefore b is a vertex of S. We may write w :::: bu + v where v consists of these elementary chains which do not contain b.
o :::: dw :::: u - b dU where bv e Sf.
+ dv
= O·-}
U :::
-
dv
~ w ::: - b dv + v :::: d{ bv)
134 This pl"loves that
o''t~'
1 and that
H(K) ~ H(·iK)
C orol!.ary!
Proof:
H( K) ,.., H(0'" ffiK ) •
Obvious.
Defi.n..i~J..9.£: Suppose f, g: K ...... L are simplicial maps. f and g are said to be £2.I!!~tol'i.a.lly ..£.lose if whenever (aO,al, ••• ,ar ) is a simplex in K then the 21'+2 vertices faO,fal, ••. ,fa1',ga o' gal"" ,gel' are vertices of a simplex in L. Theorem: f,g are combi.nato1'ially close : ? f,g are algebraical ... ly homotopic, 5..e. they induce identical homomorphisms of the homology groups. Proof: Again our task is to find a homotopy operator Q ~or which fS - gS = oQS + OdS. To find 0 for I' = 0, we observe that for an elementary O-chain ( a) , fa - ga
=
f <3(ga,fa)
if fa
i
'to
if fa
= ga
if fa
i ga = ga
ga
Thus we define (
Oa
\ {ga, fa} { . = l 0
if fa
Since f,g are combinatorially close, we know that for any S E K, fS,gS C some S1 e L. v.ie assume that for dim S ~ 1'-1, Q has already been defined so that losl C S1. Consider an oriented r-simplex S. For'm the expression
w
= fS
- gS -
G~S
It follows from the induction hypothesis and the fact that f and g are combinatorially close that Iwl C S'. Furthermore dw
= dfS
- agS ... ~OdS
= f~S
... gdS - aQaS
= oOdS
- OddS - dGJS = 0
since fx - gx = ;)Qx + OClx for dim x < 1'-1. He have shown that w is a cyCle. But from the previous theorem w 1s also the
135 boundary of something, l.oJh:i.ch we called QS, i.e. w fS - gS
= Q3S
+ ~QS ,
== ~·QS.
Thus:
for dim S = r ,
proving that f ,,-, g and that f -l~: H(K) -.;::.. H( L) and g_~:a:(K) ~ H( L) are the same homomorphis:ns. We would like to define the cylinder over a complex K. Before presenting a formal algebraic definition (for abstract complexes) let us see what we would like our definition to give us in the case of some simple geometric complexes. Let us denote the piven complex by K, and the cylinder over K by CK. As usual we do not want to violate our more intuitive and pre-analytic use of languabe. If the base, K of the cylinder has dimension n, than we want dimey: = n+l. For example, let K consist of the single O-s implex (a). C; K will be Figure 1 the I-s i::rlplex (0.°,0. 1 ) W110se space is tbe unit interval bet-r..Jeen 0.0 and 0.1 . Since ~ K is to be a cOiilplex we should say something about a tr'iangulation of this space. It is quite clear that this is a polyhedron. In the formal definition we shall choose a specific traingulation. Now suppose that K contains a single l-si~plex (0.(3). Then ICoK I is the rectangle of Figure 2, whose height is 1, and whose top Figure 2 and botton faces, (0.1 (31) and of
(0.°(30), r·espectively are
71
0.1.-::::.-------
" rep licas u of K. Similarly if dim K = 2, /C,K I is the prism of Figure 3. Thus it becomes clear a. that in defining the cylinder, we shall want the following properties to hold:
°
1
~.
"(1
'
I ~..-.~------------r---~ ~
-------Figure 3
vO I
136 i) L C K =~> C;L C CK
ii) There should be two simplices KO ,K l , c CK such that KO ~ K ~ Kl. (KO,K l are the bottom and top of the cylinder respectiYely) •
Ic,K!
iii) interval.
is homeomorphic to
IKI
X I where I
Let L be a complex which has m vertices.
is the unit In some way,
(thus our cylinder construction will not be unique) order the vertices b l , b 2 , •.. , b m•
This induces an order·ing of the vertices
of each simplex. Take one such simplex l.a o' a l , .. " an]' (recall [ ] means an ordered s i.mplex). We assume that tbe ordering of the vertices of this simplex is tbe induced one.
The rest of
the discl.lssioil "fIJill involve the construction of the cylinder over the complex K consisting of this r-simplex and its faces. The extension to L \.Jill be obvious"_ Th~ r+l-simplic~s of ~K r.1000 [1100 OJ are to be denoted "oy ~OaOala2···ar' ~Oalala2' .• a r ' 1 1 a29,2a3 1 1 1 1 0J t'r.h . . 1 LaOal ••• a 01 r . . .1 , ••• , laOal •.. ara r • .L J.S gl ves us_ r+ simplices in c;,K of dimension r+l. The face &ga~ ••• a~J of the first r+l-simpl§x will be the bottom of the cylinder and the
OJ
r.
r
°°
face ~~ai ••• a;J of the last r+l-simplex wi.ll be the top of the cylinder. Now let us see if the cylinder, as defined, really
has properties i), il), iii). Prope rty i L: 1"1 C. Ie =-'1 1.;; 1'-1 <;:. CK. Proof: Obvious. Just look at tbe definttjo!1 of the simplices in ~K. Proper·ty il}: KO ~ K ~ Kl.
~
0
e. K rOO
contains ti-JO r-simplices KO,K l , such that OJ"'"
1
r 1 1
IJ
Proof: K = ~Oal •.• a r ' K = I~Oal···ar • Proper.ty ilU: leK I is homeomorphic to IK I >< I. Proof.: In Rr + l , we may consider the canonical r-simplex as a geometric realizatton of K.
Denoting its vertices by a O,a1 , •••
a , we may wri te any point x e r
II\' I x I
in the form
137 where e
=
n-.5eros "'~---., (0,0, •.. ,0,1) and r--
°
~
i
~ 1.
Now t::.K also has a geometric realiz9.tion, and '.very point y e re,K I may be written in the form k
Y = k
'\'
L
°
r
~ L-
1
lJ..i a i ...
°
v.a. ~
~
k r
where lJ..i,v i ~ 0, ~ ~i + ~ vi = 1. We shall define a continuous mapping f:
ICKI
IK\XI
~
Then we shall prove trJat f is 1-1 and onto, and conse,quently a homeomorphism. We defi~e
a.~ + e ;
i = 0,1, ••. ,r
and extend f linearly to all other points of' !~K I. for y e le,I{' , k
fy
= L_ 0 =
r
1 ~ifai +2: v.fa. ~ ~
°
k
k
That is,
r
=L
°
lJ..i(a i ... e) +
Lk
v.a 1 ~
.
1-1013. 0 + 1-118.1 + ••• '" IJ..k-lak-l + (j.Lk+vk)a k ... vk+lak + l
+ vra r + (IJ..O+1J..1 +. ·'·+lJ..k)e
Clearly, f is continuous and fy e
IK I x I.
The mapping is cer-
ta.inly 1-1 slnce the numbers Ai' f. uniquely deter'1line the lJ..i' Vi • To see tha,t f maps leK I onto IK I X I, we must show that every point r
x
= Lo
+
L = 1,
r
whe re Ai'
f..
> 0, ~ Ai
k
Y =
L
°
A.a. + ~
~
fe
is the image of some point
138 k
where
L
o
~i
Looking at the expression for fy, we
see that: A. o = ~o' Al
= tlJ.'
!-LO + ••• + !-Lk
··;,-A.k··=···(~+-Vk)' Ak + 1
= Al
+... +
= vk+l'···A r
= vr ;
Ak -1 + !-Lk = 1.
so that f is an onto mapping which completes the theorem. The reader should not be surprised to see that the cyl:lndar construction induces a homomorphism
which is defined in the following way for elementary chains, (i.e. oriented simplices). First let us recall that in the cylinder cons truction we assumed s.n ordering of K, which in turn induces an ordering on CK and all its faces. Thus the geometric simplex {aO,al " . "a r } receives B.n induced ordering, and so may be written as an ordered simplex (aO,al" •• ,a r ) (the same subscripts are used repeatedly to avoid a cQmbersome sequence of sub-subscripts). !~re shall then write S for the oriented simplex which we associa.te with the eq1.J.ivalence class of ordered simplices to which the naturally ordered simplex [a O,al , ••• ,a r l belongs. (Remember that two oriented simplices are said to be equivalent if one may be obtained from the other by an even permutation of the vertices.) We write -3 if the order is not the natural one. The chain mapping is defined as follows:
111001110 + (aOBla2a2,··ar) + •.• ± (aOal···ara r )
As before, we may associate with (a Oa 1 ••• a r ) in C(K), two oriented r-chains over ~KJ which are the analogues of the lltopU and "bottom" of C,K, na,nely (a1Oa 11 • •• a 1r ) and (a 0Oa 10 ,· .ar0 ), If · 1 0 3 = (aOa1 ••• a r ), we sball denote by S and S , the top and
139 bottom respectively. A straightforward, but long and tediQus calculation usi:ng t.ha. . . definition of ~S = ~(aOal" .a r ) gives us the following .'. Lemma:
deS'" CJs
= SO
Sl.
shall perform the calculation for the very simple case (aOa l ). The idea is the same for the general r-chain.
!AJe
of S
=
C(aOal )
= (a~aga~) o
-
(a~aia~),
d(aOal )
= (a1 )
- (a O)
0
1 0 1 0 1 0 1 0 1 1 JG( aOal ) = (aOal ) - (aOa l ) ... (aOao ) - (alaI) + (aOa l ) - (aOa l ) = sO _ Sl C;~(
1 0 + (ala.o) - (ala l) o0
1 0 a Oa 1 ) = (a 11 a 10 ) - (aOa O)
1I\Te may extend the definition of ~K to the case where K is an arbitrary complex. Let K consist of the simplices Sl,S2""'S and all their proper faces. (v,ie assume that no Si is a proper face of any other S .). tl: If is the complex which is J the union of all the (PSI. With this, we .'(lay generalise the lemma sta.ted directly above. The proof is left to the reader. !heorem:
If x e C(K,G), then
= xo
- x
1
Definition: Let f,g, be two simplicial mappings of the complex K into the complex L. Then f and g are called combinato~11-x ~toEi£ if there exists a simpliCial mapping F: ~K ~ L which coincides with f on the bottom and g on the top. This means that FK ,FK ' the restriction of F to KO andK l respectively O... _:1.. .. .... "_' 'arethe "same lt as f and. i-in -the--following sense. We shall say
that f,g are combinatorially ~omot£Eic if there exists a simplici. al mapping B': CK ~ L for which FK : K O --;:.. L, f: K ~ L are
°
the seme homomorphism; and FK : Kl --;:.. L, g: K ......... L are the same homomorphism. . 1 vJe shall now show that if two si.rplicial mappings are com-
binatorially homotopic they induce the same homomorphism of the homology groll.ps. iNe have already shown tha.t two algebraically homotopic allowable homo~orphisms induce the same homomorphism of the homology groups (pare 129). Thus we need only prove the following: Theorem: f, g 9.1"'e cowbin9.torially homotopic braically homotopic. ~f
:
>- f ,g
are alge-
t-re must find a homomorphism
such that for all x fx - gx
£
Cr(K)
= FxO
- Fxl
= ~Qx
+ OdX
where F is the homomoI'phism of G;K into L that agrees wlth f on the bottom and g on the top. Since F is a.n allowable homomorphism and xO - xl = d ex + e...dX,
Thus we mt=l.y define Q for x e Cr(K) as Qx
= F~
Since ~x e Cr+l(~K), the right hand side is well defined and f and g are algebraically homotopic. So far we have dealt almost exclusively with homology. As the re.ader might guess, it is not difficult to derive similar results for cohom.ology. An allowable homomorphism f: Gr{K) -7~Cr{L) induces a homomorphism f : CI' (L) ~ Cr (K),
'*
such that
(fo)..~~;)x
== ';(fx)
for 13.11 x
s
C (K), € Cr(L). f* is r ,>' ~I' in the sense that 2/"f'''' == {''';/'' €
H
also an "allowable homomorphism tl (par,;e }.j..9).
\1
With this, we may state a theorem which is the ana-
logue of the first theorem proved in this chapter. The proof is exactly the same if one prefi..xes the words "chainlt, IIcycle 1t , "bound.aryll, and "ho.mologyU by "COli. Theore!!!: I~ f: Cr(K,G) --;;.. Cr(L,G) is an a1101..Jable homomorphism (and thus f"": Cr(L,G) --;;.. Cr(K,G) is alloVJable) then f induces a homomorT..)hism f~~: Hr(L,G) ~ Hr(K,G). It is also quite easy to see that
We leave the proof to the reader. -'~
Q~i:nitio!!.:
~~
We shall say that f' and g are algebraically homotopic :i.f there exists a homomorphism g{}: Cr+l(L) -> Cr(K) such that for all
s
We shall
€
Cr(L)
Sb01..J
that whenever f,g: C (K) -> C (L) are alger
r
braically homotopIc, then the induced mappings, f*,g {(-: Cr{L) -.;::.. CrUn are algebraically homotopic (in the sense of the immediately preceeding definition) and that the s arne homomorpilism
of th e cohomology groups.
f:,g:
are
Fi!lst we must
show that ~~
i) If f,g are algebraically homotopic, then f ,g
i~
are
algebraically homotopic, and then we must prove that
* *
* *
ii) If f,g are algebraically homotopic then f.;pg* are the same homomorphisms of the cohomology groups. In general this will allow us to dispense with a separate discussion of cohomology. For whenever a homotopy operator Q .,
is found, a g-'l' is automatically produced.
First suppose that
the two allowable homomorphisms f,g are algebraically homotopic. This means there is a homomorphism Q: Cr(K} -;.. Cr +1 (L) such that for all x e Cr(K)
142 fx - gx
= OdX
r *Sx -
g
+ dOX
'!il-
SX
Now r?l-,g *: Cr(L) ~ Cr(K), so that f-ll-s,g*s, e Cr(K) and
r
f?l-sx,g~-sx, e
r).
(remember (g*.;): Cr(K) ~ Since f*s and f,g are algebraically homotopic, it follows that ~~
f .;x
~
~~
g .;x
= sfx
- sgx
= ~(fx
- gx)
= s(JQx
+
= sf
QdX)
We may now define a homomorphism O';}: Cr(L) ~ Cr-l{K) for all x e Cr_l(K), ~ e Cr(L). It is clear the.t this expression is well defined. On the l.eft side, o-ll-.; e Cr-l(K) and thus (G*~)x e: On tbe right,
by requiring that
(Q~~)x
= s(Qx)
r.
r.
Ox e Cr(L), and ~.o ~(Oxt e Applying this de:finition to the express:Lon for r"X".;x - g'),.;x, we get
This takes care of i). Finally we must prove that: r,g ~~ s.re algebraically ,~ ..:~ homotopic -; ).f?}Jg~~ produce the same homomorphisms of tpe cohomology groups. Bllt the proof is exactly the same as for the homology groups. To conclude this discussion of cohomology, we shall say a few words about the cohomology groups of a O-complex and of a complex consisting of a single n+l-simplex and tts proper faces. First suppose that K consists of a single O-simplex (a). If e CO(K,G), then s(g(a) = y where g e G and y e c. is a homomorphismj so ~(g(a) + gl (a)) = y + .'(t. Therefore, we see
*
r.
s
that C:: G ~
r
1~.3
is a homomorphism . .~ . "'\'~ 1 l~ ow c' r:, £ C ( K , G) .
But Cl(K,G) is empty since dim K = O. Therefore (;) E. = 0 for all c.. e CO(K,G), proving that all O-cochains are cocycles. Furthermore no O-cochain is a coboun..!,}
dary since there are no (-l)-chains. Thus the O-dimensional cohomology group is the set of all homomorphisms of G into which is wri tten Hom (G,
r,
r ).
Now let us assume that K is a cone, b(a O'" .,an _l ). lrJe shall show that its cOhomology groups are the same as those of a complex cons"t sting of a single O-simplex, which we shall denote by L. In order to establish an isomorphism between the Hr(K) and the Hr(L), we must construct two allowable homomorphisms
such that f~~·g~~,.,..... 1, g~i-f~;·::: 1.
E9.r1ier in this chapter we com-
puted the homology groups of a cone using a homotopy operator. The operator Q, constructed in that proof, automatically defines a Q-l~ for this problem so that f:.g!: = 1 and g~:f!: HO(K) ~ Hom (G, and Hr(K) = 0 for r:/ O.
r)
= 1.
Therefore
144 Chapter VIr Homology Theory - Part III Untl1n:-ow' weha-ve--dealt e.x.elusively· 'with oriented homology and cohomology theory. That is, we have defined oriented simplices, oriented chains, etc. For some pur'poses it is convenient to use the .£Lq~~£ homology and cohomology theor·y. b'ortunately both theories give the same results about most things. As a matter of fact, we shall establish a natural isomorphism between the oriented and the ordered homology (and cohomology) groups. Defini ti.on: Let K be a complex. For every orEiered collection of vertices [aOa l •.• a r ], such that the a i are vertices of a simplex in K, we define an ~lementa!.y int~gE!! ordere~ chain to be a function that a.ssigns the value 1 to this ordered collection and 0 to every other collection. ~-Je shall denote this elementary chain by [aOa· l •· .a r ]. Remark:
We allow a vertex to be repeated any number of times.
That is, [aOalaOa08.2J is an ordered chain of dimension
4-
Definition: An. r-JJ.mens:t.ona1 ordered chain over K with ooeffioients in a g.!2u:p 52, is a linear combination -_._-----
x where each gi
£
=r
~------"""""#
."--
- -
-
-
---
g.S . ~
~
G, and each Si is an. elementary integral ordered
chain. Definltt..£.!}:
An elementar'y order-ed chain [aOal ••• a r ] is said to
be ~~~~r'~~ if for some i I j, a i = a j , e.g. [8.0al8.0aOa2]· Clearly, the ordered r-chains over K with coefficients in '" G form a group which we shall denote by Cr(K,G). Just as we did for oriented cbains, we may define a horno0 morphism ~: Cr .... Cr _l • For el~mentary chains, 0;)
145 ~[aOal •.• a ] : r
r
2 k=O
(_l)k [a a
0 1'"
where ~k denotes the vertex to be o~itted. o morphism, the eytension to Cr is obvious.
~~:
62
Aa
k'"
a] r
.
Since J is a homo-
= o.
Proof:
Same as for oriented chains. Def:tning cycles and bounda.ries in exactly the same manner as for oriented chains, we may form the r-c.irr..ensional ordered homology group by taking the factor group
Definition: The cone over [aOal' .. a r ] with vertex b, written b[aOal, .• a r ] is defined as [baOal, •• a r ] provided that [baOal ••• a r } exists, i.e. provided b,aO,al, •.. ,a r are vertices of a simplex in K. Just as for oriented cbains, we may prove the follo"Jing Theorem:
If x is an ordered r-chain, then dbx = x - bdx
provided that bx exists. The only important theorem about oriented chains that we can not state for· ordered chains is the Spar-ner Lemma. The ordered cochains may be defined in the natural way: Let Cr(T<:,G, denote the group of ordered cochains of dimension r. An element ~ E Cr(K,G, defines a homomorphism of Cr(K,G} into a grolJ.p Defining the coboundary opera.tor, cocycles, and coboulldaries, J;lSt as we did for the oriented cohomology theory, we may form the factor group
r) r.
r}
etc. We shall now establish a canonical isomorphism between the oriented homology grou.ps and the ordered homology groups.
146 Theorem: --Proof: --
We are gotng to define two homomorphisms, o
a.: C
~
r
C
r
.
{3:C
J
o
~C
r
,
r
show that they_ are both allowB_ble~ and that the--induced mapping's -1 a,~ and {3 i~ o f the homo 1o~y- groups are inverses. ( I, e. 0.* = {3~!o and
a:? = (3*, ) ox
For elementary ordered chains we define r )( aOal , •• a r )
t0
if some a i
= aj
when i
-:I
j
-:I
j
Since a i,s to be a homomorphism, the extension to arbitrary ordered chains is obvious, (Note that a is canonical.) We must now show that a 1s a.llowable (do. = a,,). Consider (aOa l • •• a r ) • Case 1: a[aOal·,·a r )
~[aOal"
Suppose a 1
I
a j for i
I
j.
Then
= (aOal·,··ar ) r
.a r ] =
ad(a Oa 1 ···a r )
o
(-l)k[aOll l ,· ''S'k'' .e'r]
= L (-l)ka[aOal"'~k· .. ar} = L (-l)k(aOal"·~k·.. ar)
Case 2:
Suppose some a.
.- -
0.[ 8. 0 a 1 • , • a r ]
L.
~
= a.J
for i
I
j •
=0
oa[8. 0 a1 ., ,a r ] :::: 0
~[aOal" ,a r ]
= 1:
ad [8. 0 a 1 ,· .a r ] ::::
L:
(-1)k[aOa1 ,·
'~k" .ar ]
(-1)k a [a 0 8.1 ," '~k" .ar )
= <-1)j(a Oa 1 ,·,1ir
.·a r ) +
(-l)i(aOal·'~i···ar)
147 Suppose j > i. By a series of consecutive inter-changes with its left hand neighbor (first with a j _1 , etc,) we move a j to the i-th pls.ce.
This means performing j-1+1 permutations
on the second torm of the expression that we obtained above, With this, and remembering that a. = a., we get J
1.
ad [aOa l •.. a r J ::;:
(-1) j (a Oa l • .• a- j . , . a r )
i+ j-i+l( 1'~ ( -1 ) aOal, •• ai_lai' .. aj ..• a r ) ::;: (-l)j
{(aOal":~j".·ar)
+ (-1)(a 0 9· 1 ":;;:-j···a r )}
.- 0
The r'efore ad::;: dU for all ord.ered chains.
Vile now show that U
takes cycles into cycJes, boundaries into boundaries, and consequently hon:;.ologous cycles into homologous cycles, thus indu0
0
cing a homomorphism of HI' into Hr' Sllppose that ()x ::;: 0 where = O. Slmilarly if y where E C e r . The n a
x "
?
ax : ;: adx
x : ;: a
x
r
I,
Y e C r+ l ' tb en
..'
A.nd finally, if x
ax"
- o.y "
,-
y i-·-;> X'" .:~
'",J
-
<)
Y
- y)
= a. ( x"
::;:
.d"Z J
.- a()z
then
"" " O'uz
::::
which proves the assertion. 1J11e now deflne the homomorphjsm
13: Unlike a,
('
C
I'
~C
I'
f3 \'\Iill not be canonical.
tices of K in some arbitrary way.
First let us order the verhTe
define
taking the + sign if aO,a1, •.. ,a r agrees with the ordering assigned to K, and the - sj.gn if it does not. Clearly d!3 = !3d. Reasoning as we did for u, we see that o
f3 induces a homomorphism of Hr into HI"
,
148 Now let us consider the mapping af3: C
r
~
C
r
/3(a Oa l ",a r ) :: [aOal,,·a r ) af3(a Oa l ",ar ) ~ a[aOal···a r ]
= (aOal,,·a r )
Thus a.f3 ~ 1. We shall now show that f3a ~ 1. This will establish the o isomorphism between Hand H. This means that we must consr 0 r C truct a homomorphism G: Cr ~ Cr +l such that for all x e Cr' (I
~ax
- x ~ JQx +
Oa x
Having shown this, we shall observe that if z is a cycle, then {3az -
Z ::
{3az - z
or
tV
~ Qz
0
which will complete the proof of the theorem. As usual Q is constructed by induction. Let x = [a] e o' then f3alal :: {3(a) = [a] so that·
a
{3a[a]
l a]
=0 o
Therefore we define Ox :: 0 for x e Co' Now let us assume that for- dim T ~ r;-l, QT has been defined and I QT I c IT I, Let S = [aOa l ,. oar] e Cr' Form the expression W :: f3aS - S - OJS If the a i appearing j.n the expression S :: laOal, .. a r ] are all distlnct, then !3aS:: S. If some a 1 = a. for i:/ j, then /3aS=O. ~ J Furthermore, since ~S e Cr _l we know that
IQdS I c I ~S I cis I Therefore I f3aS I, Is I, IOds I, c Is easily seen that w is a cycle:
I,
so that Iw I c. S.
