TIFR pamphlet on Algebraic Topology

8/11/2019 TIFR pamphlet on Algebraic Topology

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All communications pertaining to this series should be addressed to:

School of Mathematics

Tata Institute of Fundamental Research

Bombay 5, India.

c

Edited and published by K. Chandrasekharan for the Tata Institute of Fundamen-

tal Research, Bombay, and printed by S. Ramu at the Commercial Printing Press