It is also
149
But lwl C S, where S is a simplex. Therefore w is also a bounda.ry, and we may 1:lI]rlte w = ~S, proving that {3<1-:::: 1.. Thus ,as _ homomorphisms of the homology groups, a and {3 are inverses, and so define an isomornh:t.sm between HI' and Iir'_ ~ The homotopy operator 0, constructed in the previous dis~r cussion, defines a 0 for which
.>
the ~ being cochains, of course. This defines a natural isomorphism bet:.Jeen the oriented and ordered cohomology groups, Hr(K,G, and lir(K,G, Using the ordered homology and cohomology theory, we shall obtain some rather importa.nt results. The isomorphism theorems will allow us to state the results for the oriented homology and cohomology too. Let G ;::; GO' and l€1t be any group. rllhe ordered rcochain [ is a homomorphism such that
r)
r).
r
Since ~ is to be a homomorphism, we need only talk about the effect of 1; on an integral elementary ordered. obain. 'I,,11e may t~ink of ~[aO,al, •.• ,arJ as a function of r+l variables, deflneq whenever 8.0,a l , •.• ,a are vertices of a simplex in K. If 4 1" r * 0 r+ 1 or !!, £ C (K,G O' then d ~ £ C (r.,G O' ). And since (~i"l;)~ :;c d(~x), we have
r)
r
it11e would. like to define a multiplication of cochains. Since the r-cochains may be considered as fun.ctions of r+l variables, a product of cochains will be just a product of functions. Therefore we must be SUI'e that the values taken on by
150 the two f'unctions 11e in a set in which a multlplicatlon has been defined.
Up to now we have assumed that '[,: Cr(K,G o ) ~
,-
where- 'rlS· anY'group ~····I]o·o.e. -able ·to talk about the product of two cocha1ns, we must now require that
r
be a ring.
(Although
we assume the rea.der to have some familiarity with rings, we shall state a few definitions which wj.ll be used in the discussion to come: Definl tion:
A
L.l.~
i.8 a set of elements which forms an Abelian
group under addltjon and which is closed under a right and left mul t i plic at ton, that 5.s as s oci ative, and distributive with re spect to addj.tion.
That is, for elements a, b, c in a ring,
13.( bc) = (ab) c ;
Defill~;_:t;.l..czg:
that
1'01'
+ ac ;
(b+c)a
Ie,
= ba
+ ca
is a subring such
J L' a e R., we have aa e.J L •
A !"~.8E.! j.d~§;l
·:iR in a ring R is a Bubring such all a e "J R' a e f(, we have aa e I n "
A -two-sided ideal J is both a left and a right ..... -_.Th8. t is, for all a. e J, a e i, we have aa,aa e J.
Definition: ------ideal.
= ab
A left iq~_~..,1L in a ring
that for all a e
Definl~12.l]:
a(b+c)
Definit..iQg:
-~.,-.-
Let l; e Cr(K,G o '
r), 1 e
c.:s("T C h, ('to'
r) .
Then the
exteri. or, or '£'~E. E£.9.'luct of f, and Yj will be a function of r+s+l variables defined by:
(The reader may recognize thjs as an analogue of exterior multiplication of differential forms.) Theorem:
Exterior multi.plication is non-commutative, and dis-
tributive with respect to left and right multiplication. Proof:
Let?
€
Cr(K,G o'
r ), Yf as €
(K,G o'
r ).
(~ A 1)[a Oa l ···a. r + s ] :;: ~[a.Oe'l •.• ar]~[e'rar+l •.• ar+s]
(1
A ~)[aOal···ar+sJ :;: 1[aOal •.• as]~[asas+l ••• as+r] Clearly ~ A ~ ~ ~ " ~ Now let ~,~ e 5r , ~ e
BS.
If we define
(~+?)[aO···ar] :;: ~[aO···ar] + 1(ao ···ar ], we have ~ 1\ {s+1)[a o " .a s + r ] = ~[aO"
.as](~+il)[as·· .as+rJ
:;: ~[ao···asJ~[as···as+r] + ~(ao···as]7[as···as+rJ
=~
A ~[aO···as+r] + ~ ~ 1[ao ··.as +r ]
:;: (~/\ ~ + ?;; 1\ ')/) [ a O ' •• a s + r ] Similarly
IJ:1hus the ordered cochains form a non-commut.at'1ve-·ring.....--~ Theorem: Proof: -!~
d U; 1\ r()[ a Oa 1 • .. a r .+ s +1] :;: U;;!,)'/)[ a 1 ... a r + s + 1 ] r+s+l( . ~ 1\ - (~,I\~)(a.Oe·2· .• ar+s+1J + •.. + ( -l)
1)[ a o ' .. a r + s ]
:;: s[ a l ••• a r + l ]11 [a r +1 ••• a r + s tl ] ... ~ [9. 0 a 2 ••• a r +l JYl( a n + l ... a r +s +1 ] + ••• + (-l}r~[ao···ar_lar+l] il[ar+l,··ar-+s+1J
+
(-l)r+l~;[ao" .a r ]'1 [a r a r + 2 •· .ar+s+1J + •• ,
+ (-1)
r+s+1 ( ]. ( ~ a O'" a r r( a r ··· a r + s
J •
d *~ 1\ 1[ a O " • a r + s + 1 ] :;: o~~t;( a O ' . • a r +1 ) ~[ar+1' .. a r + s +1 ] :;: ~[e.l···ar+1J1[ar+l···ar+s+l] - ~(aOa2···ar+l]}([ar+1···ar+s+l] + .•• (-1)r+ls[ao···ar]~(arar+2···ar+s+1J •
~_
152 l; 1\
d""'~[ao'
.. ar+s+1J == i;[a o " .ar-lar+l]'i[ar+lar+2· •• ar+s+1J
- s[a O " .ar]1[arar+2···ar+s+l] + •.• + (-1)S+1l;[ao···arJ1[ar···ar+sJ Matching terms, we see that
where r == dim So i) t;, r; are cocyc les :=-..:)
T1:2: e ,Qrem:
c: /\ 'f
is a cocycle
ii} s is a cocycle; Y( a coboli...'1.dary -- ~ s 1\ ~ and 'i are coboundaries. iii) :r:, }'( are c oboundari es t) ~'1 are cocycles
Proof:
l~(s I\'~)
== ;/r; 1\
y/:t
1\
S
)- t; Ail is a coboundary. ..~.
~f.
~ '; e/l; == 0,
t; 1\ d~\l == 0 ~
:> sA '1(
I)
~ ==
° ======-??'
is a cocycle. Thus the ordered cocycles form a subring of the ring of cochains.
~ = d·~~.
..
s
~
is a coe-ycle, yt a coboundary (; ~r~ S = 0, Form d-l~C;I\~) = d·;}s.t\~~t;;\d~l-~. But 6)-l"sl\s= 0.
ij)
Therefol'e sf,"\. = t;1\ a~~s == J..J~(sf\~) is a coboundary. Simile.rly d~~(s/\s) = (/~c;,!\s ! s (\d~l-r; = '.·:fc;,Ai; so that ~
A s is also a cobolJ.r.dary. It follows immediately from this theorem that the coboun-
daries form a two-sided ideal in the ring of cocycles. Defin:i.tion: ---,_ ... ring
(i·a,
R.
The ~otient is the collection of equivalence classes of elements Let
be a ri ng, .;
8.
t1tJO -8 ided ideal.
in f(, where two elements a,b in rk are called equivalent, written a IV b if and only if (a-b) e j. Addition and multiplication are defined by the addition and multiplication of representative elements of each class. of elements eqni v9.1ent to e..
That is, let fa} denote the class Then we define:
153 Theorem: The addition and multiplication described above are well defined. ~f':
Fe must show that if we choose different representative elements the roul t"i.plica tion and addition remains the same. Let a' e {a], b' e tbj. Then a' a + a, b' b + ~, a,~ in I. We have
=
at + b'
=a
=
+ b + a + ~ e {a + b}
Similarly a'b' = (a + a)(b + ~) = ab + ab + a~ + a~. Since I is a. two sided ideal ab, a(3 e I. And of course a~ e I since I is a ring. Therefore a'b' e fab}. T.ITe have found that the coboundaries form a two-sided ideal in the ri ng of cocycles. Therefore we ma.y form the quotient ring of cocycles/cobounda.ries, whose elements are equivalence class~s of cohomologous cycles. That is, if s is a coboundary then s + d ~~l;' e {s} for all sI • (Of course dim t; = dirt;: I • Otherwise th3 addition would not be defined.) Thus we have de~ fined a mul tj.plica. tion for· cohomology clas ses • It follows from the isomorphism between the oriented homology groups and the ordered homology ;-;roups that a !1mltiplication is automatically defined for oriented cohomology classes. This allows us to state the following: Theorem: Suppose that f: K ~ L is a sImplicial mapping and is a ring. Then f :induces homomorphisms
r ) ..... Cr(K,Go' r) Hr ( L , GO' r) ~ Hr ( K , GO' r )
f~~: Cr(L,G o'
f::
where f~~ and f~: are ring homomorphisms. to the read er. )
r
j
(The details are left
154 Chapter VIII Homology Theory - Part IV };i'or many purposes, it is convenient to elaborate on t:'lC algebraic structure of the homology and cohomology groups. That will be the content of this chapter. The reader may omit this section without ~ny significant break in the continuity, since it will seldom be referred to in the subsequent discussion. Defip.i~j.~.£:
A = (-£) An
An Abeli.'3.rt gro 1.lp A is called a p:raded group if
(Cj)
means
j
irect sum} v1here each
An
is an Abelian
group. The elements of' A are sequences a = ( ~ , a2 , ... , ak ' 0 , 0 ... ) with only a flni te number of non-zero ter·ms. Defini tl.£!l:
An element of a graded group is called a 1!c:mogeneou.s
e l~l~ of dimens iOD.;.. ..1 :11" all a j = 0 for j
=I f.
and
0,1.
:I o.
~lliLt~.£!:!:
A grad~,L.!'in& is a set of elements which form a graded group lmder addition, and for which multiplication has
been defined. Ex~~ples
Then C
of GradeJ Gro~ps. i) Let K denote a complex, G an arbitrary Abelian group.
= (V
Cr,(K,G) is a graded group. ii) For '"K- and G as in i), and any group, C·~· = wnK,G, is a graded group.
r-)
r
iii) Grad~~~!!o,UP of Differentlal Fo~: Let ~ be a domain in R. For convenience, we consider only infinitely
n
different iable func tlons defined on b". Definition:
The exterior differential form of degree r,
1 ~ r < n over Jris given by the expression
where f.
~l
over Lf.
i is an infinitely differentiable flli~ction defined . .. r A differential-fo r~ of degree 0 is just a function
155 f(x l ,··· ,xn ) where (Xl"" ,xn ) e./.1. For r > n, an exterior differenti al forra of degy·ee r is de fined to be identically 0. Let Dr denote the additive group of differential forms of degree r, in which addition is defined in the natural way:
L.
i 1 < ... 1. r
fi
i dx. 1 • .. n 1.1
= Then D
= (£)
;-
- .1 r i. 1<.,.
1\ ... /\
[l' 1..
dx. lr
.
1 " .lr
+ g~ .1.
. ] dx~ /\ ... /\ dX i 1 ' ··1. r -1 r
Dr is a graded group.
iv) Let S be any set, and let F
r
be the set of functions
of r+l variables, f(xO'xl, .•. ,x r ), where the Xi e Sand f takes on values in some Abelian group. Clearly Fr is an Abelian group in which we have a.efined addition of two functions f l ,f 2 e F'r as follo1rJs!
1".1e may form the graded group F ==
'''\ (-I)
Ii' r'
Examples of Graded Ri~. We shall define a multiplication in each of the last two examples of the previous section. In both cases we must assume that the functions take on values in a rin,g. i) dx. /\ dx. is already defined for i < j. :1.
•
J
ltTe shall
require that dx. /\ dx. == -d.x i Adx., This immedj.ately implies J 1. J that dX-f /\ dx. :: O. We now make the obvious generalization ,~ l for the r-dhnensjonal form. That is, we have already defined dx.
/\ dX i /\ •.. /\dx. for i1 < i2 ... < i . r 2 J· r tation of these subscripts, II
dX j1 /\ dx. /\ ..• /\ dX J. J2 r
For any permu-
156 If the permutation is even the sign is +; if it is odd the sign is -. Again this implles that dx. 1\ dx. /\ •.• dx. = 0 if 1.1 3. 2 1r • • !J ..i some 1.e~ = l.j' r- r j. The extension of this definition to the multiplication of two general differential forms
,. IV
6.= is obvj ous. ii) Let f e Fl' , g € F'. f is a function of r+l variables ~ S g a function of s+l variables. iple shall define f f\., g to be a function of 1'+8+1 variables. That is, f 1\ g is in F + ' We l' s hope the rea.der has already noticed the resemblance to cochain multtplication which has already been discussed. In fact f /\ g will be defined so that the cochain multiplication will be a special case of this.
We define:
This multiplication is certainly well defined if the functions take on values in a ring.
In the case of cochains, S is just
the collection of vertices in K, and vIe require f(xO'x 1 , ••• ,xl') = 0 if x O,x1 ,· .• ,xI' are not vertices of a simplex in K. iii} Grassman A1p:ebra ---.,... over a vector space: . --
Defird tion: A vector space V over a field F is an additive Abelian group for which a multiplication by elements of F is
-----
defined which must satisfy the following axioms: a) For any v € V and a e F, we have av e V b) a(v 1 + v 2 )
= aV l
+ aV 2
c) (a l + a 2 )v ::: alv + a 2 v d) a 1 ( a 2 v) ::: (9.1 9'2) V
e) 1 • v ::: V
157 (Actually F need only be a ring witb. an identity, but ~ve shall assume F is a field.) Let V be a vector over F.
Suppose dim V
= n,
and that
e l ,e 2 ,·· .,e n form a set of basis vectors for V. Let us define an additive Abelian group GO whose elements we shall call O-vectol's.
An: O~vector will be nothing but an element of F.
Similarly Gl will contain elements called I-vectors which are just linear sums of the basis vectors, i.e. a typical element n
of Gl
is
La. e.. 1 l l
The addi tl on in Gl is component~1ise of course.
The typical elemelJ.t of G2 , a 2-vector, is of the form a ij 6 i /\ ej' PEd ,3jmilarly for Gr , r < n. We may define
f;:1
Gr :: 0 whenever r > n.
r+;
'1'hus G::
1..._"
Gr
is a graded g!roup.
make G into a graded ring we must ciefine a multiplication.
To Just
as in the case of exterior differential forms, we define e t /\ e j :: -e j 1\ e 1 , concl 11ding as before that e i f\ e i :: 0, and in general, that e~ 1\ eo; /\ •• • f\ e i :: 0 if some i. :: i(;o, ·'·1 ... 2 r J .z.. j 1-1.. '1'11e remainder of the discussion is the same as for differentlal for-rns.
r
Let A = (f) An be a graded group and let be any J\belian group. Cons ider all hOIDomorp:lisms c: ~ A ~ 1-. We shall wr'ite Hom (A,
r)
to denote tlds collection.
diately from the fact thet
r'
is
Hom (A,r) is also an Abeli.an.
[l[J.
It follows imme-
Abelian group, that
In fact Hom (fl"r"') is a graded
group. For, if we let An denote tb.e n-dimensl.ol1al homogeneous elements of A then it is quite clea.r that Hom (A, 1-) = (i) .- Fom (A:n ,
Hom (L,
r)
r) .
We haV8 8.1r6 ady enc ount ered an example of in the discussion of the cohomology groups of the
cone. In .fact, let A = C = = C r ( K , G, Irhen
r ).
G-)
C (K,G) and Hom (P..
Derini tion: A is called a Fl;!§.ded operator tlet" if
r
f!1:£''-:l~th
a
r
,r)
diff~tial
158 i) d: A -...;:.. A is a homomorphism ii) d(A,) C (A,,+ ). I..
11
"-
s
This means that for all non-negative
integers A:, the homogeneous elements of dimension
,l. are mapped
into homogeneous elements of dimension 1+s, wher'e s is a fixed inte3er which is called the shift or degree of d. 2 -----iii) d :::: 0 • ~xampl.£s__ ....2..t.2rad~ G~ps
i) Let A:::: C::::
with Differential 0Eerators:
(±> C r and d ::::d.
Surely
a is a homomor-
phism and dC r e e r-l ' so that s :::: -1. Furthermore.;;2:::: 0 , pr'oving that;) satisfies all three properties of the differenti al ope rator. ii) Let A :::: C-l~:::: (£) Cr and d :::: ;j<-. Everything is the same as i), except that s :::: +1. D~£t~~iti££:
A is a graded X'ing with a differential operator d, 1.) A i.s a graded grou.p with a differential operator d,
if:
ii) whenever x e An' y e Am' then d( x,A y) :::: dx/\y + (-1 ) n x /\ dy . Ex.~'npl~Lg.f_Q.!'~9..~£""_Bl.nf!s...J.ii th.2ifferential
Operators: We shall 1.) Let A te the ring of funct ions F:::: PI"
(V
define a differential operator d whose shift, s :::: 1. these functions
131"6
tically make C";<-
=
operator.)
(If
interpreteJ as cochains, this will automa(0 Cr into a graded ri.ne with a differential
For & homogeneous element of dimension r; i.e, if
f E F r , we define:
Thus, if f
e
F r , then df e Fr+l' Clearly d is a homomorphism. The proof is thA.t d 2 == 0 is exactly the same as the proof that"
d2
=
0, except that the x. are vertices in the latter case. l
This proves that we have a graded group with a dlfferential operator. d (f
l~ie
1\
must now show that for f g)
= df
/\ g + (-1) r f
e: Pr , g e: Fs'
1\
dg •
1.59 We have already defined:
Using this and the definition of df, we have: d(f /\ g)(xO'''''x r +s +l )
=
(f /\ g)(xl, ... ,x r +s +l )
- (f !\ g ) ( X o' x 2 ' ••. xr+ s +I) + •.• + (-1) r+ s +1 ( f!\ g ) ( x 0' xl ' •.. ,xr+ s ) ... xr+ I ) g ( x r ·-\- I'" xr+ s +1) - f ( Xo' x 2 ' ..• xr+ I ) g ( xr+ 1 ' ... ,xr+ s + I ) r+s+l + ••• (-1) f(x(.) ... x I' )g(xI' , .• x I' +S ) •
:::: f ( xl
Showing that d(f /\ g) :::: df /\ g + (_l)rf /\ dg is just the same as for cochains, where it was proved that o~r(~ 1\ r{ ) :::: d-i"l; 1\ 1; + (-l}rs /\ (/1. ii) Let us consider the graded ring, D, of differential
forms, defj.ned over a do.main e <:': R. We define a differential n operator d, whose shift is +1, for homogeneous elements of D: a) For a O-form, which is a function of n variables,
b) For' an r-form, 1 d(
L
1 1 <, •.
.f.
r
~l
. dX i 1
... ~r
A ... /\
< r
.s
n,
dx. ) ~r
/\ dx.
~l
/\ •.• /\ dx~ , .... r
(Of course, for I' > n, D is tri¥ial.) Again it is evident I' 2 that d: Dr ......;:. Dr+. I is a homomorphism. To see that d :::: 0, we . may us e the fact that dX i 1\ dx j :::: -dx j /\ dX i or dX i /\ dX i = O. The proof parallels the one showing that d 2 :::: O. CPhis proves that we have a graded g~ with a d1.fferential operator. To prove that we actually have a graded riEE with a differential operator we must 8hmv that if .6.s e Ds ' 6.r e Dr' then
160
As the proof is long arjj tedi0 11S, and not unlike the previous ones, we give the reader the option of accepting it on faith or working it out himself. It might be helpful to see what this all means for n = That is, let
Suppose g, f l , f2' f'3
E
DO'
Then
dx dg = 'dg dx + f1F.lax d.x 2 + dg .f)X1 1 ~x3 3 2
( 1)
(2 )
J)c R3 ·
3.
C?f 2 ~fl d(f1dx 1 + f 2dx2 + f 3 dx3 ) = (JX - ~X2)dXldX2 l
e>f 1 + (--,&x 2
(3 )
d(f1 dX 2dx 3
i·
f 2dx 3dX l +
Expression (1) leoks llke taking the gradient of a functio:l; (2) corresponds to t).·'l8 .£2Fl of a vector; and (3) is a divergence expression. Using t):',e Poincare relation on expression (1), we get d(dg) = 0 or curl €:I-ad = O. Similarly, from (2) we get div cU1"'l = O. Just as we introduced the coboundary operator
i:<-
by re-
"
quiring tbat a"'~x = ,;dx,we B.hall introduce the transposed differenti al operator d';" wi th the same :dnd of requirement. Let A
= C-t.)
Ar be a graded group wi th a differential
r
or-e rator d and let be an Abelian group" Consider the group A = Hom (A ,r) of homomorphisms of' Ar into He shall de fi.ne r
d*,
r.
the transposed differanti al vperator on the gr'aded group Hom (A, = CD Hom (A r , by defining d~' for the homoge::i6ous elements (Le. for elerllents ~f Ar = Hom (A r , Letf,: Ar ~
r)
be a homomorphism.
r)
r).
Define d ...· be requiring that
r
161 for all £, e Hom (A r" Theorem:
r) .
i) If the shift of d is s, then the shift of d * is -so ii) d*2 = O.
Proof: i) d{A r ) C Ar+s - ~ ~d(Ar) C ';A r + s ~ S € A r + s • Reversing the steps, C:A + ;:) ';(d(A ) = (d-l~s)A so that if r+ s oJ*, r I" s I" * I" ~ e A , d ~ e A and the sruft of d is -so ii) d *d ~~~ d *sd ~dd O. Having generalized the boundary and coboundary operator we may, without difficulty geneT'alize the notions of cycles, cocyclss, boundaries, coboundaries, homology groups, and cohomology r;roups.
=
=
=
0
Let A = An be a graded group. Let us call the group d(A n ) Bn+s C An+s the group of homogeneous exact elaments of dimension n+s. It is clear that B :.:£) B + is a graded sub-
=
=
group of A.
n
n s
We shall call B the subgroup of exact elements of
=
A. The subgroup Z == 0) Zn of A, for which d(Zn) 0 for all n, 2 is called the set of closed 0, Bn C Zn ,--_. elements. Since d for all nand B C Z. Since A is Abelian, all subgroups are normal so that we may form the derived group H(A) = Z/B. Similarly we may form the derived groups of dimension n, Hn (A) :: Zr/Bn' lrJ,e state without proof (see any text on group theory) the following
=
(-0
=
0
Theorem: Hn(A) I:±) Zn/ Bn = z/B. rived group H(A) is also a graded group.
'l'herefore the de-
Just a.s for chains, we can Introduce a multiplication of elements in A. A closed element in A may be written as a sum of closed hom.ogeneous elements. 'l'hen, exactly as before we may show that i) the product or' two closed homogeneous elements is closed; ii) the product of an exact element and a closed element is exact. For example, if x,y are closed (homogeneou.s), then
>
dx = dy = O. d{x A y} :: dx 1\ y + x /\ dy = 0 (: X 1\ y is closed. Similarly, if dx :: 0 and y :: du, then d(x j\ u) :;: dx /\ U .± x A du :: x I\. du <,..:-> .x /\ u is exact. The same argument shows u /\ x is exact too.
162 Thus the exact elements form a two sided ideal B = in the ring of closed elements Z = cD z , so that ~4e may fCl'~il n· the quotient ring H(A) = Z/B, thus defining a multiplication of the equivalence classes of closed elements. As usual, everythtng that exists Ilco-exists". That is, using the transposed differential operator, we may define co-'<. ... exact and co-closed elements, and form the derived group Hn(A ) whose elements are equivalence classes of co-closed elements. (Two co-closed elements are equivalent if their difference is a co-exact element.) The construction parallels the one for the cohomology groups. Naturally, we can also define a ring multiplication, show that A* is a graded ring with a differential ~~ * operator d':J and extend the multiplication to H (A "~). We leave the details to the more ambitious reader. In the previous chapter we extabli shed an isomorphism <:> between the ordered and oriented homology groups, HI' and Hr' We stated a similar resv.lt for HI' and Hr. Natur·';tlly, we should try to extend the 0'roup homomorphism to a ring hmr,omorphism. This will come out of the more general discussion that follows. Let A, B be graded groups with differential operators. We shall say that f: A -~ P is an allowable homomorphism if: i) f(A n } C Bn (f preserves dimension) ii) df' = fd. lITe see immediately th'lt an allowable homomoT'phism f, has the following properties; iii) ~_is closed_--'> f~ i~~losed: x is closed ~.... ) dx= 0; f is allowable -~ dfx = fdx = 0 ~ I' fx is closed. iv) x is exact =) fx i~_exact: x is exact (--)x = du => fx = fdu = dfu fx is exact. Thus f induces a dimension preserving homomorphism ~~
t,
<-)
f.,: -n- H(A) ....;.. H(B).
To extend f -l~ to a ring homomorphism, we as sume that A, B, are graded rings with differential operators. We shall say that f is an allowable. homom<;?F2h1sm wi$ respect to the ring structure, if~ in addition to properties i) and ii) above, f has the following property:
163 v) for all x,;;r in A, f(x A y) = fx /\ fy. Since the produe t of clos ed elements 1s clos ed, etc., :i. t is clear that f induces a ring homomorphism f ,,: .H{ A) ......;:.. H( B ) "i~"
•
Thus we have s bown that if A and B are rings, a homomor-
phism of the derived f-ro'.lPS of two graded groups may be extended to a rlng homomorphism. This allows us to extend the group . ,. . or r .. . lsomorprnsm of Hand H to a rlng lsomorphlsm. Before concluding the general discussion, let us look at a special case of a ring homomorphism. Example:
Let A be tbe ring of different 1al foY-rns over a domain
D C Rn' and B the ring of differential forms over a domain
.6 C
Consider a mapping ¢: ~ -....;:. D. (For convenience, as sume that is inf1n1 tely d if ferentiable . ) Let us denote Rm'
points of D by x == (xl, ••. ,xn ), and points of ~ by Y == (Yl"" ,Ym). '1'hen ~(y) == x mei3ns that there al'a n functions ¢i(Yl""'Ym) == Xi' T'lie may assoctate, with this mapping, a mapping of funct:i.ons of x l ,x 2 , ••• ,xn into functions of Yl'Y2'" 'Ym in trH3 follovJing way:
f(x l ,·· .,xn ) == f(~l(Yl' ""Ym),··,,4 n (y1 ,· ··,ym)) == ($f)(Y1,···,ym )
(Note the resemblance to chalns and cochains.
Let f be an alIow-
ble homomorphism of Cr(K,G) into Cr(L,G}. ':Chen f(a O'" .,a r ) == y(bO, .•• ,b ), 'where (aO, •.. ,a ), (b,.." ••. ,b ) denote elementary r r' v r chains over K and L respectively, and y e G. f also induces a homomorphi.sm f":-: Cr (1,G) ~ Cr(!Z,G). That is, for any
s
e Cl"(L,G), y[r;(bO, ... ,b r )) = ';f(s.O,· .. ,a r ) == (f';~s)(aop··,ar)") It is not difficult to see that ~ is a ring homomorpbism~
Indeed
164 !(f+g)(Yl""'Ym) =
f(~l(Yl""'Ym)""'~n(Yl,··"'Ym))
+
g(dl(Yl""'Ym)""t~n(Yl""'Ym))
= f(xl,···,x n ) + g(xl,···,xn ) :::: ( }f) ( yl' • • • , Ym) + Cis )( Y1 ' • .. , Ym) Also,
f(~l(Yl""'Ym)""'~n(Yl""'Ym))'
!U'g)(Yl""'Ym) ::::
• g ( ~l (y l' • • • ,Ym) , ••• , ~ n ( Yl' •
=
0
•
,ym) )
f ( xl' •• • ,xn ) g ( xl' •.• , xn )
:::: (3E f )( y 1 ' • • • J Ym) (lQg) ( Yl ' • • • ,ym ) Th:ts takes c are of 0 ... forms • form of degree r.
L:
wr ::::
fi
i 1 < •.• < :l'r
1'"
Now consider a differential
i (Xl'···,X )dx ..
r
n
~l
/\ ••• j\ dx.
~r
Ll
Corresponding to the mapping ~: ~ D, we have a mapping of r-forms over into r-forms over D, which is given by:
Ll
L
f
(
xl' • •. ,x ) dX i
/'.\ ... /\ dX i r
of
of
:: L
fi 1 . .. i 1" (4l(Yl""Ym)""'~ n (Yl""'Ym.l. )d~.t l
"'1 •• • ... r
n
m
where d~i
j
= r:: k=l
cl4 i
1
1\ ... /\ d~ir
.
,--J dYko oJ
Yk
Now let us return to the general discussion. Here we sh~ll be very brief since the arguments are quite similar to those used in the special case of homology and cohomology groups. If f: A ~ B is an allowe.ble homomo:phism, and a.n Abelian group, f induces a homomorphism f"'c; Hom (B, -> Hom (A, by requiring {r'~~}x ~(fx) for all x in A , t; in ~, "n Hol1l. (B n , f' is also allowable (page 49) since fX'd'X' = d f'X'.
r)
r ).
'*
=
r r)
'*"
165 Here A and B are graded groups with differential operators; d * is defined so that (d-l~)x :::: ~(dx) for all s e Hom (An' and x e An - s' or all s e Hom (B n ,r-) and x e Bn-s where s is the shift of the differential operator' on both A and. B •. That both differential operators must have the same shift for this discussion to make sense, follows frloID the fact that f preserves dimension and that fdx = dfx. Fer both A and B we form the factor groups of co-closed elements modulo co-exact elements, which we denote by H~~(A, and H*(B, ( ) respectively. As before, the allowable homomorphism f "* induces a homomorphism
r)
r}
*.. : H ~1- (3, 1' .,,,
r
) ~ H .;~ (A,
r
)
•
If A-~ B -E~ C are both allowable homomorphisms, then gf is * ~. = (g *f *)*, an allowable homomorphism, (gf).;~ = g~_f.;p and g*f* just as before. Now suppose that f,g are both allowable homomorphisms of A into B. When do these homomorphisms produce the same homomorph:tsms of the derlved groups H(A) an.d H(B)? The answer is the same as before: If there is a homomorphism G: An ~ Bn _s (s shift of d) such that f-g = dQ + Gd. The proof' is also the same. Similarly, if f, g are algebraically homotopic, the homomorphisms f:~, g:~: H~I-(B) -> H1*'{A) are also the same. This is done by defining a homomor::hism G": Hom (Bn; ~ Hom (A n + s ' such that (Q""s)x = s( Ox) for all x in An' ~ in Hom (B , (page 141). If f: A ~ Band g: B -;:. A are both n-s allowable homomorphisms, then
=
~
~
r)
~
r) r ),
i)
gf ::: 1
V
so that gf ~ 1 and fg ~ 1 7 H(A) ~ H(B). Indeed, gf '.::!: 1 ,~) gfx - x :::: dQx + Odx. If x is closed, then gfx - x ; dQx so that gfx ~ x proving that S*f* ; 1. Similarly for ii). The same arguments hold for ...~ . 0),",
i)
g f
,.J
1
?
~...
~~
g.;~f~~= 1,
ii)
f -~g *
=:. 1
=)
t *~** ;
1 •
166 Chapter Homology
Th~ory
~X
ofa
Polyh~dron
In tr:.1s· chapter we will prove the main theorem of co:nbina-tori.,'al topology which states that the homology (cohomology) of ?
.co~plex
,
depends only on the s.pace of the complex.
To
:pr?~ePt1is tl!~
we will associate with each continuous Yilappinp' f of space of a complex K i:nto the space of aD-other co;nplex L a
hpmqmbrr:hism f~:_ ofR phism f"'" of Hr(L,G,
C(,
r)
into Hr(L,U) (and Jually a homonorinto HrU:,G, r)).
This honlomorphism is obtained by appro:ximatins, in a cer': tain sense, a comtinuous ma'.:;ping by a simplicial rna-ping. ;'··e will prove next that this "induced hO;1'lOmOrphism ll f -;~ is unchanged if the cont inuous mappi:n~ f is d.eformed cOl1tinuously. In technical 19.np:uape, the induced homomorphism depend s only homotopy class of the continuous ma"·pL1g.
0
i
the
Fe will also shoW'
th 9.t th e induced homomorphism of a compos i te 0 f two ln8.PDill€'S is the composite of the corresponding ind.uced homomorphisms. Finally, the identity manping induces the identity hommi'1orT,' hi sm. Using these facts, v-Ie can conclude that if we are ;"iven two complexes and continuous mappings between their spaces f'
IKI
-=+ -E-g
lLI
such that the mappin0:s f~~ and gf can be deformed into the identity mapping, the corresponding induced homomorphisms are inverse of each other.
In particular, if' f and g are inverse
homeomorphisms, the induced homomorphisms are isomorphisms. Once this result is obtained, it will be easy to associate homology .s..,,"1.d coho :1.o1ogy proups with any polyhedron. In what follows i, Land M denote complexes. Defini t~~l. Let g be a simpl ie 1. al mappin.? and f a conti'nuous mapping of' 1;< I into /L I. g is said to be a s implieial "l.pproximation to f', re lat i ve to K and 1, if' for every x every simplex 8 in L
I::
II I
and
167
f(x) e S
~
g(x) e S
Definttton 2. g is said to be a simplicial approximation to f if given any vertex a £ K f(St(a)C St(g(a) Definition 1 ~ ~ Definition 2
Theorem. Proof:
1) ASSll..111e definition 2.
x e
being realized in RN • r
The refore x £
(\
i=O
£
f(f!
°
!KI
)
r
x =
Lo
L
Aia i , Ai > 0,
Ai = J.
St ( a. ), and we have
r f(x)
He imagine both complexes as
St(a i
J.
»)
c.
nr
f(St(a i » c.
i=O
r,r
St(g(a i »
.
i=O
Hence 1-1.~ > 0, '1' > 0 . 1.
where b i co.:nprise tbe other vertices of the simplex containing f(x). But g is a simplicial mapping and so r
g(x)
= L::
A.g(a.)
1=1
~
1.
Thus we have that g( x) lies in a si.mplex containing f( x) • 2) We show that definition 2 is Ivrong -:::.--?- definition 1 is wrong.
Assume defini ti on 2 does not hold.
Then there is a
vertex a e ~ such that f(St(a») q~' St(g{a)) • Thus there is an x e '3t(a) such that x e St(a)
and
f(x}
I
St(g(a»)
.
l'iow
x
£
'St(a) --; x ::: Aa +
Since g is a simplicial mapping
L
Aibi
A,Ai > 0
168 g(x)
= Ag(a)
2:
+
A.g(b.) 1. 1.
Now if definition 1 were to hold ( i .e
f ( x ) ~ S e L --~
•
g(x )
e S e L )
we would have that g(a) is a vertex of S and hence f(x) = ~g(a) + .•. which is a contradictlon. As an example of a simplicial approximation consider; til.e Sperner mapp ing (1. e. a s h:1plicial map of kmL I into IL I which takes every vertex x of ({mL into the vertex of the Hsmallest" simplex in L containing x).
This is a simpl:tcial approximation to the id.entity mapping for we have (see p.
57)
for every vertex x
Let gi be H simplicial approximatlon. to fi' i
= 1,2
where
Then g2 g 1 is a simplicial approxime.tion to f 2 f l : since f 2 f l (8t(a)
=
f 2 [f l (St(a»]
c
This is clear
f 2 St[gl(a)] c Stg 2 (gl(a»
= S t ( ( g 2 gl ) ( a) ) Defin~.!..~o:£.
A continuous mapping f:
IKI
~
IL/
is said to satisfy the star £.onjJ..ti~ if the image of every star of K is contained in some star of L.
169 If f 1.3 .a continuous mapping, f: IK I ~ IL I which satisfies the star condr·tiont"hen·-tb:e-re-ex:b~·t..s..a simplicial approxima tion, g, to f. (The converse is trivial in-view--or---defin:i..tion 2) Th~~~.
0
Proof:
For any vertex a e K there is a vertex b in L such that f(St(a))CSt(b) •
Choose one such b for each a and define the vertex mapping
g(s.) If (a O " _
l
tg(s.o)" mapping.
0., a
'l is rJ} •. ,g(a )
r
=b
0
a simplex in K we will show that is a simplex in L and hence g is a simplicial
Let x e {a
o, ... ,ar 1.-
.
Then x
I' ()
E
i=O
St(a i ) and we have
which implies (po 35) that the g(a i ) span a simplex. is a simpli.cia.l approximation to f. L~~.
m :3 f:
IKI ~ ILl is continuous 1c:rrl:K1 -.> IL/ satisfies the star
If f:
Clearly g
then there exists an condition .
.E!:o_~!:
Consider K and L as realized in Euclidean space. Since a continuous mappi.ng of a cOlnpe.ct space is uniformly continuous there exists a modulus of continuttyw{o). That is If(X l ) - f(x 2 >/ ~ w(tx1 - x21),
limlA1(o) = 0 • o~
The distance between two points in St(aJ, where a is a vertex in K, is at most 2 (mesh K). Hence the distance between two points in f{St(a)) is at most L{)(2 mesh K). Therefore in ordel: to make f satisfy the star condition we must subdivide K enough times so that Lv( 2 mesh dmK) < 0 1-Jhere 0 = the Lebesgue number of the stars in L.
170 Let f: lKI ~ L be continuous. Subdivide K so that there will exist a si::nplicial approximation g: kIDKj ~ ILl.
Definjtion. ----.-.-
Let g also denote the l:'8sulting allOtvable homomorphism g: Cl'(o,mK ) -> Cr ( 1) and g~} the induced homomorphism of the homology groups. Define
f.;} is called the induced homomorphism of Hr(K) into Hr(L). 1/Ile will show that f and that if
-l~
depends only on f and not on m and g
where f I , f2 are continuous then
We note that a similar construction could be performed for c'obornoloi?Y groups and in this case (f 2f 1) ~~ == f~"f~' f ~I- depEn~js only on f:
We want to show that if
--.-
,.,.,m and.j
ml
are
m1
two subdivisions of K then (g<jm) ~z. = (gld ) -l:.' ~1Je first consider the case m = illl . Hence we must show that g.,z. = gl.,' But g and "A"
gl are combinator-ially close and thus (p. l3}.~..) they are algebraically homotopic. Hence g.;} == 8:1 • m .p m d Now let ml = m+p, p > O. ipie note that d P .,z.+ = o.;}d~}an thus it is enough to prove that g* = gl.::~' By the first case
*
"#'"'
it does not matter which simplicial mapping is chosen for g1 • Cons ider the ma-opings ".
/oPkmj;\.)
I
ider:tity~
-
..
'f.
»
m
k KI kiLl
is a Sper-ner mapping, i.e. a simplicial approximation to the identity. We cnoose gl == g~. Then
where I
g1
a'~.
...~~
== (g'n
.:"o'~
= g*'" * d ~ = g~.C{; a'P) *
171
By Sperner's lemma and thus
We now show ::: f2 "f1 ii"",
IMI
f2
~
--
ILl
-,,~
Subdivide L as many times, m, as necessary in order to construct a simplicial approximation, g2' to f2 (relative to ~mL, ~l), '\iIje have that
Subdi vide K so fine (n times) that thex'e exists a simplicial approximation hl to fl (relative to cfmK, o-IDL). Let 7.,: IdmL I -;:.. IL I be a Sper:r.Ler· mapping. Define gl ::: 7,h1' ':Chen flol~ ::: (gt"n) {Eo =:-i!-hl d~ and we have c{"
f
f 2" 1, 1"" "i\"
-3f-
=
But and we are through.
--
Definition. Let X and Y be topolo~ical spaces and let f and g be two continuous mappings of X into Y. Let I denote the unit interval. Then f and g are said to be homotopic if there exists a comtinuous mapping F( x, t) of X y.. I i.nto Y such that F(x,O)
= f(x)'
F(x,l) ::: g(x) F is 0;:,':"J..ed the homotopy connecting f and g. In looser language we say that f ~ g (where ~ denotes nis homotopic toU) if f is continuousJ.y deformable into g.
172 It is easily seen that this is a~ equivalence re12tion. To show that it is reflexive choose F(x,t) = f(x). To show that it is SJrm.metr-ic-·w.e note that F(x,t) = F(x,l-t) will suffi~_e. We now wish to show that ·the relation is ·trans·ltive.· f ~ g, g ~ h there exl.sts F(.xt), G(x,t) respectively with the properties stated above. To prove that f~h it is sufficient to define o < t -<12 (F(x,2t) H(x,t) :;:: i.G(X,2t-l) 1 < t < I
--y
2 -
We note that if f
~
fl ,
g~
-
81
..:<- ...f'
-X
-E-..f'
~l
We may choose G(F(x,t),t} as the homotopy connecting fg and flg l , i.e. fg ~ flg l " t4ai.n Theorem Let to and fl be two continuous, homotopic m~ps of IK I into /L /. 'l'hen the tnduced homomorphisms of the homology groups are identical (and of course, similarly, for cohomology).
Theore~.
Proof: Since fO ~ fl there exists a continuous F( x, t): IK! 1... I -""" IL I. Let 6 be the Lobes p:'.le nu.rnber of the stars in L. Since F is c.ontlnuo·us on Ire I y... I there is an yt > 0 such that
Ix
.. XII
<
ii -7 IF( x, t)
- F( Xl' t)
Divide 1 into N equal parts, where N is Let I l / N denote [O,~], Now choose m so Let ftlN :: F(x, tiN). Then fO ~ f l clear c ince we may choose F( x, t f ) , t f = connec~t:i.ng fV/N and f v +l / N'
I
< 6 •
i
fro
chosen so that < that mesha·~. < liN.
/ N ;;~ ••• ~ fl. Thia is (t+v) IN as the homotorY
Ferm the cylinder M, 11111 f :;:: IK I 'j... I 1 / N • 'Jlhen mesh M < liN + liN:;:: 2N and thus the dlameter of the stars in M < 4/1~ < )l' Since F( x, tiN) maps 1M I ~ ILl we see that F satis_ 1'ies the star condition and hence there is a Simplicial
173
= = W
9.pproximat.:i.on £[ to F. Clearly for t 0, ! is a simplicial approximation to 1'0' while for t 1, is a simplicial approximation to f l /N • 'tt1Te denote these simplicial maps by ~d and ~l respectively.
~O and ¢l are combinatol'ially homotopic (p. 139) and thus induce the same homomorphism 01' homology groups. But ~o and 1'0 induce tbe s rune homomorph.ism of homology groups, and similarly for ~l and 1'1" Thus we have shown that fa and f l / N induce the sarae homomoY'phism 01' the homology groups. We may continue this process, i.e. sho1.1 that this holds 1'or f l / N and f 2 / N , etc., and we get the desiree result for fa and 1'1" '1'he proof of cohomology follows similar lines. Let l' be a homeolllol-phism l'
.>
~-
/LI
1'-1
Then 1'-11' - 1.
Cons irier the induced homomorphisms
Therefore f.~ .." is an isomorphism , i.9", homeomorphisms induce isomorphisms of the homology groups" Jnstee.d TNe could bave assl.lIued gf.:; 1 and 1'g ~ 1, and the same thiuf;!; would hold. In this case IK./ and ILl are said to be of the sawe homotogy ty·pe. Let X be a polyhedron. A cOinplex K together w:i.th a homeomorphism 4: X ........ li\:! is called a triangulation (K,~) of' X. By defini tion, every polybedl"on has at les.st one and, of course, infinitely many triangulations. Let {K,~) and (K' ,~I) be triangulations of the polyhedron X. Two homolocry classes x e HI' (T{,G) and x' e Hr(Kf ,G) are called Q 1 )*(x). It is clear that this equivalent (x a Xl) if Xl 1.s in fact an e qui valence !'elation. lrle note also that .--~~,,-,,-,-
= (4'4-
174 x
;;0:
Xl
,
;>
y:;:: y'
X .;.
y
.=
X,
+ y'
Thus we can defi.ne 8.n addi.tion between equivalence classes of homology classes.
In this way we obtain a group, called the
homology group of a polyhedron and
denoted by Er (X.,G).
Coho-
mology groups of polyhedra are defined similarly. Using the results obtained above, it is eas:! to a.ssociate wi th every continuous mapping f of t:te polyhedron X ont 0 the polyhedron Y an induced homomoIThism f,,: Hr (X,G) ....;,. Hr (Y,L). '>~
Let (K,~) and (1,*) be
The precise definition is as follows. tri£mgulet i
OllS
of X and Y I'espec ti vely:
Let [xl be an element of Hr(X,G), represented by x e Hr(K,G). Set y - (~f4-1)*(x). Then the equivalence class of y, [y] e: H r (Y,L) is f '>~., ,,( [x]). T11e homornorp11ism f
-:,~
:
similarly. As before we },ave that 1 ( fg )
~~
~~..;~
::: g f
and f ~c oil.
= f !'" 1\
f
-3f-
'l
?~
= f
= 1 t~~
~~
= 1, (fg) ~<- = f
if f
'-V
g.
-l..g-)i-'
175 Chapter:-E Homotopy Groups I Definit'on. Let X be any topolog~cal space. Then a is said to be a _sur'''y~ in X if a is a continuous mapping, a: I -;:.. X. Definit-lon. The 8UEP(~:;~'t of a curve a is the point set composed of the valhes a(t), denoted by lal.
I al.
Since I is connected and compact so is
Definitton. If' att) and f3(t) are two curves su.ch that a( 1) = P( 0) then 'fr. e def5ne the pr-oduct
.(a(2t) !
0 < t
\
a(t)p(t) = < lf3(2t-l}
-
--,
1
2-
1 < t < 1
~
-
-
il,Te note that the prod 1.lct is neither 8ormnuta.tiV8 nor
associative. Definit:ion.
a-l(t) = a(l-t}
tTe shall nov.] defh~e another equivalence relation between
curves which we sL:1Jl af9.in call homotopy. Definitjon.
Two curves a and
P
't-Jlth the same endpoints are
said to be homotopJ_c (relati VB to theL:' end:ioints) (denoted by :;:;:) if there ex.is ts a comtinuous lTWppL1P' (cf:.l.11ed a bo~notopy mapping) Q( x, t), of ~~)(. I into X such that
3)
O(O,s) = a(O) O(l,s) = cd 1) O(t,O) == a( t)
4)
O(t,l) == [3(t)
1)
2)
Pictorially we could represent thi s li1u.;:ping by the figure, and we see that two curves are hOl'l1otopic if one can be continuously transformed into
for all s for all s
1'76 the other without changing the endpoints.
In the same way as
for ordinary homotopy we can show that this is arJ. equivalence relatjon. Theor~.
If u
"".l
01, 13
P'
'.J
and
0.13 exists then alp' exists ar...d
up ::::: a' pl. Proof:
Since 0.(3 exists a and
P r:::
~'
-)
P
have a common endpoint.
a I and 13' have a COIT'.u"Uon endpoint and hence
alp' exists. Let 9 1 be the homotopy mapping for 0. an:'!. at and 9 2 be the homotopy mapping for (J and (3'. Then a homotopy mapping for
up, alp' may be c.efiw-l by (Ql(2t,s) 0 =
-
,)
\Q2(2t-l,s)
2 - !.)
(0 < t <
(12
< < 1) -- t
Si
:
i
1_._=:::.' {c )..... ~ .. I
I
I
"'\'/11)1
.i
I
_
Theor~.
Proof:
i . ____ ._ .L __ ,__
If (a(3)y exists then a((3y) exists and (aP}y;;.; a«(3y). To show this we define 9{t,s) as
l7? 8+1
1)
a(hl_)
0 < t
::: -~-
2)
f3(}+t-s-l)
s ,-1 -4"- _.
s+2 t < 4-
3)
Y "2-:;-
:;:-+1
<..
(L' t-s -?)
s'}2
4
-
-< t -<
1
To visualizE) this we look at
the dI'e.wing and V.1e c uuti:lue the j.r. . dica ted mf.ppings tnr'u I I linearly an indicated
,... J.r:.es. "
Let a br, El.:..'pport is a., thc,t l~
=a
Lemma.
Let a( 0)
Proof:
1) To she-
We define E a
E Doir-t.
&nd a( 1)
the curve wh0sa
~s
= t.
a .-.., ea. e.
a'~b'~ a Jet
0
G(t,s)
,"
s+:
2"
J.
L
c-;-l
<
2-
- t -<
1
<:
and let
2) to show
=
9(t,s}
rea
(t)
ta(?t:!il) To
SAe
this we look at
o
< t
<
l,·s
-y
. . t -< 1
178
1)
2)
<
'I !
I
k--'><....:.· ._, I\
I
I l ...;;-.-t-J._ - - I . - ._ _ _ _ ~ -, it.;. I t i
\
\ \
_ _ _'1---:-",'--',______ . _._..;,
~.-"'.
~J
t:
Lemma. Proof: Q( t , s)
<.§.
/a.( 2t)
0 < t
,
s 2 ::. t < s
I
= ) 0.- 1 ( 1-2t )
la , \
- 2
-
s < t < 1
-
-
Definitton. Let X be a topological space, x € X and consider the curvos which beg'in. a"1d end at x. Let [a] denote the class of curves homotopic to (1 and define (0.].[,,] = [a.{3]. By our lemmas we see that tbis l1lultiplic'1tlon is well defined and associative, [ex] is the identity element and [0.]-1 = [0.- 1 ) is the inverse of a. Hence we h'1.ve a mul tip;tic"ltj.ve group which is called the !~~a.men~al p;rou~; of _X at t4~o~nt x and is denoted by F(X,x) or TIl(X,x). Irle see by our defini tion that the fundamental group depends on the point at which the curves s ta.rt and end. We wish to know when F(X,x) ~ F(X,y). Definiti.on.
A mapping h of a group G into itself is called an
inner au!-,?morT2h~I!! if there is age G 3 h( x) =. g-l
g
Theorem. If there exists a curve A between x and y then .. -..F(X,x) ~ F{X,y). This isomorphism depends on A, but for any other curve A' the isomorphism will differ from the previous one only by an inner automorphism.
179 Proof;
Let a be a curve which
begins ane. ends at
f(o) a
Le-t T~en f(~) is
= AQA- 1 . t-:hi~h
C1..1.x've
ends at y. f([a])
begins and
We
=
J1:.
ca~
define
[f(a)] •
In orde:::" to show that
\.
~his
'.
definition 1s good we must
""~-'"---
prove the.t ~ :.:..:. a ~''''~ f~ (3) ~ f(a). BQt this 18 c~eer sinc~ f(f3) il/ote
a~~:-)c
t~lD.t
f(a)
= AaA- 1 •
f }):;,'ese:"ves multi:?Jlcation 01' homotopy classes! ., ,
=
f(a)1'(f3)
f(a)£(t3)
AU\. ··,J..At3A - . j .
Aa;3i.- 1
f(af3) -
and he:r.we
= A~A-l,
/.1
f(up)
homom::..-:cr'h.:>T , . cf 11om3topy c las ses . Since (a] = [i, -"::,( r... ) A] the mapping is 1-1. Take ar.l.Y curve y whi eh ber~:.'~~f ~.i:}t ends a:; ~). Y .'::;:.' AI. -l YAA -1 and thus the
lJ.'hus f
jr,
?
mapping is onto ace: f is an iso'TI(JJ·::-hi.:m. Let I-L be a.ny otl:u::.:. ..:.;urvG)f:;l..\-'!e.:F1. JC g([a]) ::: [~a~-1]. Now
AUA -1
~ ~~
- } , -1
')aA . ~~
8:i.!'~
.J,
arJ.d. dafine
-1
and -1] [ AaA'
., Let fL·· .. A ::: v.
=
r
l ~~
-1 AQA - -1 ~tL -1]
•
Then v begins and ends at x
Thus
f{[a]} ::: g([vav- 1 ]) ::: g«(v](a][v]-1) Define h([a])
= [v][a][v- 1 ]. f ::: gb. ,
Then h is an inner automorphisre: fg- 1
=h
•
180 If X is arcwise connected the fundamental groups at all ooints of X are iSOJlOr'ohic and we can talk about the fundarnental group of the space.
~
~
',N'e shall nO~-J describe the HU1"Wfcz homotopy groups which are the n-dimensional analogues of the fundamental group. Definition. Let I 11 be the (closed) n-dimensional cube. if a is a continuous tiJap
Then.
such that '1 takes the :':!1.lrface of t.he cube into on€: point, x, then a~ is c F.tl1ed an n'<~Ul've at x. As sn example of an n-CUI've TVe have the constant mappiLlg €x which takes the whole cube into one point. In the following dis'cussion x is assumed to be fixed and hence we can write € :::: ex' If 0.( t 1 , ... ,t n ) is an n-cur,re, where tl J • • • : tn are the n-jimensional Euclidean coordinates, then a(tl, ••. ,t n ) = x if ti == 0, or ti :::: I fox' any i, (since for ti :::: 0 or I we are on the sur'f ace of the .n-cu~e). Definition.
Let a. and;:> be n-cux'ves at x then we define
(~(2tl,t2,···,tn) =\
,~(2tl-l,t~,
I
c.
••• ,t ) n
o
<
<1 tl -2
1 < < 1 2- tl
-
\
We can visualize this in 2-space.
If a. and f3 map T.1..2 like this,
~L--
@
t
' - -_ _ _ _--:"'.l.--
a:
*--'---
s
1
then a{3 maps I2 like this.
I
1.1--_ _-...._--:-r---.
o.{3 '-----::-'r::'---"""'"4.:~- t
t
181
Definition. Two n-curves a and ~ at x are called homotopic if there exists a continuous O(t 1 , ••• ,t n ,s), 0 ~ s ~ 1 such that s =0 s = 1 if ti
=0
or 1 for any i, for all s
That is two curves are homotopic if one can be continuously deformed ilito the other while the boundary is always mapped into the same point. It can be easily seen that this defines an equivalence relaticn. And exactly as in th~ case of the fundanloncal group we r··Ci.Ve 1)
a ~ ~, at -::: ,,',
2)
(a,,)y ~ a("y) e a ~ a ;::: e.s
3)
cr."
exists ---:.') a"
c::
at,,'
and we can prove that
4)
aa -1
-1
'va';1
,...J
e
We derine thd in.verse of a
8.S
Hence we have a group of homot0PY classes which we call the ndimensional homotopy group with base element x, and which is denoted by nn(X,x). Theore~.
The n-dimensional homotopy group is commutative for
n > 1.
Proof: Instead of letting just the boundary of In be mappE'.d into x by a and " we thi.cken the boundary and let the thicken3d portion map into x. '!J-Je can easily show analytically that these mappings are homotopic to a and ~ respectively. In two dimensions the picture looks like the figure on the next page
182
t
where t'1.e hatched area ~!nd.·l_·.~a+.e~ t1-.'.l~ port'';.... on to b e mappe'd ~n . t 0 x. . NOt-J we pusb these two mappings togetrJ.er s
and Bq1:.ee::;e
t
s
and Ie c t.he bottom of a and the top of
&.nJ. float the
~
thicken
~mhatched
areas past each ot:;.lAj;'.
1:'
~./Ti~ l1/'/ D1
r-1/ __:~~
t
1
if we reverse the pro cess we will c orne out wi tb !3a.. That these mappings can actually be accomplished so that they remain homotop5.c to a. and {3 is left as an exercise to the reader. Since the n-b.omotopy group is cO!1lIllutative the operation 'Illill be written as addition rather than multiplication and a. -1 as .. a.. ~ow
Let A be a l~curve with initial point Xo and terminal point x. Let a. be an n-curve at x. IJlhen we define Aa. as follows: Definitio~.
S II' By the core of In we 1 mean the closed cube of side length one-half with the same center as In and edges parallel to those of I • n The closure of the complement I - -_ _ _ _ _ _---'........~ t of the core we call the rim. 1 Now AU maps the core as a maps In (with the proper change of scale). The mapping of the rim is defined by considering the rim as the ur.ion of straight segments lying on lines thru the center of In. AU maps each of these segments as A maps 11 , with the prcpe~ change of scale, and with the point on the core as the endpoint. Again: as in the case of n-curves if A and ~ are l-curves, U and ~ are n-curves we have
1)
i\
2)
If
"V
AI A~
u::: a I >Au ~ AI a I exists (i\~)a ~ A(~u). ,
In addition we now have a distributlve law: This can be seen by cons i .dering
A( a+j3)
S
1...,.1------..
t
1
AU
iI.[3
These two maps are pushed together b a s 1 ~
:)
t 1
2
1
0 1/2
1
N
AU + A!3.
181+ and since at the points a and b the mapping takes on the same value we see that the deformation is continuous. Let [;\] e; '](l(X,x), [a] eTC (X,x). Define [A.J[aJ n
=
[Au].
It is easy to show that this definition is j.ndependent of the particular representative chosen from each class and thus we see that TIn(X,x) admits TI 1 (X,x) as a group of operators. 1{,1 e shall now mention tt..e extent of the dependence of 'Jt (X,x) on x. If there is a curve A joining y and x then j.t n
can be sho"t-Jn, in the same way as for the fundamental group, th·).t there exists a canonical isomorphism ~ = ~A such that
1)
¢:
2)
~ is onto
3)
~([AJ) == [\a]
4)
TI n (Y"" ".x)
-~
TI n (X' Y)
If ~ is another curve joining y to x then 0 ~-l~"\: 'IT (X,x) -> TI (X,x) is the identity if Ik is homotopic to Ik ~ n n A and is an inner automorphism other~1ise.
185 Chapter XI ~omotopy
Gro:!.?:£s II
(At this point the reader is urged to review the algebra at the beginntng of Chapter V, particularly the statement of the homomorphism theorem on p. 5' • ) In this chapter we discuss the connection between homology and homotopy groups. We shall prove that Hl(K,C O) determines the fundamenta.l group of K and we shall state (without proof, due to lack of time) Hurviczfs theorem on higher dimensional homotopy groups. Defin:ttlon. be written
A commuta.tor of a group G is 8n element which can
a,b e G . In an Abelian group the identity is the only commutator. Deflnition. The ?o~utator subgroup Gf of G is the smallest group containing all th8 commutators of G.
<>
Detini tion. g € -as g e G, g = product of 2s elements . -1 ai' a i i::: 1, •.. , s taken in s.ny order where s is any integer.
A
G
=
.
A
V G •
Lemma.
s It js clear that G is an i . . lvariant subgroup of G.
1i:::
Gf .
-8 :,:)
Pro.2.f: It is clear that Gf. We will prove by induction that everyG (;; GI. (}l ::: {l} C 0 1 • We assume that the inclusion S ' A holds for all S' .:5. s-l and prove it for Gs ' Let x e G s " 111]e may assume that any element in the product of elements which form x is not followed by its inverse: for if this were the case x e Gland our induction assumption would give us the sdesired r'esult. Thus x may be written as x ::: awa
-1
v
where w U v contains 2(s-1} elements.
186
= awa-Iw-Iwv =
x
(awa-Iw-l)(wv)
But (awa-:'w- l ) e G' 1:)"" definitior., wv e Gs _l C Gt, by i.i.1duction. HencE; x. r. G-I. C· ~.b
n:::>I'r.le 1
:and th~ refo:,"e G' i8 norIi.al.
Thus we can
i'O:'r.1.
G/GI. Le~~.
GIG I
is abelian.
1Je ml.lst show that the com..l1ute.tor s'Ubgroup of G/G' Lst 11 -:.: :aJ, B (b]. 'rhen ABP'-lB- 1 [abe.-l-::,-I]. But a ~ a .. ] b - 1 (. GIS 0 t ha t ABA -IB -1 = [I]. ~!:£.?f.;
=
a/Lt'
=
is called "G abelianized".
,>'3t K os ~. C·)Ill~eC tl3d c::>mr-.lex. Then !~{ I i~ a::-c"Iiise can1.-. :·,r! 1::_",,-,'3 fl. IlL.l.s.men 'd t L I grcup F = F( I'K I) • Le t net;:. t ·e(,~ anc... :..dnce :n., HI ::.; H.L (K,GO) dan0te th~ one d::i.zr..en3iona']. hom?logy group f)f K, where G~ denotes ~ae gro~~ of integers. 1rJ~ wis~'l to prove that F/P! '~'5. HI' We note that since F/F' is t cpo l0giclllly inver' ia.;:lt, this identity contain::; a proof for the top :-:'.ogical Ll'var'::!:-i")ce of l-dimensione.l homology groupE!. Def1nit.'.on. Two C11..:.'V63 r.1"e sajd to belong to (or to determina) the same path if ont) 1.3 obtain&d from the other by a strictly monoto~'le parame tor ch9....Yl,3E'l. If t'tol0 curVCl3 belong to ths S~{le pat;::- they aro homotopic and we the!"efore could nave dei'iJ.1ed fU':1.dE'.Ir.!€":-_':;al groups in terms of homotcpy classes of paths. The definit.i.on c.f multiplication of paths 1s obvious and this m1..;::''ciplication, unlike multiplication of c~rves, is associative. For the rest of thts chapt eX' we assume that we have in some
~a~~er
Definition.
ordered the vertic6s of K: a O' aI' ••• , as. Let
Gik
be a curve such that
o.ii:
I ...... a.
Q.1k:
I - ..~ '( a1 a k )
l.
,
if (aiak ) e K ,
in such a way that Q. ik is a linear mapping with Ct1k(O) = ai'
187 Uik(l) = a k • If (aiak ) I. K the u ik is not defined. called an elementary curve • •
ai1r:: is
!
Definiti2..g.
A special curve is a product of elementary curves.
Defini~.
A special path is a path determined by a special
curve. Definition. A simplicial curv~ a is a mapping of I into K where a is a simplicial mapping of dmI for some m.(We consider I as a complex with two vertices 0 and 1 and the one I-simplex (0,1)). Definit1.on. A simplicial path is a path determined by a simplicial curve. Derin:!. tl on. If a special curve is a product of 2 m elementary curves it determines a simplicial path which we call its simplicisl equivalent. Now let the special curve a be the product ~; elementary curve~ 2 m- l < r < 2~ and let al be the initial point of a. Then the simplicial equivalent of a is defined to be the }J atb determined by the simplicial curve m
(a . • )2 -ra II
Lemma. Every special path is homotopic to its simplicial equivalent. Proof:
Obvious •
Let A be a special path: A; Al, •.• ,A s where Ai are elementary paths. 1rJe define a mapping ~ of special paths into chains by ~(AlA2) 4(A l ) + 4(A 2 )
=
~(aii)
=0
~(aik)
= (aiak )
for i ,. k •
v.Te see immediately that a special path and its simplicial
equivalent have the same
4 image.
188 Lemma.
Let f and g be s :i.mplicial mappings f:
g: such that g is a simplicial appr-oximation to f. Proof: Let p points of :smr.
= 1.
Then ~(g) = ~(f)
=
Let 0 to < tl < ••. < tm be the division The division points of ~m+lI aI'S
o = to < 8 1 < tl < ••• < t m• ~\ie note that since g is a simplicial a~proximation to f we have for every v
, and either Hence Len~.
~(f)
g( s
'Y
) = f( t"_l)
or
f
= ~(g).
Let A and ~ be special paths with It ~~.
Then
~(A) " .. ~(~). ~t:
e'ince eve;:-'] special path has a simplicial equivalent we can assume tha.t A B.nd ~ ~re simplicial paths. using the previous lemma we can fur-ther assume that It and ~ are maonings of I subdivided the S2me mmicer of ttmes. Let,b' denote the homotopy connecting A and 11. Since 1<' :LS co:ntinuo'1s we may subdivide I2 a suff'icient number of times so th9.t tb.eI'o will exist a .
s
1
I x ___
~~
189 A
simplicial approxi>nation F to F. Let x, xl' y, Yl be the cbain.s alonp" the bottorr.>.,t::p, r:tght-and--l"e:f"t OfI 2 - j 'les-pect:tv-ely. We-'--note tha.t along the upper (or lower) edge -t.' is a simplicial approxima tion to A (or u) 1.rJe denote these approximations by ~ and 11. Now ..... ~(~) = ~(lJ.) = F{x} 0
~()\)
"B'( y)
A
= q(i\) = F(xl ) = ~(Yl) = 0
A
Consider F(X+Y-Xl-Yl) x+y-xl-Yl is a bour~ary and a simplicial mapping takes boundaries into boundaries. Therefore since 0
A
1'.
F(x+Y-XI-Yl)
= F(x)
A
- F(x1 )
= ¢(lJ.)
- ~(A)
we have that ~(i\) - ~(~) is a boundary and so q(A) ~ ~(~). Lemma. Every curve A beginning and ending at a i is homotopic to a special curve. Proof: i\ is a. map of [0,1]. If we tB.ke the barycentric subdivision of [0,1] often enoup A will satisfy the star condition and thus there will be a 31·1v11cial approximatj.on, f, to A. A(O) ;:: a. where a i is a vertex of A. f(O} lies in the sarne simplex of lowest d.imension as a i and hence f(O) ;:: aio Similarly f( 1) ;:: a i • Hence f is a sped.al CUI"IB end. since every speci a1 curve is homotopic to its s implici al equivalent we ar'e done. We are now ready to define a mapping x of F/p' into H. Let [A] e FI/F' and define ~
.
~([i\])
=
[¢(A)]
The previous two lemmas tell us that this is a satisfactory definition. Now we want to show that x!
1) 2)
3)
is a homomorphism is onto has the commutator subgroup as kernel.
190
W((A][~]) = W([A~]) = [~(A~)J
= (¢(A)
+ ~(~)) = [~(A)J + [~(~)]
= ",([AJ) + 1/1((~]) • Li~.'tq,:
To all a i assisn a special path a i which leads from
a O to ai' To all [a 1 a j ] assign a special path ~ij which 1s precisdly the edg3 a i a To all [a. e . P... J a 3 sign a spe clal path Yo Ok 1 .~ :{ lJ TrJe wish te, ,-1;)-,)1-.' t; ~Ht a-n:r elementary cycle
r
=
(a i
1
~n.
J. ~ .-.
) + .,. + (n.
J., .~
'~i )
1
is the ~ imag6 of some pa.th.
But
we see imm.ediately that
tet
Lemma.
~
= special
path then
Cj~(,)') ~ 0 (: --')
[A]
e
GO
Pr22!: If (aia j e K foT' i < j we define 01j = Uit3ijujl. We claim that [Olj] .form a system of gerJ.srators for the fundamental group. Any spec:i.al yatb., (), is
8.
p..i.'oduc ti of
~.
s.
TheI·efore () is
homotopic to a specls.l curve of the form
and
thus
Hence ~(d') = 0 -=) 4Tf{0i)
and 0i3
sin~e
0ij
= 0jt
=0
this implies that 0ij e TT(Oi} -
e l'T
nUl i
) eGO-
9
191 Since the image of every commutator is zero we have the convel'se. \jJ has the comm~tEttor.~u.bgroup as kernel: [io,!") suppose ~((5) :: .3:x. :: d ~ Y) • • • .
can arrange i~ < i2 < i3" ~b (0) =
L 1i i i
1 2 3
~\ll1.
Then
"123
(ai.a. a. ) where we 1 1.2 1.3
[( a i a. ) - (a. a. ) + (a. a. )] 2 1.3 1.1 1.3 II l2
But
and thus
However 8.nd so since we h8.ve
I( --1 162)=0.
Using our lemma
but
(5' ,,}
1 and thus
state now without proof the theorem of Hurwicz: Let .r:: be A. connected complex. Assume IKI is simply connec ted and. that the homotopy groups 'its ( !xi) are trivial :for s < r. Then the inter;ral homology groups Hs(K,G-O) are trivial :for 0 < s < r and Hr (K,GO) ~ ,1Cr (IKI). ~;Je
Corollary 1. Assume K is connected snd IK I is sj.mply connected. If the integral homology groups Es(K,G O) are trivi8.1 for s < r then the homotopy groups 1Cs (Ili!) are trivial for 0 < s < rand Hr{K,GO) ~ 1C r ( /KI). Corol:J..a!:y 2. If n > 1 tbe homotopy groups 1Cs (8 n ) are trivial for 0 < s < n-l and 1Cn (s 11. ) is a free cyclic group.
192 Chapter XI! Cant :l.nuous Ma:0pj.np:s of the n-Sphere into Itself Consider an (n+l)-simplex S = {ao, ••. ,an + l }. Let K be the complex consistin~, of 6 and all its faces and let S be the ' n complex consisting of all the proper faces of S (i.e. all faces of S except S itself). i~!e note that for n > 0 3 is connected n and is homeomorphic to the n-sphere. Hence from now on we will talk of the n-sphere n > O. Theol~em.
1)
HO(S n ,G)
r"
2)
Hr (3 n ,G)
=0
G 0 < r < n
N
3)
H (3 G) G n :a' Proof; We have al:!.'eady proven 1). !'-1e now consider r f. 0 and we know (p. 130) that Hr(KG) O. Therefore every cycle on K is a boundary_ Let z be a cycle on S , dim z = r < n. ke can n consjder z as defined on K and hence z = au. Thus every cycle of dimension < n on a sphere 5.s a boundary and hence the homology gr0ups are t~ivial. (This is quite clear if we realize that r chains and r+l chains of 3 n are the same as those of K forr
=
= gaO"."
and thus u (v is called cycle on the gv canr:ot be Hence g'v hi 0 Exerc:tse.
(
T
..... (
)
.
a n +1 ) where g e G. .wet 0\ a O' •.• "an + l :;: V. the fundamental cycle of Sn.) Therefore every n sphere is a multiple of a certain fixed cycle v. a boundary since it is 9. cycle of highest dimension. only if g :;: 0 and thus Hn(Sn,G) '?:' G.
Compute the cohomology groups of the sphere.
We shall now compute the homotopy groups, TCm(3n.), of the sphere n < m. ille just discuss the cases m :;: 0, m :;: n. It can be shown that for 0 < m < n TCm(Sn)::: 0. Defi!}ill2!!' A simply connected space is an arcwise connected space in which the fundamental group is trivial.
193 _1'heorem·,. __ ~(s.n'X,!_~. [x] (i.e. consists only of the identity). Proof: Take a closed curve on S. \1ithout 'loss' of" 'generality n we can talee the south pole as initial point. If the curve omits the north pole it can be contracted to the initial point. (This may be done with the aid of stereographic pro,jection or we can draw the nieridians and let the curve move with unifor'tll velocity to the south pole.) If the curve passes through the north pole we can triangulate tr:l8 sphere so that the north pole is an inner point.
'1 hen
we can use a. si ;'''r.licial approximation and map the curve into one which doesn 1 t p~ss thru the north pole. 1"e now in'.;x"oducc the lmporta.nt notion of degree of a mapping. To eat an jnuitj va idea of what the degree of a mapping is considar a continuous mBp~ing of a sphere into itself (or equivalently into an identical sphere). The degree of this mapping
:j
s the number of times the firs'::; sphere is wrapped around
the second.
Now formally:
Definitjon.
Let f: S
- - - - - -~- . -
n
~
S
be continuous.
n homomorphism f~~ of Rn(Sr.,G O ) i":::lto Hn(Sn,G O • therefore f';l- is a homomorphism of GO .-;::.. GO'
Then
r
induces a
N
But Hn = GO and If t'~.(l) = k e GO then g is called the dep~~of th~~E?1Eg f (denoted by 6( f»). We note that 0(1) = 1. Consj.der t:b.e fundamental cycle v of S n (p. 192). Since eve ry n-cycle on Sn is
8.
m1).1 tiple of v we
h~.ve
that
where k e GO Therefore k ~ degree of f. Assume 3 n is triangule.ted (not necessarily the same triangulation on both spheres) a.nd let v be tbe fundamental cycle on 3 n and w be the flmdamental cycle on f( Sn). Suppose for the moment tha.t f is a simpliCial mapping. Let Si be nsimplices on. Sn' ~i n-simplj.ces on f(Sn) oriented so that they appear with a positive sign in
Y
B.nd w.
194 w
--
= > "s.
Then
].
f(v)
wi th e. l
:=;
+ 1 or O.
That is 1'(v) = , - e.S. L:C ]. Ji
But we know that f(v) is an n-cycle in f(Sn) and every such cycle is a multiple of w. So
= kw
f ( v)
=
L
A
kS . J
Therefore t::e degree of f is equal to the number of simplices which are ':i1o.p:r;.ed onto any simplex preservi.ng the oI'ientation minus the number of si''Uplices wbich are mapped onto the simplex reversing the orier:.tation. Now let f be any continuous map. First triangulate the sphere into which 1:~e map and then subdivide the first sphere until there exists a si'rlp1icial approxim9.tlon g to f. Since the degree is defined in terms of induced homomorphisms of homolog:? E'T'OllPS we ha7e that 6( g) = 6( f) . If f is a constant maFPing then 6(f) = O. If l' ofO.1 ts a point then 6( f) = 0 s hlee then the mapping can be continuously deformed j_nto a constant mapping. :c:.~~i':'19.:i t_~on.
A mapping f is oa1181
to a mapping which omits a POi:rlt. If l' is inessential then 6(1')
jnessel.~.tial
= O.
= 6(g)O(f).
Leinma.
6(g1')
Proof:
He are concerned with maps of S
----
if it is homotopic
n
into itself. .6-
clearness. we shall denote the image space by 3 n . 1':
S
A.
n
.. 'S n
g'
-0;:..
S
--;;:.
sn
n
For
195 Lemma.
'lihe degree of a homeoli.iorphism f is +1.
Proof:
ff- l == I
) O(ff- l )
::=
C(f)O(f- l ) :;: 1 thus
6(f):;: 0(f- 1 );:: +1.
If o(f) = +1 f is called sense preserving. f is said to be sense reversing. Theorem.
o(f)
If
o(f)
== -1
For every inter:er k there exists a mapping f with
== k.
Proof:
11,:e
81:.all constl"Uct a reap having degree two.
Cons ide I' two splvn·es S a::1d 3 v-.Jith centers on the axis n1 n2 where
of S
n
and '·.'n Q are tangent at the point which is the 2 north pole of 8 n, and the south pole of 2. 1)
3n
1
n,..L
.J..
S a n d 3 n are tangent at their north poles and S nl n2 and Sn are tangent at their south poles. DraiN the horizontal tangent plane 1. Let g: 3 n -::. S U3 2)
n1
be the mapping which sends each point p of S on 8 n U 3
1
to
n2 /'-
~ 3n , i ,f i' .:::> ni pole). Let f: S
f:;: f 2 g below,.~. ~\'e
into the point q
closest to p and such that the line pq is parallel
j. and meets the diameter of S • ,..,
n
n2
= 1,2,
n
perpendicular to ,.i.
be such that fl(south pole) ==
n .....>~gn be such that f == fIg, above 1 and
We shall show that 5(f):;: 6(f l ) + 5(f2 ). may as sume the f i to be simplicial maps. Triangulate
3 n and consider the projected triangula. tion of S n. 1
If 3 is a
196 I::
simplex in 3 n
irJ6
ask which simplices will be mapped onto it.
We 'seeclee.rly t.hat these will be precisely those simplices
which fl maps onto" Splus···thos-e"Simpl-i-c-e.s whi.oh-1' 2 ma.psont.o-,S.and, r~oting that fl and f2 cannot act on the same simplex, we are through. We have the identi.ty mapping, I5(I) :;: 1, and since the condition on the north and south pole of fl and f2 respectively is not restrictive (it; can al't-Jays be taken care 01' by a rotation) He have shown that there are mappings of degree r for t:my integer r > O. NN-J all
~le
is to constru0t a mapping of degree -l. 'trIe shall construct a map with d.egree :;: -1. Let 3
and g be define1 as above.
/'-
ni
n~9d
Let fl map S
3 n be trmslation and dele,tion arJd let f2 :map 3
trru1;'31ation, delation G.nd reflection. before.
into n
by
Let f be defined as
Then we have
Since fl 6(f l )
n2
n); into S
= 1.
j,g
~omotopic
to the iCentity we have that
We deform f into a
constant mapping as follows. We shrink Sand 3 in the n1 n2 ratio 1 - t : 1 and by translation arrange that their poi.nts of tangency with 3 n remain the s arne. Then we map the region between the horizontal tangent planes
11
and f2
onto the south pole of Sn' other points are mapped as before. F'or t
=0
we have the original map and for t :;: 1
TtJ8
have the
const ant map. Hence 5(f) :;: 0 and thus 6(f 2 } :;: -1. As an application we shall prove that there exists no retraction of the closed ball
197 onto its boundary (which is the equivalent of the Br-01.lvJer fixed point theorem; p .. 87) .
-
Suppose there were ~cDntinuous f(x), Ixl < 1 such that /f(x) I = 1 and f(x) ,= x for Ixj : : i:-ConsiderF(x.t) =: f(tx), -"- ___ 1
o .::
t ::: 1. For e8 ch t, f (tx) is a mapping of the bd'Zldary of a ball into the ball. The degr-ee of F does not depend on t since F is continuous with respect to t. Let t = 1. Then F(t,x) Let
= identity
o( F) :::: 1
:::: 0
of : :
F(C,x) :::: constant
C
and we have a contradiction. ~fill.J-t~
Let S be any set. An (n+l)-dimensional vecto~ field ~ is a continuous mapping f of S ~ Rn + l • If f take s a point of S into the origin then this point is called a singularity. If there are no singular points we say that the vector field is ~-,i_np;ularlty free. If at every point on. the sphere Sn an (n-t-l)-dimensional vector is defined we say that this is a vector field on the sphere. If we assume. that tbis fleld is singularity free we can divide by the absolute value of f(x) at every point. Let g(x) :::: f(x)//f(x)
I.
TheD. g(x) maps the sphere into itself and
thus has a. d.egreewh.ich we call tile ~~_.9! the f5e ld. If the vector fie ld is char.i.ged continuo"J.sly into another vector field so that no singularities arise, the degree remains unchsnged.
This will be true if the original and final vector
at each point make an a.'1g'le a, 0 :: a <
'IT
;
i. e., if the two
vector fields are at no point opposite each other then they have the same degree. lrIe note that the vector i'jeld of exterior normals has degree 1 since f(x) ~ x. For interior normals we have y = f(x) :;:: -x or in coordinates YO:::: -xO'."'Y n = -x a?:ld thus n n+l f(x) is the product of (n+l)-reflections, so that o(y) = (-1) •
198 Th.~~.
(Brouwer) Let n be even and let l' be a ccr,tinuo-.l.s mapping of '3 n into itself. Then either the mapping has a fixed point or it takes scme point of the sphere into its opposite point. P~f:
As sume the theorem is fal se. Then the vector field l' is not oppos ite to the field of' the exteri or (interior) normals.
Thus the vector f:i.eld has degree -1 (and 1) which is j.mpossible. Theore!?!. (Brouwer) On an even dimensional sphere every tangential field (f(x) x) has at least one singularity.
.1
Proof: Theore r :l.
= 1,
Otherwise 6(f) (Hopf)
= -1:
6(1')
contradiction.
Let n > 1 and let f and g be continuous maps
of Sn ->- 8 n vJith 6(1') = 6(g). Then l' 01 g. (Note tha.t this is the converse to the theorem tha.t the degree is a homotopy invariant.)
!£oS;£!
Ta.ke a point of 8 n , say the north pole and take
an n-ball whose radiuB is less th~n the dia~eter of 8 n about this point. fIhe ball
L
cuts out a cap sphere.
We show that there
is a continuous
S
n
into
from the
itsel~
m~pping
of
which is homo-
topic to the identit~ which maps the complement of :;:- into the south pole.
Let
to define this map a.long the meridians of 811.' 1';"
+lj
=
t :; Y/
'Ie
-
It is sufficient
a
t
l·=-t
Then the mapping is given by
l"
f(x,t)
I
::;:<.
Ii \
1£ a + E, x
'It
.
o
a +
~
for a + l; < x <
11:
for
< x'::;
-
-
199 Consider the boundary Cn of an (n+l)-cube. Cle~r'~I 2n __i~ ___ _ homeomorphic to Sn~ _ ._I,-~t t.here be--.gLv.en..-an n-ba.Tl -in Cn , By the'-lyrev i::UU'S'-remark there is a continuous mapping of Cn into itself which 1) is homotopic to the identity
2 ) takes the n- ball into the c omplementof a given point .x and takes the complement of the ball into x,
'l'he boundary of a cube c an be tri ang:.:tlated. L'et f 1 be a A continuous map of C into itself. (C ). 1r.Je can deform f I into a -
n·
n'
mappin[::; f which :1.s hor:::.otopic to f' and which maps everything except the bottom, of the cube {i. e, the points for' 1-.1h1ch the (n+l)-st coordi:i.l~::.-~8 is zero) into a :3~_ngle point x. Assume 6( f I ::: k. TrianguJ.c~.:;e (~ so tbat x is a vertex of 11 a simplex. Then 7.riang·ulate C so t;:'I-.t thF::'6 exists a simpli,.1 cial approximation to f. r.rLis approximaticn is homotopic to f and has degree k and trGrefore we can assur:e that f is a simplicial map. Pic:,.: an
1\
in the bottom of C and choose an open n be,ll in the s imp16x. Look at all those n-sililplices Si in Cn which are mapped by f onto this simplex. These Si must be in n-simpl:~x
Z-
the bottom of Cn , Since these mappings are linear we know t1:1at the unIverse, images of L are ellipsoidal ballo (solid open ellipses) in the Si' " I,~ :Set ¢ map On into On jon such a TtJay that
200 ~(2=) C complement of x 4(complement of
L} :.:
x •
Then ~ is homotopic to the identity. 'I'hus ~f :::; f. cPf takes a point which is not in one of the open ellipses into x. Introduce :r'ight-handed Euclidean coordinates into our cubes
f
i.)ottom of C
>
EB
bottom of C'"
n
n
t8.ltes . the radii of ~ L- parallel to the positive coordinate axes into conjugate semi-axes of the ellipsoidal balls. If f preserved orientation of the ellipsoidal balls then we get a right hrulded coordinate system in the ellipsoidal balls. If f reversed orientation we obtain a left handed coordinate system in the ellipsoidal balls. IJet p be the number of ellipsoidal balls with a rieht handed system of axes and n be the number of those with a left handed system. Then p - n = k. We note that !,.};.s completely known__if we know. the ellipsoidal balls and their f-l
..£on.iu8a~~9:ii
s
By a similarity t:rs.nsformation we rr!ay shrink the ellipsoidal balls into small balls and the new mapping of Cn into "C n will be homotopic to f. Then we may rotate these balls and move them and provided we dontt allow them to touch one another the new mapping of en into 6~ will still be homotopic to f. Therefore the mappinQ; is completel! described by p ~£....!!.' except for homotopy. If p n 0 then f constant. If' p 0 then n -k rud if n 0, p -k. Thus the theorem will be proved if we can deform f into a mapping with p, or n, or both equal to zero.
=
= =
=
=
=
=
201 To do this move two oppositely oriented balls so that they are tangent, as shown below. Push these balls together, mapping
the--int-e-rs-ect'i-on-uut-ox ..and the reJ:naining part as before. Let d(C,Ol) l-t. Oont5..nu5.ty will hold on the dIviding line since the ma~pings at a ~1d at are the same. The ma~ping depends continuously on t. At t = 1 everythjng goes in.to x. \ve there,-,._ fore can reduce the numbers p and n by 1. We continue this process until either p or n or both are equal to zero.
=
202 Chapter XIII
Up to now we have restricted the discussion to the socalled "simplicial homology" which is applicable only to polyhedra. In this chapter we shall define the homology (cohomology) groups of arbitrary topological spaces. The inter-esting results, as we shall see, will be for ~rac1 spaces. 1111e v major part of the discussion will be devoted to the eech homology (cohomology) theory. froward the end, we shall define the Alexander cohomolog:)T e-rO'..lps. As can be expected, all three theories gi va the s r:U118 1',",81,."..1 ts for polyhedr·a. v' In order to define the each groups, we shall need to introduce some new definitions. Definition: A set S, whose elements we denote by a~ ~, y is said to be partially om.ere~ if there is a relation ol, such that
(It is not necessary that every pair a, ~, of elements in S are related by ..t.. IJ. hat is, a and f3 may be two distinct elements of S such that neither a ~ ~ nor f3 ~ a holds.) 1
Definitio~:
A set S is called a directed. s~!, if i) :i. t is partially ordered i1) given a,~ E S, there aye S such that a
~
y,
~
,
...... y.
of Directed 3ets. (1) The integers form a directed set. (2) A countable set is a directed set. We write a ~ p, if a <-> n, f3 <--> m with n < m. (n,m are positive integers; <-> is to be read llcor:cespond.s toll.) (3) Suppose X is a compact set. Let S denote the set consis ting of all finite open coverings of X. T,~!e ca.n introduce a partia.l ordering into S by saying that a..( " if f3' is a E~~les
203
refinement of a. The relatton ot. so defined 1s certainly transitive. For clearly, if ~ is a refinement of a, and y is a refinement of 13, then y is a refinement of a, so that
a ~ 13, 13 ...f
y
=>
a """
y
To see that property ii) is satisfied, just let y be a c~umon refinement of a and 13. In fact we may choose y to be the covering consisting of all possible intersections of sets in a with sets in 13. Direct System 0£_~·!E2"ups. Let S be a cJ.re0ted set. Suppose that to each a in S, we have associated an Abelia:i."l group Aa' The collection of Abelian groups indexed by the set S is called a direct sys~ of grouEs, if whenever a ~!3, there is a homomorphism
fAa)
f a !3..
A
a -
... A
13
which satisfies the following consistency condition: a ~ j3 J, y, then
If
Inverse System of GrouEs.,' Let S be a directed. set which indexes a collection {Ba) of Abelian groups. The collection {Ba} is called an inver~ ~ystem of sroups, if whenever a ~ 13, there is a homomorphism
fpa!
B!3
->
Ba
satisfying the consistency condition:
If a. ~ 13 ...{ ,,(, then
Let X be a topological space, and let S be the directed set of finite open coverings of X. lv-i th ea.ch open covering a of X, we may associate the nerve of a, N(a) (see page 62). If a ~ 13, we shall define a mapping
204 1':
N(!3) ~ N(a)
_l' will be called a projeetiop map£,~.np;... ---'1re·-I'eci:{Ir--.tIrsYthat
"N( (3)' 1:s -the-c.onipl;; whose vertices are the open sets U. in t3. ~
(We shall not den.ote sets B.nd vertices by Ui and u i respectively, as was done in Chapter III, since it is an unnecessary pedantry.)
{Ul ""'Ur
1 is
a simplex of
N(t3) if and only if
I'
n Ui
1 f( Ui )
.:J O.
= Vj
Propert ies
Denoting the open sets of only if Ui C V j '
0:.
by Vi' we define
f is certainly not uniquely defined..
!?!.~ Pr~ jec...;;~'~:..:;;;i;..;;o_n~Mapp.l£g:
( 1) l' is s t m12.!:!.ci ,$01:
Supp:se that {Ul , ••• , Ur } is a sim-
n
'l'his means that Ui -I O. If f(U i ) :::: Vj then r 1 i Ui C V ; ' so that f' Vj :J 0 implying that SV. "", Vj ~~ is a vi i=l i 1. Jl r) simplex of N(a). ( 2) If 1', g :_§.( (3 ) ~ N(a) are projection maps, then consil--.. .. __ plex of N(
P).
dered as sim.E~icis). rn~J2.~, they ~ combins.t9rially.......£.!~: We must show that if Ul, .. o'U r are vertices of a simplex in N(f3), then the 2r vertices f(U i ) :::: Vji ' g(Ui ) Vk1 , i :::: 1,2, •.. ,1"
=
belong to a simplex of N(a). Ul"",U r are vertices of a simplex in N( P) <=) Hi ::j. 0, i. e. there 1s a point x e: () U i • But slnce U i C V ~ , U ot c. Vk ' i = 1,2, •.• ,I', the point x is in
n
both
ni
0)1.J. 1 V. and f\ Vk ' so the. t Ji i i
n (n
Vk ) ::j. 0, prov1ng iii are vertices of a simplex in N(a).
«() V j ) 1
that the V. and Vk Ji 1 (3) The product of two pro ,iection maps is, a projection ~E.: Suppose that a.:o{ t3 ...(.y, and that l' and. g are projection maps such that f: ~le
N( y) ~ N( P);
g:
N{ (3) -"'"' N( a)
must show that the composecl mapping gf:
N( y)
.~
N( ct}
205 is a projection map. Under f, a set Ui of y is mapped into a set Vi of (3 which contains it. Under g, Vi goes into a set Wi of a which is such that Vi C Wi. Thus gf takes every setUi of y into a set Wi or a with Hi C Wi' Let r be a fixed int-e-ger~ __ .a.nd ___ l.et-G;--
.... be Abelian groups. For convenience, we sball write H (N( a.)) instead of H UH a),G), r '. r r r and H (N(o)) lnstead of H (N(a),G,r), whenever no confusion can ar·ise. If a -{ (3, t£-,ere is a projection mapping N«(3)
f:
~
r
N{a)
Since f 1.s simpl:tc tal, it induces homomorphisms
Although the projection mapping f, corresponding to a ..l., (3, is not unique, any two projection ma.ps f,g: N(j3) ~ N(a.} are combinatorially close (property (2»), therefore inducing the same homomorphisms of the hO"llology and cohomology groups respectively. That is, the pair u, j3 with a ~ (3 induce ~~ homomorphisms f(3a and f
---
a(3
•
r,
Theorem: For f:i.'xed r' G, i) fHr(N( OJ is a d.irect system of groups ii) {Hr(N{a))} is an inverse system of groups.
H
!!£2£:
Let S be the directed set of tinite open coverings of a topological spa.oe X. To ea.eh a in S, we associate the Abelian group Hr (N( a) ). For a. .:;.i\ (3, there is a homomorphism
which is induced. by a pro jection mapping of show that fHr(N(a.))} is a direct system, we consistency relation. That is, for a ~ (3 ~ if fay = f(3Yr a (3. But we have already shown
,.
N( (3) into N{ a.) • To must verify the y, we must show that that if
206 are pro jection maps I then their product f 2fl = f is also a projection map. Furthermore, since f l ,f2 , are simplicial maps, ~(.
_''-
... ~
.Ho.
f" = (f f )... - f "f .2 1 - 1 2
As a result of the uniqueness of the induced homomorphisms of the cohomology gr'oups, we have, for this case, f = fay, f; = fa~ and f{ = f~Y.
*
ii) Here, for a ~ {3, there is a homomorphism
Again, arguing that pr'o Jection n1aps compose, and that f*= (f 2 f 1 ).;:.= f2*fl~.' we get fya:: f{3a f y(3
1
proving that tHr (N ( a) ) is an inverse system of groups. We would like to be able to talk abo1),t the homology and cohomology groups of a (compact) space X. (The definitiomr " will be for al'bitr'3.ry topological spaces, but the theorems to be proved are for compact spaces), We might guess that some kind of limiting process on the Hr(N(a)) and Hr{N(a)) will give us what we want. F'oI' this p~rpose we shall i!ltroduce the idea of a
li~
L
lJ
=
of a £.ir.!:ct ~".st~ a.r."1d a lliit of an inverse system. Suppose that {Aa1 is a dlr·ect system. Form the set Aa' We shall define an eq'i.~i valence relation for the
a
elements of
2:.
Definiti.on: Let a e Au' For any !3 ~ a, consider the element f a.~( , a) b e A/3' \~ie shall call b a succes~!:.-Ef a.
=
L
Definition: 'I'Wo elements a, at of will be called esuiv~.le~,:l, written a ; at, if they have a common successor. We must show that :.:, so defined, actually is an equivalence relation. Clearly i) :::' is sy-mmetr!..£. il) == is reflexive
--_....---
207
We shall prove that "') j t rans~' t '~ve: ~1~ =.s To show this, we use the consistancy property of the tap. Suppose. that a E Au' b e Ap' c'~'Ay' and that a : b, b ; c. The reader may .findit· -convenient to follow the argument with the aid of the schematic diagra.m below: a
b
c
r
a :; b, b == c =--) there ;.8 a 5 a,{3 such that fao(a) :; d = f P5 (b); and theTe is an e 'C' (3,y such ths.t fPE(b) = e :; fYE(c). Since 0, E are elements of a directed set S, there is an element 'r) in t S such that ">7 '1"" 5, € • This give s us two hommnor·phisms fOri: A(j
-?;o.
A~
fe:'l: Ae ~ AY"[
Let us call f(j~(d) :; u, fe~(e) :; u',
Since ~ is transitive, (3 relations that
~~.
We have
It follows from the consistency :; U
so that u == u' , proving that a and c have a common successor. That is tart(a)
:;
f5>(fao( a)
r'(~(c)
:;
ter/tye{c)
=u = u'
...,.
u
.
208 In this manner
L
is split into equtvalence classes.
We
shall show that th5.s collection of equivalence. classes forms an additive l\belian group.
Let
'...lB ..denote
by [a] the class of
L
elemeL.ts in which are e qui valent to a. We should expect to define addition in the usual way, i.e. [a+bJ = [a] + [bJ, where a,b are arbitrary representatives of their respective equivalence classes. However if a and b are not in the same group Ay , . the expression a+b is meaningless. ~.Te mus t alter the definl tiOD somewhat. Suppose that a e Au' b e A~, u I~. Since S is a directed set, there is aye S wit:i:l u,~ ..I, y, e.nd two homomorphisms
.
f
I
Now fUY(a) ively.
= a', r f3Y {b)
Thus [a] :: [a 1
:: b
),
l
13Y. A • '13
are successors of a and b respect-
[b] ::: [b'}.
'Ive may now define
[a] + [b] :: [a 1 + b'J for a', b
l
in. the same Ay and at in [a], b' in [b].
this procedure as man,v tj:n8s
Repeating
necessary, we ITiay show that for
$,S
any finite number of equivalence classes, thel'e is a group Ay having an element fr·orn each class. From this, and from the fact that the Au are Abelian, we have ([a] + [bJ) + [c]
=
(aJ -+ {[bJ
i
(c]);
[a] + [b) == [bJ + raJ
(associative law) (commutative law)
To find such an element [b] such that [a] + [bJ :: [aJ
for all [a], we must solve the equatton
a' + b f
=
a'
U'y is an Abelian group containing sucin an appropriate Ay ' cessors of both a and b, ~hich are denoted by a f and b' respectively.}
b t :: 0. Solving, we get .
€
Ay so that (0 J is the zero
209 element.
cont~jns the
(0]
0 element of each Au.
may also contain other elements. that for some ~ ~ ,(, and f[3Y: A(3 with d f-
In fact, it is quite possible
~
A y ' we may have f(3y(d}
o
in A[3' so that d 5 0 for 0 in each Au" to prove the existence of an inverse, we solve [a] + [b]
:=
However, it :=
0,
Similarly,
[0]
by going to an appropria te group Ay and solving the equation at + b'
:=
01
This completes the proof that the set of equivalence classes in
I:
is an Abelian gl'OUp.
Thls group, to be deno ted by A, will be
called tbe ~...i.!~.~).i~ of the £iT~.£.t.System of grou-es {Au} which 1s to be written
Defitl;t~J.2':£:
The !_:-d;~~e~1.£.D.3!..--.£9E..?~logL~?UP of a topologic~l
space X is Hr(X) = lim Hr(J(a») -.~
The natural question t '~) ask now, is II-',hat does Hr (X) look like? n To ge an element of HT(X), we cboose any open covering a. of X, Hr(N(u). Now pick form N( a), and theYJ. Hr (L ( a.) ) • Let l:r := u an element (a cohomology class) in HrO'J( u)). This determines
Lt
an equiv8,lence class of elements in Lr.
It is this equivalence
class of cohomology classes that is an element of HreX). To define Hr(X), we shall introduce the inverse limit of -~---
an inverse system of gro·ups.
-.--.
--~~.-~~
Defini.tioE: set S.
A
Let {Ba} be an inverse system indexed by a directed
stri.~
such that if u
<><:
is a collection of elements, one in each B a , (:3, the elements a £ Ba.' b e B[3 are both in the
same string if f[3a(b)
:=
a.
210 . -The 'first question that artsesis~ "Are there any strings?" The answer is yes. Take the string which consists of the 0element of each Ba' Clearly strings may ncross". An element a e Aa may be in several strings. It 1s not difficult to see that the collection of strings, which we shall denote by B, is an additive Abelian group. ~Je def1ne adel it ion as II coordina te -wise 11 • That is, let cr'l be a string with elements aa e Ba for each a in S; 8,nd let ()2 be a string with elements b a e Ba' Then 6 1 + 6 is to be the string with elements (aa+ba) E Ba , For a ~ ~,
2
The first eq\lality sign follows from the fact that fj3a is a homornorphlsm. It:is quite clear that the associative and commuta.tive laws hold for string addition, The inverse of the
-°
st:l:'i.ng ,:..\, wh1.ch we will denote by 1 , conta:T.ns elements -aa e B a' for all a, tn S. Thus the strings form an Abelian group B.
It is t}'lis group B that we call the
the i'!lY.t?rse sys,te,r.~ ~Ba}'
j,n~~~~,2,~mi.!
of
It will be written
----_
----
14e now define t:ne r-dimens_.-,'----,-tona,l homo 10(,,:'7 fIroup of a ..
Hr (X)
Theorem:
=
Iir (N ( a) )
Suppose X is compact and dim X
= Hr (X)' = 0
= n.
Then Hr(X)
for r > n.
Proof: Dim X = n <-..,......:.> every a. has a refinement (3 whose order ts not greater than n. Thus HrOJ( (3») = 0 if r > 11., proving that every element of L r is equivalent to the zero element. That is, if a E Hr(N(a)), then f
f3 is a refinement whose
211
Furthermore Hr (N(f3)) = 0 for such.B._rei'inement, so that for··-e: ...,),..{3, f~a( 0) = 0 and__ ~r-{Jll.a:H-= O. But this is true f'or all a in S. ThuslYXr-: o. After a few preliminaries, we shall prove that if' X::; IKI, v then the eech homology (cohomology) groups are the same as the simplicial homology (cohomology) groups. This will automatically furnish a.n additional proof that the simplicial homology and cohomology groups are invariant under homeomorphism. Indeed, it is quite clear that the eech groups are topologica) ly invariant since their definitions contajn only topological concepts; i.e., these groups are defined in ter-ms of open coverings only. directed set, an.d let S' c: S. ~is said to ~~~ofinal~ith S, if for every a: in S there is an a' in S' such that a' r a. That is, SI extends out as far as S. ~finit~!!.:
"8xample: integers.
Let S be
9.
Let S be the d irec ted set of j.nte;5ers, and S' the even
It follows immediately from the definition, that if S' is cofinal 'Vd.tn a directed set S, then S' is itself a directed set. Lemma 1:
Let S be a directed set, and let S' be cofinal with S. supp~ also that {Aa:1 is a direct system indexed by S, and that {Aa ,} is the direct system indexed by S' which ts such tha.t Aa = Au' whenever a' = a. Then lim Au ::; lim Au'
-->
Proof:
--.;..
Let us write
.
1
see immediately that A I cA. Indeed, each equi va.lence c 18 S.'S of AI is contained in, and. thus determines an equivalence class of A. Now suppose that [a] e A, and that a is a representative element in some Aa, for which a if s ,. Since S' is cofinal with S,
~le
212 there is an phism
0. 1
e 8 such that f aa,
Calling few' (a)
=:
I..
A
0. 1
a -;:..
...". 0.,
A
0.'
aI, we have [a.J
and therefore a homomor-
• =:
[a'], proving that every
element in A determines an element of A', i.e. A C AI. Lemma 2:
Let 8 be a directed set and let 8! be cofinal v.Jith 8.
fBal}
If {Bal ) is an inverse system indexed by S, and the inverse system indexed by Sf which is such that Ba , =: Ba if a =: at, then lim Ba
lim Ba l
=:
-«-
Proof:
~-
Let us write
B = lim Ba <.-'---
We must establish an isomorphism between Band Bt.
j,'here is a
natural horr-omorphism F' :
B.-;:. B'
Take a string 0" in B.
Consider only those coordinates b a e Ba which ar's such that a =: a' in S 1. These coordinates uniquely We define Fa = a
determine a string 6' in B'. is a homomorphism.
:B'urthermore it is onto:
show that every strjng
<)'t
in B' can be
II
l •
Clearly this
In fac t we shall
con tlnued" into B, thus
proving that ()' is the image of at lea3t one element of B.
Let
us call this continued string 6, (whose existence we shall establisb) and denote its elements by b o. for a in S. () must have the property that if a at e 8' then b a = bat e cr'. Let us
=
take some
f3
i
S'.
Since S' is cofinal with S, there is a y
Co
8'
sllch that y )<> (3, and consequently a homomorprdsm B "(
-,>
Be I
=
We shall of course choose f y (3
21 3 does £~t depend on the particular r ~ ~ that we use. This follows immediately fr'om the C01l8is tency properties of the fyf3. Deta.iled verification is left to the reader. Now "t-Je must show that F is }.-L._ Let ($'t,-6 2 be two distinct non·-zero s'trings in·B. Since B-is a group 1 -62 == 03 e B. iATe must show that F()3 i O. If F63 = 0, 0"3 must contain th4 zeros of all sets B a , for a l e S'. Suppose 0 ~ b a represents a set Ba in the string Clearly a i S '; otherwise b a = 0 would follow. But S' is cofina.l with S. Therefore there is aye S' such that y )0> a, a.nd the r·epresentati ve of By in ~3 is by = o. Irhere 1.s a homomoI'phism fya: By -0;:. Ba' and from the definition of a string fya{b y ) ba. But by = O. Therefore b a = 0, proving that 03 is the O-string and that = 2" Th5.s completes the proof that F is a 1-1, onto, homomorphism, and therefore an isomorphism.
°
(53'
=
6i.
°
Le~-l=
Let S be a countable set (S is also a directed set; see page 202.), and let {Aa-} be a direct system indexed by S. Suppose that A is an Abelian group such that for each a in S there is an isomorpbism
For every pair a,~ e S, (a~ (3) we define an isomorphism
ra13
= (f~)-lra:
(faa is the :i.dent:i. ty).
Then
for all f3 in S. Proof: LeIn£1.~.2:
Obvious.
Let {Bo,"\ be an inver'se system indexed by a countable set S. Suppose that B is an Abelian group such that for each a e S, there is an isomorphism
214
(faa is the identity).
Then
11m B a ~ BAS:- B 4--
I-'
for all [3 in S. Proof! ---
Obvious. can prove the main -theorem of this chapter.
N01.J ~le
If X ::
The orern:
H (X) r
~
JItI H (K) r
t.hen
.
I
In other words :If' X is the space of a complex, then the eech and the simplicial homology (cohomology) are the same. (As a matter of fact we shall even construct a natural isomorphism.) For all non-negative integers m, let us consider 0 illK, and vli th it the cov8r'lng of X:: IK I by barycentric star·s in
Pro~f.:
clllK, which we denote by am. Clearly am 1 am+ l , and the set {a i is cofinal 1Adth the set S of all finite open coverings of X. Furthermore r.1( urn) ~ cmK (see pap.e 64). 1,\7e have already defined this natllra.l lsomo!}1hisrn. Form,ing t~n.e prJojection map
3
f: N (a ) ~ N ( a,
m
Lij.,.p
)
it is easlly seen that the mapping
induced by f is a Sperner mapping. ~--
Thus we have a natural
isomorphism
I
Hr (N( am) ) ~- Hr (<j~)
'v
Hr(K)
Hr( N{ am) ) ~ Hr (6 mK)
=r
Hr{K)
215 which is true for all m.
II
From
H (N(a )) I' m'
N
nr (l\f ( am»
!Y
I~
we get
HI' (N (an) ) HI' (N( an) )
for all integers n, m > O. Since follows f rom lemmas 1 and 2 that
(a.l ~. is cofinal with S, it t
1 im Hr ( N( an) ) 4--
III
= 1 im Hr ( N (a )) n
From I, II, III, and. le'Umas 3 and
4,
we have
J\ s in the simplic ial homology theory, we shall see how a
continuous mapping of one topological space into another induces nomomornb. isms of the homology and cohomology groups respectively. Let X,Y be topological spaces, and let f: X tinuous. (3 :::
X.
f-l~
If
f
13
== {VI"'"
\j J is
::: ~-lV l ' ••• ,f-1V).} ::: Take some element of Hr(X).
~
Y be con-
an open covering of Y, then
~Ul"'" ~r..}
i s ope~ covering of It is a string 6, which has one
and only one representative in each Hr(N(f\)).
The map f induces
a simplicial mapping
a.nd consequently a ho:nomorphism p1~f.,:
t-'t-'
~~
/',
Hr (1',;(B» ,-
-;.. Hr (N({3)
If b p e; Hr (N(/3»), let us denote its im~ge under {313f.~ by (3~f "~b (3 ::: bf3' This defines a mapping f -l} = (3E f }; such that for
f
<S:::
{b(3}
e Hr(X),
216 We must show that ~ e Hr(Y), i.e. ~e must prove that ~ is a string:. Suppose that b a , b[3 are representatives of (J in Hr(N(a)) and Hr (N(!3)) respectively, with a ~~. Then there is 9. homomorphism
We f(lay define a homomorphism
in the obvious way; 1. e. let
f"~ab,!3':::; b.a'
The consistency
rela tion fyo. :::; f'~afy~' follows from fya :::; f ~afyW Thus &- is a string. It is also quite easy to see that f.~: Hr(X) -> Hr(Y) is a homomorphism. \rie leave this to the reader. Now choose some element (a13 J e Hr (Y) • It is represented by a~ E Hr(N(~)).
TbB simplicial mapping
f:
N{!3) -> N(~")
induces a homomorphism (3!3f-1": Hr (I'J(I3)) ~ Hr OH!3)) Let us wr~,.te ~~f~l-8:'{3' :::; a!3 e HrOH (3)). Again, this defines a mapping f" which takes the equivalence class [a13] into a set of elements {a~J' where each B!3 E Hr OH!3)). We must show ~hat.A,. '" {B/3} E :ar(X). Suppose a~, a-o: e [aj3}' Then there is a Y 't' a,f,3, and homomorphisms A""-
fay: Hr(N(cL)) ..~Hr(IHY)) ..... "
f/3y: Hr( N(~)) ~ Hr( i.j"(Y}) /" .... ;...
.... ""-
"'
with faYa~:::; f(3Ya~ :::; a y e [a~J. homomorphisms
We may define the obvious
217 ray: Hr(N(a)) ~ Hr(N(y)} r{3Y: Hr (I\f
un) .-;:.. Hr (N ( y) )
with raYa a = r{3Y a {3 = a y e Hr(N(Y)) .. Again the consistency relation holds, so that the set {a(35 is an equival"6nce class of homology classes proving that [a{3J e Hr(X). Thus r*: Hr(y) ~ Hr(X) . It is obvious that r"" is a homomorphism. Before closing this chapter, we shall say a few words about the ~41_?.~ld~ cohomology ;~ro'lps. Let Co be thG add:Ltive group of' integers, group, and X a topological space. r-d1:~nsJonal
functi~~
r
any.Abelian
For r > 0 we define
on X,
r.
with xO,x1 ' .. +,x r e X and y e (For r < 0, we shall say that f == 0.) Cleal~ly the r-dimenai onal func tiona on X which :bake on values in denote by F,r. Definition:
r,
forro an Abelian group, which we shall
A hmilornorphism
defined by
is called a _coboundar'T operator. • ~_-J.:_~
-
Definition: .-
= dim 'J! = r.
Let a be a finite open covering and let dim! Then we shall say that
I
= i with respect to a if .....,
...
218 Defini tion:
I
1,r.Te shall sa~T that
= W locally if there is a
finite open covering a. such that!
=~
with respect to a.
~inition:
A function (ji is called a cocycl~ if d} = 0 locally.
Definition:
Two cocycles tp,
~
1V'li,
iIi
are c8.11ed cOhomologous, written
if there is a function Q such that ~ Clearly, for all Qi, dQ) . . . O. (d 2 = O. ,J.
'If = dQ locally.
S3.me proof as before.)
Having defined addition of two r-dimensional functions, we observe that the sum of two cocycles is
8.
cocycle, tbe sum of
two coboundaries 1s a coboundary. lJ:'hus the r-dimensiona.l cocy .. cles forill an Abel"i.a.n?!'011p containi.ng the Abelian 'J;roup of r .... dimens1. onal coboulJdaries. i~re form the factor group of cocycles modulo coboundaries.
r
This is the r-dimensional Alexander
.£.2.!:l£-
mology .q:r~..E.' If is a ring, we may define a cochain mul tiplication, which allows us to talk about a cohomology ring. Exactly as before 1:Je may show that a continuoas mapping f:
X
~
Y
induces a homomOl'l)h:i.sm cf the cohoInolog:v groups (rings) of Y into the cohomolo::;y groups (rings) of 7... carry this all out agaLn.
There is no need to
Finall;y we state, without proof, the
following Theorem:
v
For a compact space, the AleX9.nder and the eech coho-
mology groups are the S8.me. It follows immediately from this theorem, that if X =
IK I,
the
Alexander cohomology groups of' X are the same as the simplicia.l cohomology groups.
219 9h~J~jie r
XIV
Local Degree an§. ]ndex - Part I (Finite Dimensional Spaces) In Chapter I (page 27) we defined a normsl SP8C~ X as ona which satis~ied the ~ollowing property: i) I~ A, B are closed disjoint sets in X, then there exist two disjoint open sets GA, GB such that GA 4 A and GB ~ B. It is sometimes more convenient to de~ine 8 nor·mal spaes as follows: ii} If A C Be X, with A closed, B open, then there is an open set C such that ACC~GcB
Theore!!!,:
Definition i
<,->
Definition ii.
---
Proof: i i> ii: Suppose A is closed, B is open, and ilC B. Then X-B is closed and i, n (X-B) ;:: O. From i, there are disjoint open sets GA , GX _B such that GA =' A, SX_B ~ X-E. Clearly, GA is th~ set we want. That is I'I """GA '-'-x~ '-B AC. \J1\ 1;.. -"'X-B ,--
ii
=
i:
If A, B are closed, An B.: 0, then AC X-B, where X-B is an open set, FroID ii, there is an open set C such that Ace c;
Cc
X-B
Take A third characterization of a normal space is given by: Urysohn's Le11lma.: Let FO,Fl be disjoint closed sets in a normal space X. Then there is a continuous function f:
X......;;.. I
(I is the unit interval) such that f(F O) ;::
° and f(F1 } = 1.
,;
220 '1\
Proof: Consider the dyadic fractions q/2~ (q is an odd integer :1 such that 0 < q < 2£). To each y;:: q/2n, let Gy be an open set such that FO
C
Gy '-r GY r~ Gy' c_ - X - FI
if y' > y.
Define, for x E X - FI
Clearly f: (X-Fl ) -;;,.. I since 0 < y < 1. 'ifJhat is f(F O )? arbitruri1y large FO C Gy ' vJith 'Y ;:: 1/2n. Thus, for x f(x)
;::
For n E
FO'
g.l.b. y ;:: 0 XEG
Y
For x
€
FI , we define r(x}
-
1
Is f conthl",lOuS? .;>
a)
t!1
.0
x e FO (FO is the lnterior on FO):
FO is open
-.,;.) there is a n>3 fghborhood of x whi ch is contained in F'" O. this neighborhood f := O.
b) c)
In
<7
x e FI :
x
same as a)
We must show that the inverse image of an open"'Set-is~op'6'i1.":"-Let a, b, be such that 0 < a < b < 1.
€
(X-(FOVFI1 ):
Consider the open inter'val (a, b), and a point t e (a,b).
221
Since the dyadic fractions are a dense subset of the real numbers, there are dyadic fr~ctions "(1'Y2'''(3'''(4 such that
a < "(1 < "(2 < t < "(3 < G
-G
"(4 "(3
'3
< b
If x = f- 1 (t) e f-l(a,b), then
is B,n open set of X.
x e (G'V
"(4
-G 2 ) C ( G'V' - G1 ) C
'4
(,b) f-1 a
proving that f- 1 (a,b) is open. d) x c) proves this. C onver'sely,
=
f- l (0
~,
t
f.:
(PI-PI) or (Fa-Fa):
Ury~john r S
<
1/4),
GF
Lerr.l1l8 =f,-1
A combination of a), b),
I i : Jus t take / (34
1
Of course, our choice of a and 1 in the statement of Urysohnts lemma, was not dictated by necessity. We could have taken 8ny two real numt,ers a < (3 and asked for a continuous function f: X -~ [a,(3J, such that f(F a ) a and f(F(3) (3.
=
=
Tietz0 1s :,~xtension Thsnrem: Suppose F is a closed subset of a normal space X. Let
such that f
=
~
I ~ 1 for {xldo(x) .:: 1/3-!
on F, and jf(x)
a
all x e X.
Proof: {xl~o(x),::; -1/3}, are closed disjoint sets in X. Prom Drys ohn f s lemma there is a continuous function f such that (' 0 1', 1 if x e ~xl~o -< --, \-3( ) 3 l f (x) = \ 1 i 1 0 > .- l if x e lxl~ I 3 I 0 - 3 5 \ ". ,I
\
Define ~l = ~ -f o
0
(on F of course). /'
~1 ,i s cont j,nuous and
,
J
1~11 ~ 2/3. The sets ~,xl~l ~ -1/3 • 2/3 ~ , I~l ~ 1/3 • 2/3 are closed disjoi.nt sets in X. From Urysohn ' s lemma there is a continuous function f1 such that
{x
222
(-; . ~ fl(X)=J
1
if x e
2
l 3 · '3
if x. e
Define ~2 :: ~l-fl (on F'). And so on. In general
~i+l
;:
~i
- fi
~
1
~.
1-3
3)
~2 is continuous and 1421 < (2/3)2.
-
!~il! (2/3)i
J
~ fi ~ 3
lX I~ 1 -< - .~3 • .f} 3 rxl ~ >-.-~ 1 2';
L:
2i
= 31
(3)
J
Ifil <
(~)i ~
1 (1 _ 2/3) = 1
n
~ L
f n :i..S a sequence ot continuous functions which n 0 converges uniformly on X, implying that
Thus -.J":t
::
co lim ,'}: - ~ f = t n~co n - LeT' i
is continuous on X. co
co
o
0
On the set F
~ f.~ ;: ""'i"-.._ L_ (~.~
(Naturally, it I qi( x) I .! IvI for x F; F, we may find a continuous extension f, with If(x) I :5 M. Just apply the theorem to the function ~(x)/M.) Corolla. ry ------
1: , Suppose that
Let F be a. c los ed subs et of a nor!nal s pace X. ~:
is continuous and II ~(x) continuous mapping !': such that f
=~
F.....;:.R
II
n
<:£.II for all x e F. X~R
on F and !If(x)
Then there is
Ii
n
II.::: Jii lVl
for all x e X.
Proof: ~: F ~ R means that there exist n (continuous) map-n pings ~i! F ~ R (R is the set of real numbers.) Clearly each 1~1(X)
I
~ M for x e F.
Let fi be a continuation of ~i to X,
223
such that Ifi(x) I ~ M for all x £ X. Doing this for i = l,"»,m, we have a continuous mapptng f: X ~ R such that n
IIf( x) II = (f: 1
fi<x) )1/2
~ ji1
M
Corollary ~: Let F be a compact subset of a normal space X and let ~: F --=- Sn be continuous. Then If p is in Sn - 4{ F), there is a continuation of ~, call it 1', such that p is an interior point of Sn - f(X) • Proof.: F is compact ~ ~(F) is compact ==> ~(F) is closed (-} Sn_~ (F) is open j> there is a nei ghborhood of p that is contained entirely in Sn_~( F'). Pro ject the sphere stereographically onto the plane with p as the north pole. T"Te see immediately that the plane image of ¢(F) 5.s bounded. This reduces the problem to the situation in cor·ollar·y 1. Thus, there is a continuous bounded ex-tens ion of the conth-.. uous mapping of F into the plane. Napping back onto the sphere, we see that the boundedness of the plane image, implies that p t Sn - f(X). Cor?llary-l: Let F be a £l£~. subset of a normal space X, and let ~: F -.> Sn (,€;I continuous. Then if p is an interior point of Sn _ ~(F), there j.3 a continuation f, such that p ls an interior point of Sn - f(X).
_.,._.-_._-
neil2:~'1bol;bood of p. The proof is Proof: sn _ ¢( F) conta1.ns a --, .. the same as in Corollary 2. We have alread-y' defined the degree of a mapping f, if f is a continuous map of Sn into itself. Let G be an open subset of Sn, G its set theorettc boundary, and G C V G its closure. Suppose th"l.t ~: '(3: ~ 'Sn and that p eS n - 4(G).
=
-that p Lemma:
Proof:
There is continuous extension f! S €
""n
S
i)
1.1)
n
An
.....;;.. S , of
.,
- f(G). If p e ~(G), then p € f(G) ;::: ~(G). If p E -Sn - ¢( G), we ma.y use corolla.ry 2.
1
q>
such
22)~
Definition:
For p e
Sn - ~(G)
we define the degree of ~ with
respect~E-G at th~_J?s~int_E.., written deg (~,G,p), as the degr·ee
of f. \'le
have already shov.Jn that such a continuation is possible.
In general there are i1'1fin1 tely many continuations wi th the required properties. In order that the definition be me'lningfu1, If 1'1' 1'2 are two continuations of ~ • An (} --. .. fl(G), peS - f 2 (G), then deg 1'1 = deg 1'2.
we must still show that: ~n
such that peS
.f
.
.._------
i) If Sn = G, we ar's through because there is only one possible IIcontinuation". i i) As sume Sn - G "I O. Ins tead of writing do'ltJn a homotopy" we will use a picture. TrIe know that both fl(Sn- G ) and f 2 (Sn-G) avoid a neif7hborhood of p. ~.'hus both 1'1 (Sn_ G ) and l' 2{ Sn_ G) are ..,o·'n contained in the shaded part of S •
\'I]e shall show that 1'1 and 1'2 are homotopic.
w(p)
w
> ~n
Pigure 1 Let 1'1 be a continuous deformation of 3n whic.h pulls down the cap around p, so that the image of this cap (on ~n) contains at least a hemi.sphere. We shall also suppose that the boundary of the image of the cap consists of arcs of great circles. (The reader should comr:1.nce himself that this can always be done.) For each xeS n -0, we send wfl(X) into wf 2 (x), in a continuous way, along the arc of a great circle lying entirely in the lower shaded thO 1- i . a · pOl't lon 0 l' ~n i'.:) ..ine ar"c LS un~que SJ..Ilce every1ng es 1n 1"Tl-
hemisphere.
• • •
We have, tberefore,
225 which implies that 1'1 A I 1'2. The tra.nsi tlvity of the homotopy relation f'01101'JS from tbe fact that a continuous fu.."1ction of a continuous function is continuous. The definition of deg (~,G,p) tells us how to compute it. We have only to find a sui.table extension f and compute deg f in the usual way, by counting. In outlining this procedure, He shall see that deg (~,G,p) can be compilted. without even uslng the extensi on. The purpose of f, as we shall see, W8.S only to formally define deg (4,G,p). Assuming f to be a suitable extension of 4, we know that there is an Ii' > 0 such that d(fx,p) > Y{ for all x £ Sn_ G• . ......n Triangulate S' so tbat the mesh of the triangulations is not greater tha.n ~., a~1d so that p ls an inner point of an n-simplex () of tbts triangulation. Now triangulate Sn so that it satisfies the star condition. If g is a simplicial approximation to f with respect to tbese triangulations, then d(fx,gx) < '7 for all x E S11. Clearly no simplex intersecting Sn_G is mapped onto
a by g.
Therefore, to find out how many times ~ is covered, we do all our cou.nting iI1;.dde some c·;:,mplex K for which JK leG ( Remember, any n -s i.mpJGY whtch will be mapped onto Ci' will lie entirely in G.) 1;'1e see hmnediately that f was unnecessary. Indeed, s j.nce d( cjJx:,p) > ri for x 8 G, we triangulate '3 n so that its mesh is not greater tban >"i and so that p :LS an interior point of an n-simplex 0- of this triangulations. Then we triangulate Sn so that the sts.r' condition is satisfied for a subcomplex K in G, which sattsfies the following properties: i)
IK leG
If L is a complex, and IL Ie: G, th.en Letting W be a simplicial approximat ion to ¢ 11\ I ii)
II, I:· IK I (~IK I is the
function ~ restricted to IKI) we have d(4x,wz) < 1( for all x € JKI. To get deg (cjJ,G,p) we count the number of simplices of K that are mapped onto ~perties
by
~J.
.
of t!16 ~£!~~ 0r~ M~ppiEB
ProPt::rty (j): Proof:
c1"
:J p for -~ /Sn be a
If ¢x
Let f: Sn
any x in G, then. d(~,G,p) = continuation such that
o.
226 r-n peS .. f (Sn ).
n.....,
'l'hEm- r-"is·-a---c-ontinuous· mapptngof""'Sinto'·-i-t-se-l.f·'
which omi~ a EO,igt.. fr')
:;::.~:
Property
cPU}),
1
p
e
3n
p
e
cP(G)
- cP('G)
then p e -Sn - f(Sn) and deg (~,G,p) If p e ~)(G), let f be the
deg f :;: 0, since f omits a point.
Tr.len deg (~,G,p) = deg f = 1-
identity mappi.ng.
p
(0
If P e Sn ..
· .....
.
i cfo(G) ,
deg (~,G,p).
-
=\
D f ~-2:E-1:.~l
=:
G -->- An S is the identity, then
1
q>:
deg (~,G,p)
!!££f.: =:
If
Ther·efore 0 :;: deg f
If 1 . -G ~ "'S'rl are con t lnuous . . - - d'0,411' mapplngs suc h th a t 41 (8), and. if there is a homotopy
Gi: IT 't. I such~that ~(x,O)
= ¢o(x),
.......
"Sn
j{x,1)
= ¢1(x)
with !(x,t)
i
p for all
x e G, t e I, then ¢o and 41 are said to be homotopic avoiding p.
Prop~'nty deg
(3).:
(do,G,p)
Proof:
If ~o ar~d ~l are homotopic avo:i.ding p, then =:
d eg (4\1' G,p ) •
Let}: G '/.. I
W( x, t) -I
-:;.>.
o
'S-11
p for x e G, t
cPo
be a homotopy of
e I.
and ~l such that
Let
F: be a continuation of
Sf
such that p
i
P{ Sl1
'j-.
I .. G )<. I).
Clearly,
F(x,l) = fl(x) are continuations of f 0 ~ fl'
dO
and ¢1 respectively which avoid p, and
Therefore
deg (cPo,G,p) :;: deg fa :;: deg fl :;: deg
(4 1 ,G,p)
·
h?Eerty (W.!. (Rouche) Suppose that deg (cP1,G,p), deg (~2,G,P) are defined, and that for some pO$itive number -~ < 'Tt, d ( cP 1 ( x) ,p) > ~l for x e
x e G, we have
G.
The n if d ( cP 1 ( x)
1
~ .2 ( x}) < '11 for all
227 ~f:
,since deg (¢l,G,P) and deg (~2,G,P) are defined, ~l(x) i p and ¢2(x) i p for x e Further~more, since d(¢l(x), ¢2(x)) < ~ < ~ for x in G, we may continuously deform ¢2 into ¢l by moving ¢2(x) along the arc of a great circle toward $1 (x). Choosing the shorter arc, ~2(x) will move a distance that is less than~. Thus ¢l and ¢2 are homotopic avoiding p. Using property 0), we have
O.
deg (¢l,G,P) ~ deg (~2,G,P) Pl'0:2 - ert ,--"y
G.
(rQ: ,
Let )G.), t J• .l be a sequence of disJ"oint open sets in Thsn, q~'iC =J P for' x e IT - UG i -~.
=! ~f:
F'irst we shall show that cl>x ~ p in at most a finite number of the Gi , so that deg (,G 1 ,p) ~ 0 except for a finite number of Gi' proving that L deg ( ,G i'P) makes sense. Suppose this is not true. Then there is a sequence {xn} of points in G wi th xn e Gn and Gn. i Gm t'or n f m, and such that 4Xn ~ p. Since G is compa.ct .(x \ has a linli t point x e; G. Suppose that '. n. co x e; G. Selecting a subsequence (x ) converging to xco ' we have, co ,. l niJ by contlnui ty, QZoo ~ p. Thus Xoo is in some Gm• But GIrl is open 1
and every neighborhood of xco con.tains infinitely lUany xu" This contradicts our assumption that the Gi are disjoint. Now suppose that Xoo e G. Again, by continuity, ~we get ~xco = peG. But deg (4,G,p) is not defined for p e ~(G). Thus the value p is taken on in at :mos t a finite number' of the G1 , so that 1 t suffices to prove the theorem for two sets G1 , G2 • The rest will follow by induction. Assume that p t [IT - (G1 U G2 )]. Since
228
Triangulate Sn so that the star condition is satisfied, and let g be a sim.plicial approximation to f. Argui.ng as before, only s implic es lying entirely in Gl J G2 can be mapped onto o·~ Counting these s :tmplices, we get
!22.f:i.1yttl..9!l: Since 'Sn - ~((}) is open, we may partition it into a dem.1.mei."'able n 1.ill!ber of dis joint open sets /\i in such a way that two paints x,.V9.re in Lii if and only if they may be joined by a cur've lying eiltirely in i • The 6. 1 are called the
6
(co~t'?dU£~.!l.~~l t !?2f In._.:-M~.l.
(Re~£k:
T'bat there are only a del1UIllerable number of components
follows from the fact tll.at the rational numbers are denumerable. The proof is left to the reader.)
tt'?Ee ~t:r_i.§l. :
If p, pIe
6.i
then
deg (¢,G,p) ~ deg (4,G,p')
r
!:roof: Let p e be trle set of points in 6.i at i and let which the degree is the same as at p, l.e. pi ere 1\1 ( > deg (~,G,p) :-= deg (q,',G,p'). To calcu18.te deg (¢,G,p) we count the number of times the simplex (;- (which conta.ins p) is "covered". All points in the interior of (i are covered just as
6.
many times as p, so that the degree at anyone of them is the same as the degree at p. Thus is an open set. It is very
r easy to see that r i.s also closed in l~i' (Show that it contains its limit polnts.) But r can be both closed and open only if it is equal to
6i
itself.
Proper!Y_(7)..:. Let cb: G -.> 3 n , \II: '3 11 ~ n be continuous. SUp-;;~n pose that for peS· , deg (w~,G,p) is defined. Then if ~i are ""-n (.. the components of S~ - ¢(G),
1
229 Proof:
An argument similar to the one used to prove propertyC'?)
allows us to conclude that deg (~,G'~i) = 0 except for a finite number of the /\., proving that the sum at least makes sense. l deg (W~,G,p) is calculated in the usual way, by first ~n 0 tria~w:ulating S so that the mesh is less than d(W¢(G) ,p) and so that p is an inner' point of an n-simplex of this triangulation. Then 1,\Ie triangulate 3n and Sn so that the star conditions are satisfied, and replace ~,W by simplicial approxirna.tions f,g. (f: IK I -s;.. -Sn, TAhere IK leG) • Prom a previous the orem, gf:
IKI --...~n
is a simplicial approximation to \jI4>:
IKI ~Sn .
F'inally we count the ncmber of simplices of K which are mapped. onto C"! by gf. This is deg (W~,G,p). A simplex in G (or in K) wh1ch
j.
s mapped ont 0
@),
From property
(l
by gf' is first mapped into some
6.1
by f'.
ever'y simplex of t~i is covered the same muu-
bel' of times; call this number n 1 . The degree of \jI vdth respect to 1 at p, dag (1jJ ,Ll ,,P) is the nml1ber of times g covers 6'.
.6.
61
Thus, for e80h
1tJe haTe a contribution
dag (~,G'l~i) deg (\jI,L~1'P) Prom property
@.)
dag (1jI~,G,p)
=;.-
dag (LG'!~1)
dB@'
(\v,Lii,p)
tN'e shall extend the concept of degree to a certain class of functions ~: fA ~ . .
Rn ,
A an open set of Rn •
Definjtion: Let A be a subset of R , and let ~: A ~'R be of n n the form <J,(x) = f(x) + x, with f(x) continuous and If(x) I < Ivl for
-----.
all x in A. Then (x) is called an ad~mj. ssi ~l~ mapping. An admissible mapping (by means of' an inverse stereographic projection)m8.y be considered as a mapping c/J s : G ~ -gn for which the north pole is a f1xed point. That is, if fA (in. Rn) is u...'1.bounded, (Hx) ::: f(x) + x, for x in A, is unbounded. frhus points very f
9I'
from the origin. to co.
from the origin are mapped into points very far
On Sn, bowever, the north pole corresponds
230
,.-'<.
Pr2.l2~rtY__ ,\a~
: Let A be a compact subset of Rn' is continuous, then ~ is admissible.
If ~: A
-i>
R n
f!0of: A is compact <.=) A is closed and bounded in R n ) Ixl < Ml for x in A. Also, ~(x) is continuous on a compact set 9 ~(x) < M? on it. Thus, if rex) ::; ~(x) - x, we have If(x) I < Ml~+ M2 "
!!:£E~ty (p'): If ~: A -> 6. .1,: R -> Rn such that '+' n i)
",(x)
ii)
itn
is admls si ble, there is a mapping
= ~(x)
for x e A
'" is admissible
Proof: ¢(x)::; f(x) + x withf(x) continuous, and If(x) I < M for all x in A. Using Corollary 1 to the Tietze extension theorem, we may extend f continuously and in a bounded way to the whole planej call this extension g(x). Clearly \It(x) ::; g(x) + x is admis si ble and. IjI ::; 4) on A. Proper!y (Pl: Let F be a closed set :l.n Rn. admissible, then d(F) is closed.
If~: F
--;..'Rn
is
Proof: ~(x)::; f(x) + x, where /r(x) I < M for x e F. If .(F) is not closed, there is a sequence \Y n } of points in ~(F) which converge to a point Yo i ~(F). We may form the associated sequence {Xn} in F by writing ) + x Yn ::; qIlexn ) ::; f(x n n If \Xn '} has no finite limit point then xn .,~ 00, and since ~ is admiss:i.ble Yn -~ 00 too. Thus {Xn l must have a. limit point xo' J . to xo. S'~nce F :LS Let :x ) be a subsequence which converges ~ n. J ' t, J closed, Xo e F. By continuity y
nj
= ~(xn
j
But {Yn\ converges to Yo'
to Yo'
That is
) ~ ~(xo) Thus ~very subsequence also converges
231 Yo
= 11m
Y.
.0.
= lumyn'j = ¢(x0 )
and Yo e ~(F), provirJ.g that q,(F) is closed. F'irst we shall define the degree of an admissible mapping ~: R ~R . n
n
D~fi;:~tion:
Let~: Rn -o>Rn be admissible.
Mapping Rn' 'Rn onto S , S respectively, by means of an inverse stereogre.phic pr'ojection, ~ becomes a continuous mappi,ng ~s: Sn ~. "Sll which takes the north pole into the north pole. We define dep;.J. to be deg ~S. 1 Definition: Let A be an open set in R , 4': A --9:0- /"R an admissible ..~.--A () n n ma.pping. For any point p in fln - ¢( A), we define
-
deg (~,A,p) = deg (~S,As'PS) where AS ,PS are images of A anq, P respectively under' an inverse stereographic projection. (Slnce the stel'eographic projection is a homeomorphism, /\n 1 (;:> AS is open and Ps e S - (JIS AS)' so that deg s ,AS 'PS) is defined. )
(4
We ba ve defined tLe degree of admlssible mappings of Rn into i tse If in t enns of e. ~rti~l:l:~.!. kind of continuous map of Sn into 1 ts elf (north pole go es into north pole). 'l'hus ~rgper
ties~
- <1?.-h2.1W2Lsuch maEEinp:~. The restriction on the type of mapplng of Sn into itself, allows us to state soroe addittonal properties.
l:
Prop~i?":LiID~:
Let A be open in R. If ,.,., ¢: A -+ 'i~n , A -;.. Rn n are admissible, and if deg (~,AJP), deg (¢, A,p) exist, then, 1,1
4(x) = ~(x) Proof: -
0-
for
x e A
~
deg (q,A,p)
= deg
i""_'
(~,A,p)
An admissible mapping, ¢, is homotopic to the identity:
In fact a homotopy is given by
I(x,t) = x + tf(x)
232 with Sk(x,O) •
!(x,l) = ~(x).
= x,
••
<.1:"'
I
-
l'hus, taking inverse stereogra-
phlC pr'o Jectlons, (JJ"", <.p,.,: As. ...;:.. :j v Indeed, if F(x,t) is a homotopy ,-..., = ~s(x) for all x € AS' s.nd all and the definition of degree of deg (¢,A,p)
= deg
(~S,AS'PS)
An
ar'e homotopic avoiding PS'
S
/',J
of ¢S and ~S' F(x,t) = ¢s(x) t e [0,1]. From property admissible maps
(j)
"'-'
(~S,AS'PS)
= deg
= deg
PropeE.ty,__(2l~: (s trong form of homotopy invari8nce) that ~, ¢: l\. ~ are admissible mappings. Then if c:> n homotopic, on A, avoiding p,
R
KT:'?:?;~~:
Suppos e ~, rare
,....;
= deg
deg (4,A,p)
rJ
(~,A,p)
(~JA,p) •
I'-v .An 1 'Y On the n-~phere we have ¢S'~S: AS ..,..;;:. S where <.pS'
~wmotopic
on AS' avoiding PS' But we have also seen that ans" two admlssible mappings are homotopic (see proof of property
8XS ,,.' '\
~.~
1"'-'
l'hus we may extend the homo copy of ~S and. ¢s to all of AS' Fr'om property (J) and the definition of degree ~ of admissible rIB J:p ine: s,
(g) ") .
deg (~,A,p) = deg
(4 S ,AS 'PS) = deg
r'-'
(¢S,AS'PS)
= deg
(4,A,p)
~rope ~!y(l.9),,~: (:;)trong form "-Iof P.ouche t s tbeorem) If, for t X e A, d(cp",X),p) > €, d(~(x),¢(x)) < e, then degree (¢,A,p) ~:
= deg
I
(~r,A,p).
,....,
Proof: Just as in the proof of property (~, we show th:~ i'~ -t-:l" arehomotopic on it avo~~1ng p. This fact, and property give the desired result. Property 1"9)* sts.tes that the degree depends only upon
C?)"
'--.-
what happens on the boun.dary.
This suggests that the concept of
degree may be further extended to admissible mappings defined on Let·~ be an admiss:i.ble mapping of a
9.rbitrary closed. sets. closed set FC R
n
into
tfl. n
Since R
n
- F is open, we may talk ..
about the components, Di' of Rn - F (pe.ge 228). Clearly Di c: F. Since }' is closed there is an admissible extension \jf, of 4, to the whole plene.
For p
i
J(F), we may talk about deg (,jl,Di,p).
233 Definition:
For the situation described above we define
= deg
deg (~,Di'P)
(~,D1'P)
(0
From property of admiss ible mappings, it follows that ~n - ~(F) is open. Letting ~i denote its components, it C~~ easily be shown that
6
for p, p I in the same j • The proof is the same as before. lJ.'his allows us to use the expression deg (~,Di,.6j)' Le t F be the ct2C'L;.::lJ'erer~ce of the unit circle, and those portions of the lines y = .:!: 1/2 which it intercepts. D1 , D2 ,D 3 ,D 4, are the components of R2 -F, as shown tn fig. 2. Let ~: R2 ~ R2 be given by ~ = w ::: z2 = (x_iy)2. cp takes lzl = 1 into Iwl ~ 1. y = ~ 1/2 is mapped into 1; i . 1)2 ::: x 2 - '4':" w = ( x:!:. ~"2 x. Ex§.mpI~:
I
~R2
H~---Y=~ _ _
.-t-.. ._I~_
1
v--
- -" - 2
Figure 2
We may calculate deg (~,Di'~j) for i,j ::: 1,2,3,4. We shall leave it to the reader to verify the following results: deg (¢,Di'~j) :::
0 if i
=I
j
if i
=
j
f-z
1
Fi~re 3
shall now prove the Jordan-Brouwer Theorem, and as a oorollary the Jordan 'tATe
234 Curve Theorem. lJ:'he P!'oo.f given is due to Leray. \lifi th a few minor changes, it can be carried over into a Banach space. (See Chapter XV) Let PI be a compact subset of Rn , and let ~: F I -;;.. F 2 C /R~ be a homeomorphism. Then the number of components of Rn Fl is equal to the n1.L.'11ber of components of 'R - F 2 < n Theorem:
Proof:
From the countability of the rational numbers, it .fol-
lows that Rn'- Fl has at most denumerably many components. Let us call ~-l = \)I, D. the components of R - Fl , and Li the com~.
n
l
2.
ponents of Rn - F
Since FI , F'2 are compact it follows that both ~ and ~( are admissible. (Prc;perty~») Therefore, for all pairs i,j for
6.
which D., exist, the expressions deg (~,D. ,~.) = u .. and l J l. J lJ deg (~I has as k ,D"t ) == }3,lL_~ are defined. The matrix A = (u;;.) .... J many rows as ther-e are components of R - F l , and as many
,.6.
L'
n
columns9.s there are components of Rn - F2 • ., B = (!3kk)'
ble.
\jI~ is tIle identi ty map, which is certainly admissi-
I't folloHs from \jI(~(Di)
= Dj .
and Di
deg (W,¢,Di,D j ) where
And vice versa for
(0 .. ) lJ
=
=
n Dj
== 0 that
0ij
I is the identlty matrix.
Using property
(7) \,--
1-
u. i3 • == 0 .. , or, that j\·B == I. The lS s J l J same ar::ument results in the conclusion B-A ::: I. A well known
We have just proved that
the.orem of linear algebra says that: If A.B :::: B·A ::= I, then A and Bare ~quare matr1.ces. may be infinite of course).
(A,B
But r.scall1.ng the relationship
between the number of rows and columns of A and B, and the number of Di and
6.. j ,
'l.-le may conclude that the number of Dj. is equal
to the number of~~ • .;
Definition: A (closed) Jordan curve (in R2 ) is a homeomorphic -1 image of S •
235
Corolla~y:
A Jordan curve separates the plane int~ two disjoint connected components, one bounded and one unbou.."lded;. Proof: 'Ille separation into disjoint connected comytonents f;tlloil'Ts .from the general theorem. 8 1 is compact
.J) >
,~ 4(Sl) is compact =9 ~(Sl) i.) boundedL~ .
>.r
there is a 0 such that all points z in R2 with l,z I lie in the same compo~ent L12 ( the upbounded componeht). Then all z in C1 are such that Iz\ <
fe
(It can al~o be shown that the unbounded compon?nt is simply, connected,) 1'he corollary as stated above, obviolsly ho1dsJ for Rn' n .:: 2... Howeve:>, it carmot be shown for n :f 2 ihat 6 1 simply connected. In fact, for R3 , a counter example ~ the Alexander horned sphere~) The ideas developed in this chapter allow us to p~esent a second proof of the I~~rian~e of Domqin for finite dimelsional spaces o (See page 88.)
+:
/'
The ore!!!.: Let A be an open set of R and let A ~R be 1-1 ~ n ~ h and cont.inuous. Then ~(A) = A is an open set in Rne To prove this we need the following e
I:..emma: Let B donote an open ball in Rn' B its set theore~lcAl boundarYQ If~: B --3lI> ~ CB) is a b,omeoHlorphlsm, then :> (B) is ""open in the topology of Rn 'l.' a
Proof:
B
~
is compact ~¥(B) is compact.
6
Figure
4
2
236 o
6
Let Dl , D2 be the cort.lponents of Rn - B, and let /\1' 2 be the components ofR' - ME,). l;']e know that since c/>(B) is compact, n
and hence bounded, one and only one of its components unbounded. As jn the previous theorem, we may wri te
6.2
is
(In talking about the degree of C/>, we are treating c/> asar1 ado missible mapping of a compa.ct set, B, into R. See page 233) n It is quite easy to see thet
2
~;--a. (3.:::: -'-1 :1.S sJ s=
Since
62
:is unbounded and.
<1>(13)
a'
~
j
5s bounded,
there is a point p
j.n /\2 such that d(p,c/J(B)) > M > 0 for any t~ > O.
(property
(k,')).
Then
Therei'ore,
In a similar manr'.el" we can show that i3 12 :::: O.
Equating elements
of the two equal matrices, lvB ::: I, we also have
1 :::: a 11 (3:n + a12~22 :::: ul1i311 so that all :::: 13 11 == +1. Prom thi s i t follows that Lll C ¢( B) • F'or, assUJue that there :is a point a. E 6 1 such that a ¢(B) ::: (D 1 ), 'l'hen
i
+ 1 == all ::: deg (~,Dl,Lil) == deg (¢,D 1 ,a) :::: 0
(the last equality folJ.ows from pr-operty
(?))
which is a contra-
diction of course. In the s arne way, l-Je :may prove that B .=: Dl C c/>- l l ), which would imply that q,(B) C 1 ' Thus
.6.
C6
(B) ==
61
237 /".
But by definition, WI A is open in R , so that this proves the n lemma.. Returning to the main theorem, we have.a 1-1 continuous . onto ~ ~ map qJ: A .:> Awlth A open in R and A CR. TIJe must show that cI>-l is continuous, or th9.t
Since A is open there is an open ball B around p such that B is compact.
B <.: B CA.
A 1-1 con.tinuous map defined on a
compact set is a home omol''Phism. open, which proves that ~ is meomor·phism. Before
tUT'~!Lig
f?Jl
From the lemma, the set cI>( B) is open map and consequently a bo-
to tb.e concept of the degree in a Banach
space, we shall defi:!.l.6 t.he i!ld~~ or trmultiplicitylt of a mapping Folloi~d.ng
at a point.
the di801).sslon of the index we shall make
a slj_p~ht digress ion into tbe r·e s.lm of Function 1'heo1'Y. rI'he purpose of this dig~esslon will be to prove a few well known results of Function Theory using the ideas we have developed here. We sh9.l1 trJ to indicate, by means of a few specific examples, the na ture of tlle dit'far-once between the _ topological and _ _ .a _ _ _ ·pronerties ... the
fp!2~~:~on...!.~l~2reti~.al
and
di:
pl'operties of analyti c functions. In the following dJ.scusston A Hill be an open set in R , A
~
"Rn
Defh}2.t i
o
n
will be an adDJ'Lssible l1.iS.Pphig.
•
A point Xo E: A is sa.id to be a .!~.(71!1~!:.....p..s:!nt of t,? .2" :If the:e8 is 8.n E: > 0 such that whenever
< d(x,x o ) ~ e, (x)
i) ¢(xo )
=0
t
o.
That is, either
-I
0 for 0 < d(x,x ) < E, or o ii) cI>(x) 10 for 0
their equivalence. Let A be an open set of Rn , and ¢! A ~ Rn be admissible. If x o is a regular point of ~ with respect to 0, then we define tr.le index of
!?efiniti~,l:
-~.----~.----:p
by ir(¢,x ,0), as deg (~,B (x ) ,0). (B (x ) is a closed ball of o r o r 0 radius r around x which is such that (:x:) -I 1'01" o X E B (x ) - fx ~.)
°
rot
0",
238
Definition It: ----
Let S denote the surface of Br(xo ) •
the mapping
\If
=
~(xr) ----.-: II ~ (XI') "
,
Sn-l-i- Sn-l
(x
€
This is c erta:i.nly rJell defined s ince ~ (y) f; 0 for y define i11(~~xo'O) as deg
w.
Th~~~:
Consider
Sn-l) €
S.
'tAle
i 1 ::;: iII"
P:r.~oof:
Since an n-dimensional ball in R is homeomorphic to an n n . n-simplex, we sh'3..1J. replace Br(xo ) by an n-simplex cf c. Br(xo ). Let -;..n denote an .J-:2, implex in 1{n which contains the point 0. Tr:i.angulate &n and 6 n so that 0 is the barycenter of an n-simplex o-~ of the tl'ie.ngulc..tion of ern, and so that ~ satisfies the star condi tion.
Let us suppose that the n-sin~lices ~ of the triangulation (of ern.) are so or:i.ente1 that .....dL i
n
6~i
:;:
-.:;-
n-l
..i._ 6 ;
j
~
(rrj-l a1'e the (n-l) -simplices of ern.) Suppose ths.t the same thing holds for the S·~ of ~5·n" itJithout loss of generality, we assume that 0 is the bal'ycenter of ~'n (5·1' Let f be a simplicial approxi.rnation to ~ with respect to these triangulations. Then since f is a simpllcial ma.pplng (and cor.i.seguently dimensionpI"eservtng) ,..-
f\,>
T
Remembering that 0 from the fact that deg
€
n) ~ '\ /,n ,(5'i ::;: 4.- (\..0"". j J J
8·~, we have iI(,xo'O) :;;: "'1.
(q,vn ,0) ::;: deg
( n ,0) f,6
= A.l
This follows
.
Furthermore, since f is an allowable homomorphism
____ ~n-l v k
::;: ro.l.. k
239 where m ::; iII(~,xo,O)4 This follows from the fact that the restriction to the boundary (of (5'"n) of any simplicial approxima ... tion to ~ is a sLnplicial appr-oximation of the restriction of 4 to the bOL.Ll'1dary (of .sn) • Since 1'(2: ~~)
A.ljj,
=L
A .Ja~ J J
L-
Jf(.2:: 6ni) ::;
we have
:;
m""" ';ynk-l 'L...-
Now the boundary of this expression vanishes: mOl2
'2-~-1::;
L
A
l'a B-j : ;
0
so that the "surface" ofe- n is a pseudo-manifold. that all the A. are equal. Thus we have
It follows
J.
,
"\ v-,n
~
"I.
L- l\iC)O' i
·,...n
L- I'-lei IJi
::;
'\ ~ _,An "I. ~ ~J-.n-l 1\,1""- a 6" i ::; 1\1"'- ' j ,~
::; m.£..-
,r-n-l ?} j
proving that 11.1 :;:: m and i1 ::; ill. ~ni tl"~.n:
Xo is
9.
_ _. _ '..
regular point of ~ with resps ct to a if Xo -I
._ _
is a regular polnt of ~ (x) ::
a
&(x)
- a with respect to O.
Def!.nitisn: If Xo is a regula!' polnt of ~ with respect to a, we define the index of ~ at Xo with respect to a, denoted by
_
..--..."...--'._-"'_.....
i(~,xo,a) as i(~a'xo'O).
--
Theorem: then
.........--.---....'--.---.--
If ~(x) ::; a for at most a finite number of x in A, de g (~, A ~ e,) ::;
2._
x€A
i ( ~ , x, a)
Proof: Let 4{x) :;:: a at x::; x l ,x2 , •• "xk o It is easy to see that i(4,x,a) ::; 0 if x I Xi for i ::; 1,2, ••. ,k. In fact, since Rn is a metric space, there is a closed ball Br(x) which contains none of the xi. But then 0 :;:: deg (~,Br(x) ,0) :;:: t(~,x,O). In addition, for each x j , there is a Br (x.) Br such that: j J j
=
21lO ~(x)
i) i1)
B
I' j
'I
for x
a
E;
Br • -
\""x.'t J r
J
n Br i = °
I
for i
)
j
.
Thus :i.(~,x,a) is defined for all x e A and
2_
k
:i.(~,x,a)
:.=
xe:A
L
j.(~,x.,a) =
j=l
k
L
j=l
J
deg (4,B r .,a) J
Using the 8.ddi tJon theorem for tbe degree (property (~)), we have deg (~,A,a) ;;::
2:
deg (~,B
rj
,a) =
L.
i(c/l,x,a)
Since the i:'ldex bas been defined in terms of the degr'ee, it would. not be diff}J: ul t to re~JOrd properties Q) - 0), (~-l" _ ~j)~:', EO that t11ey would become statements about the index. .Again we leave thls as an exercise for the more ambitious reader. ~:IO
conclude this chapter, we shall say a few words about
the funcJ:;ion-theoretical interpretation of the index. Let G be e. dome.in in R2 bounded by a Jorde.n curve (~ (or curves 0,1'(;2"" ,G.m) , and let 1'( z) be analytic in G U &. Assume that f(z) 'I for z e t;. 1hen deg (f,G,O) is defined. From
°
°
the identUy th.eorem for analytic functions, f(z) :.= for at most a fInite number of points z e G. Using the theorem just proved deg (f,G,O) ==
z= 1(f,z,0) zeG
For simplicity suprose that 1'(0) ;;:: O.
Then for some neighbor-
hood of the origin f(z)
;;::
13.'0 Z
= 8." z
k
1-
k~
II
a z 1
k+2 + k+l + a2z
a1z +
= aozk[l + b(z)]
...
2 + ••• 1"
8. 2 Z
We want to compute'~(f,O,O)
where k > 0 and :b(z) is analytic. and see what it
tnB8.11S.
FLt'st we note tha.t f( z) is homotoplc---t.o_
aoz k avoiding 0 (for somo neighborhood of 0). topy is given by
,
Indeed, a homo-
o~
t < 1
Thus i(f,O,O) = k is the £!Elt},£licitI, of the zero at z This means that the expression deg (f,G,O) =
2:
= O.
i(f,z,O)
gives the number of zeros that fez) has in G, each zero being counted tdth its proper multiplicity. With this interpretation of the degree many of its properties proved earlier in this chapter appear to be complete trivialities. However it is worth noting that a rewording of pro,.-. " I perty (:1..9)~'" gives the c las sical s ta ternan t of Rouche t s Theorem: Theorer~:
(Rouche)
Suppose that f, g are analytic and sinj;tle-
e.
valued 1n G () If' there is an e > 0 such that If( z) I > e and If(z) - g(z)1 < e on ,~,:, then f and g have the same number of zeros in G. Proof: We have ,lust seen tht'l.t the number of zeros, counted with the proper multiplicities, of f and g respectively are de g (f, G, 0) and de g (g, G, 0). But, us ing (LO) -;(. ,
deg (r,G,O)
= deg
...........
(g,G-,O)
'vIe know that in the Theory of Functions of a Complex Varia-
ble, Rouche's 'l'heorem is a result of the flArgument Principle". As we might conjec.ture, the argulflen.t principle is also a topological statement, which is true for functions of a more general nature than merolllorph.ic functions.
B1Jt let
1.,1S
state this prin-
ciple for a function f(z) which is analytic. and single~valued im / G U vand which does not vanish on
~
(p.
If P denotes
t~he
number
of zeros of f(z) In G, counted wj.th the proper multiplicities, then
Since
i+"~
f' ( z) iTz)
)&
t
dz = q:J d (log f( z) ), we may hope to prove the JC;
argument principle for a wider class of functions, namely those for which the integral on the right exists in the Stie15es sense, and which have at most a finite number of zeros in any bou.nded region. Setting f(z) = If(z) leH~,
~
1 ~1tT(\
)) 'a.'( log f (z -
'C Since If(z) I is single valued, the first integral an the right h,snd siG.e vanishe3. The second integral gives us the increase in the argument of fez) t·ibich must be of the form 2'JTm, m = 0,1,2, •.. if we assume e, to be positively oriented. For, as z traverses C, f(z) describes a closed curve r(C) whi6h surrounds the origin m times. If f(z) is regular, f{~) is bounded, and it is 8t least intuitively clear that f(G) 1Ieavers" the origin m times. Of course thts is precisely "what we mean when we say deg (f,G,O) ::: m A more general s ta tement of the argi,m~ent principle (which
may als 0 be extended to the wider c las s of functions) is given for f(z) meromorphic in G U (:;'(1.e. f(z) is ans.lytic in G except for a. finite number of poles.) P - N
::: .J_
/l
2'rri ~) .J
In this cese f'(Z) dz
fTzJ
(Q
where N is the number of poles, counted T..]i th the proper u1lJ.lt'i.plicity, of f(z) in G. The topologlcal nature of this theorem becomes evident when we observe that if f(C,) surrounds the origin in the positive sense, then it surrounds ro in the negative sense. Furthermore, a po1e of order m at some point z e G,
243 corresponds to a zero of order m at co. 11hus 1\J gives the number times 00 is covered by f(G), the minus sign indicating that fCC) is negatively oriented with respect to 00. It might be clearer if the re
fez)
= a o zk
g(z)
-
= fez)
...
ao
k+) + bIZ k+l+l + b0 z
In this neighborhood of
h(z)
+ alz k+l +
z
=I
0
bo
-I
0
= 0
+ g(z) = aoz k + alz k+l + ••• + (a 1+ b 0 ) zk+J
+ ( a1 +l
k+l+l + ••• + bl)z
Therefore h(z) h'ls a /';'81"0 of order k at z = O. Ii. counterexample l~ill show that thj.s theorem is not true 2 jj;(. in general for non-analstic functions. Suppose that f( z) ::: r e ' and g(z) ::: rei(k+.t)Q. Clearly f and g have zeros of order' k and k+i respectively at the origin. h(z) ::: fez) + g(z)
=
Again we form the sum . (k+ I) G
re 1
.
.
• £' Q
[re- 1
+ lJ
244 The function
o ~
<
t < 1
i(k+l)Q
as ta.b lishes a homotopy between h( z) and h( z) == re • The reader should check that h( z) and il( z) are homotopic avoiding o. Thus /'-.
deg h == deg h == 1{ +-l
245 Chapter XV ~._"'-W_""
__
~'_
Lo£..~ _Degre~ .. !ID9--:tnd~.!
- Part II (Banach Spaces - Schauder Leray lrheory) In this chapter we shall extend the concept of degree to certain mappings defined in Banach spaces. The principal ideas involved are due to Schauder and Leray. In the Proceedings o:f the I!lt~~io.!!?-].._ConE~ of IVlathe:nati.cia~ in Ca:nbri:!g~, 195c, Volume II, pp. 202-207, there is a more general discussion (by Leray) in which ideas s imila.r to those we shall develop here a.re applied to lines.r topological convex spaces. We shall begin with some prelim5.aarji lemma.s concerning the degr·ee and index in Rn' After defining the degree in a Banach space, we shall prove generalizations of the Jordan-Brouwer Theorem and the Theorem on Invariance of Domain. Lemma 1:
Yi ~
n
2:: j=1
Let ~(x) ~ y: nn -> Rn be a linear map, i.e. aiJ·x j + ~i' i ~ 1,2, •.• ,n.
Then, if (x) == a,.
...
1
=-
1
det ( o.i j )
> 0
->
i(~,x.,a) ==
det ( at j )
<.
0
--.)
i(~,x,a)
Proof: To sho rten the pr'oof J we wtIl assume that ~i = o for i == 1, ••. ,no This only amounts to a translation and wonft change the degree or index. That is, we w:1.11 prove that det (o.1j ) > 0
~
V
det (o. ij ) < 0
i(~,O,O)
+
1
i(~,O,0)
== -
1
The proof 1s by induction. Suppose that n == 1 and y == ax. If a > 0, Y is homotopic to the identity. In fact, a homotopy 1s given by o< t < 1 H(x,t) = tx + o.(l-t);x , H(x,O) ;:; ax ;:; y
.
I
-
H(x,l) ==x==y
-
246 -,,0
CI
and iI(y,O,O) ;:: deg (y,{3r'O) ;:: deg (y,{3r'O) ;:: 1. is homotopic to a reflection mapping. That is,
H(x,t) H(x,O)
;:: ;::
-tx + a(l-t)x ax I. H(x,l)
,
=
0 <
If a < 0, Y
t < 1
-
-x
and iI(y,O,O) = deg (Y'~r'O) = -1. Now we assume tha.t fOl~ linear maps in Rn _1 , det (a ij ) > 0 > y is homotopic to the :i.dentltYi det (CIij ) < y is homotopic to a. map Yi = !SiXi' with an odd rlu.mber of the ~i ;:: -1, the rest equal to 1. That ls, '5; is a reflection of an odd nUlflber of the xi' keeping the rest fixed. To prove it for Rn J we
° =>
first C0:1s::'d6r a special J.inea~ mapping Y1 = l;:: 1,2, ••.
,n,
o
0 ••• 0
0'22 •••••••
Suppose that det (O:ij) > cas e 1:
all > O.
$
a .. x., J. J
J
\
CI2
..............
an2 •....•
n
L1
~
n
)
ann /
0
Then
/CL22
I .: \
\0:11.2
•••
n
From the induction hypothesis homotopic to the identit.y. a homotopy
with Hi(X,O)
= I: ~ijXj
y =~
'UijX j , i ;:: 2,3, .•. ,n, is
That is, for :t ;:: 2,J, ••• ,n, there is
and Hi (x,I) :;:: xi'
H1(x,t} = tX1 + aI1(1-t)x. the identity.
F'o;r: l ;:: 1 we use
n Thus y ;:: ~ aijx j is homotopic to
247
2:
case ,2: all <,0. Tben det (~ij) < 0 and y = ~ijXj is bomotopic to a mapplng which reflects an odd number of the xi' B~t Yl ::; a1lx1 is homotopic to a reflection. Thus Yi
= ~=l
aijx j , i ::; 1,2, •.• ,n is homotopic to a mapping whicb
reflects an even number of the x J..• On the sphere, such a mapping has degree 1. By Hopfts theorem it is homotopic to the identity. Mapping stereographically onto the plane we get back our original mapping. From the definition of degree and homotopy 1.n Rn' we have that ~(x) ':;: Y is homotopic to the identity. If det (a ij ) < 0, the same kind of arguments give i ( ~ , 0 , 0) ::: -1. l~le
still have to show that every general linear mapping yi ::: 2: (3~ .x,., i = 1,2, ••. ,n, is homotopic to a special one as ~J J described above. First we will show that the mapping given by /\ «(3i j) is homotopic to one with a, ma.trix «(31 j) for which [312 = O. A honotopy is given by .,~
Hl(x,t) ::: ~llxl + (1-t)(3l2 x 2 +,f3l3 x 3 + ••• + (31nxn (312 Hi(x,t) ::; (3il x l + «(3i2 - t ~ll (3il)x2 + (3i3 x3 +···+·(3inx n for i = 2,3, ••• ,n. For' es.ch t, O.~ t.5 1, the matrix of the transformation descrj.'bed above is obtained from the original one by multiplying elements in the fi,rst Golumn by t«(3l2/f3 11) and subtr'acting this from the second column. Thus the determinant of the transformation never chg~ges sign, which is what we want of course. Now we shall show tha,t Yi ::: L: f3 ij X j is bomo~,.,.., /'.. . topic to y = ~ (3ijYj' where (321 ::; O. A homotopy ~s given by /'
f n
Hi(x,t) =
f3 ij X j
(321 ::; «(321 ~ t ~ f3 ll )x1 '+ 1-'11
(321 + «(323 ~ t (3il (3lj)X j + ••• +
for i :/ 2 + •••
rr'hus ev-ery genera.l linear map is h.omotopic to a special one. Lemma 2: Let G be an open set in Rn+m, and let ¢: G ~ Rn+m be a continuous mapping given by Yl
:;:
Y2
::;
• •
·
Yn
~l ( Xl ' • • • , xn ) 42 (xl' •• • , xn)
= ¢n ( xl ' . • •
1
xn)
Yn+l :;: xn+l Yn+2 ::; xn+2 • • •
Yn +m = xn+m
where the ~i are bounded.
Assume that ¢(x) R
f
° for x
., e G, so
that deg ~$JG,O) is defined. Letting 4 n denote the restriction of' to the· SUbsp~ce of Rn+m for which x n +l ::; xn+2 :;: '"
::; x
n+m
:;:
0
'
and G n :;: G
n Rn ,
deg (4,G,O)
then
= deg
R R (4 n,G n,O)
Proof: We shall assume that ~(x) ::; 0 at a finite number of points xl ,x2 , ••. ,xl., and that j.n a neighborhood of each of the pOints, ~ is lineQr. (Replacing ~ by a suitable Simplicial approximation which does not che.nge either deg (~JG,O) OJ;' R R deg (¢ n,a n,O) will give us this property.) We have seen that if ~(xk) :;: 0, then l(¢,xk ,O) = !l depending upon whether de t (a
i j
~(yI' • . " Yn+m) 'd\ Xl ' • •• "xn +m)
)::;-:=:r:::.-
R i is positive or negative. Sim.ilarly for i( n,x ,0). is restricted to Rn ). Also
249
·.. ·.. det
a
anI
o •
· o·
0
...
~
• •• 0 • • ... 0
··
= det
(~~~ I
~\
... nn ° ...
a 0 2n
•
\~nl
1
I
)
1/
o •..
.. ..
...
a
nn/
This, together with the identity deg (~,G,O) = imply that deg (~,G,O)
I"
xeG
= deg
i(~,x,O) R
R
(~ n,G n,O)
Defini tion: Let H be a Ba.nach space with norm II ". A mapping ~: B ~ B will be called adEQis~l.bl~ it' i) ~ is continuous, ii) ~(x) = x - !(x) where
B.
Definition: nuous if
The mapping (jj: G ...;;:. B (G C B) is csmpl~tely c2E!.!.-
i ) ! is continuous, i1) whenever G t is a bounded sub-
--
set of G, theri ~( G 1) is compact. Here is a brief outline of what we shall do. We shall consider mappings ~: G -> B (G is open in B) ,with W(x) = x - ~(x) completely continuous. Then we will extend ~ to B itself, so tha t the extension ~ il-: B -.-;:.. B is admis sible. Finally, 4-:l- will be approximated by a mapping W which takes B into a finite dimensional subspace A of B.
We shall then define
deg (~,G,O) :;: deg (~,GnA, 0) :;: deg (\jI,G(}A,O)
250
L.emma 3: -Let-F c B be-cl.osed~.sc_ Bb.e convex and compact.
If f: Io' -;:.. S is contInuous, then there is a continuous extenslon g, of f, such that g: B -~ S. Let co{e n } be a sequence of decreasing positive numbers
Proof:
such that
.2:-
Using lemma 1 (page 93), we may asso-
en < co.
1
ciate with each en' a positive integer k n in the following way: There is a continuous mapping f n ! F ~ Gn C S, where dim Gn :: k n , and
II f n(x)
- f ( x)
I
-< en
Setting hn+l :: fn+l - fn' hn(F) ::::: An
C
for all x e F B irJhere An has finite
C)
d
=L
h (x).
1
n
Furthe rmore
From Corolla!'"'.! 1 of the 'llietze Extension Theorem (page 222) there is a continnous map
such that ~n(x)
= h n (x)
for x e F.
Let
(~n(x) j
H (x)
n
=,;
~n (x)
t€n!l~n(x)jj \~
It is readily seen tbat En satisfies the following properties: i) Hn is continuous on B ii) H = h on F
n
iii) iv)
n
IIHnll ~ 2e n
co
L. 1
H == H n
B ~ B is continuous, and H :: f
on F.
irhe extension g(x) is defined as follows;
g:: f on F ii) for x e B~F, g(x) y is chosen so that yeS and d{ y,R(x») = minimum. That there is such a y in S follows from i)
=
251
the compactness of S. The uniqueness of this point is a result of the convexity of S. It is left to the reader to verify in detail that this is the function we want. Lemma /.]..: Let G c B be a bou.nded open set, ~: G ~ B be of the form ~(x) = x - !( x), where '$i.3- completely continuous. Then there is an admissible mapping ~ -l<-: B -:> B with r~ = ~ on G. Proof: ~ is completely continuous and G bounded ~ }(G) is csmpact. From lemma 2 (pag'e 95) the closed convex hull of ~(G) is compact. Applying lemma 3, we have the desired extenoian. Lemma 5: Suppose that ~: B -0> B is an admissible mapping, and that FeB is closed. Then if 4(x) I 0 for x e F, there is an e > 0 such that 1!~(x)H~ e for x e F. ~ is admissible means that ~(x) = x - (j)(x) where
Proof:
jQ(B) C S, a compact subset of B.
Suppose that no such e: > 0 Then there is a sequence n} in F such that
exists.
I im n-;::ooo
\X
II Q( Xn) II =
0
It follows that
since W(x ) lies in a compact (and hence bounded) subset of B, n and since II
l
nj
j
x
also.
But
nj
= _.i[.(xnj )
{xnj$( is a sequence
+ q{x
nj
)
-.;>
of points all lying in a
set F, so that y e F. By continuity, .(y) our assumption that ~ lOon F. L~_£:
y
= 0,
cl~~,ed
contradicting
Let ¢: B ~ B be admissible, and let G be an open set for which ¢(x) i 0 when x e G. Then there is an admissible mapping ¢l such that il(x) = x - ~l(X) maps B into a finite dimensional subspace Al and
253 'Def,inition!.............. "., Let~: B -0:0. B be admissible, and let G C B be open 0 with q,{ x) f 0-' rorJf;'€~ .G •. -.J:r-" \]1'" I's ""9.11" allowabls-- approx1mat.;ton. ..t.o "', and A, a f'inite dimensional define
~ubspace
belonging to \II, thenwe,_
Definition: For the situation, descr'ibad in the previous definiti.on, we define deg (~,G,a)
= deg
(\jiA ,G
(f A, 0)
/\,1
where W(x) is an allowable approximation to ~(x) - a. Def'inl.!.ion: Let G C B be open and bounded. If ~: IT ~ B is of the form ( x) = .x - !(x) l-.lhere is completely continuous, and ~(x) 'I a for x e G, then 'Ne define
lex)
deg (4,G,a)
= deg
(¢~,GJa)
where 4l *: B.....;;:. B :i.s an admissible extension of <1>, i.e. ~ * is admissi ble and ¢-1.. = ¢ on G. Defini~:
Suppose that ¢ is admissible and that deg (¢,Br(xo),a) is def'ined. Thea, if' there is a positive number e _< r such thr,lt ~(x) I- a for' 0 < IIx - x I < e, we define
o -
(That the index is well defined follows f!;om Lemmas 4, 6, 7.) From the constz'uc.tJ.on of the index and the degree in a Banach space, it is n~t difficult to see that all the properties of the index and degree in fln carry over. We are, however, unable to conclude that a closed set, Fc B, partitions B-F into a denu,:Tlereble number of components. In proving the Jordan ... _.. .,Brouwer Theorem and the theorem on Invariance of domain we used
(.6.
matrices (U ij ), «(3j1) where i,j indexed the sets {Dil' j} respectively. To use any theorems about matrices we had to assume that both sets were countable. We shall get around this by using a more general theorem on abstract groups. Let {Cl iJ }
254 be the (non-denumerable) set of integers such that d.eg «LDi,A j ) = ?'i" Similarly \13 ke } is the set of integers such that deg ($ -.L '~k,DJ.') = 13 k .(. Property (j) still holds since only a fhltte number of non-zero terms wlll enter in the sum. As before, we have ¢-l¢(D i ) = Di , etc. Thinking of (aij ) as the lima trixU of a tr9.ns forma.t lOll in a non-separable Hilbert space, we find (exactly as for R ) that (a i .) has both a. right n J inverse and a left inverse, namely (~k~). Thus the transfonaation is both 1-1 (because of the right inverse) and onto (left inverse) • This shows that the sets {Di5' have the s arne j cardinal n.umber, whtcb is what we want. (See Birkhoff and J.VlacLane; ~~~1f~ of M~~...!].c.££2;; 1953 edition; pa.ge 121) :r'he remarks in the pr·evious pal>agraph and the lemma that follows, will allow us to state the Jordan Brouwer Theorem.
f6 1
Let $: B ~ B be admissible. ¢(F) is closed. Lem~a:
Proof:
Same as for Pn •
Then if F' c B is closed,
(page 230)
Theorem: (Jord.a.n-Drouwer-Lera.y) of B, and let 4: iill ont.~ F2 be a pings X - 4(x) and x - ~-l(x) map compact sets, the 1) B - .f\ and B components.
Let F l , F2 'be closed subsets homeomorphism. If the mapFl and F2 respectively, into F2 have the same number of
Proof: SBme as for Rn' The requirements on x - ¢(x) and 4 - d-l(x) are to eaable us to t.alk about deg (Q,D i ,.6. j ), deg (4-1'~k,Dl)' The Schauder Theorem on Invariance of Domain may also be generalized to Banach spaces. Theorem: Let G be an open bounded set in B, 4: G ~ B a 1-1 continuous mapping with 4(x) ~ x - !(x), !(x) being completely continuous. Then $ is a homeomorphism and 4(G) is open in B.
t!..2.2f:
The proof is almos t like the one for Rn' There we used the fact that a l~l continuous map of a compact set 1s a homeomorphism. The oompact set in question was a closed ball. Here
255 we must be careful. Although tt is true that the closed ball in R is compact, it is not trt1.e that the closed ball n( } Bl = x I II x II < 1 in a Banach space is compact. Allowing our ... selves a slight dlgression in this direction, we shall offer a counter example, i. e. a closed bal. 1 in a Banach space which is ~.compact: Let B be the separable Hilbert·space. In this case
t
Bl
= {(~;""Xn"") I ~
xi
~
l}.
A sequence of points in B1 ,
from which we .£§E;}10:t choose a convergent subsequence is xl :::; ll, 0,0, • .
·1,
=
X2
to , 1,0, . • .},
•.. xi :::;
r6il , 6 i2 , ••• , 6i
"'"' j.
~r
.
Hence for this case we need the following Lemms.: Suppose that ¢: P ~ B, where F is closed and bounded in B, ¢ is 1-1 and continuous, and x - ~(x) is completely continuous. Then ~ is a homeomorphism of F onto 4( F). Proof: Since I(x) :::; x - ~(x) is completely continuous !(F) is contained in a compact subset S of B. Let us .assume that the inverse mapping
is not continuous.
Then there is a point y
{yn} in ~(F) such that
o
e: ¢( F), and sequence
llYn - :To H -> 0 We may select a subsequence if necessary (also denoted by {Yn) for convenience) such that 1I¢-lYn - 4- 1 Yo ll > E > 0 for all n. Let ~-l y
Then, since
¢ is
0
4- 1v"n
= x0 ,
1 ... 1 and x o ' xn e
:::; xn
F, we have
Yo
:::;
~(xo)
Yn
:::;
~(xn) = xn -
:::;
Xo - (f(xo )
256
a
Because the points W(x ) lie in compact set S, there is a n subsequence W( x ) \. and point z e S such that
rt-
nj ~
II -W( xn· ) - z II ...,.,.
0
J
Then x UJ-1y
n
nj
:=
~(x
- ~-ly
0
nj
) + Qi(x n .) ~ y + z = J 0
II
>
~ is continuo-us:
e, Il~
:>
0
-
x
0
II
>
-
e.
x
"IAlhere, since
0
But, F is closed
)
x
.. F ..
C'-'
~ CX o ) =l:tm'~ {x -t--= 11m y = Yo = ~ (Xo ) nj~ nj nj
x
Thus we have ¢(S{ ) = ~(x ), but =I x. This can't happen o 0 0 0 since ¢ is assumed to be 1-1. 'T'his proves the lemma. The rest of' the proof' of' the ~ain theorem is exactly the same as f'or Rn'
'.
n·
" &7 51 A
P &4
8;
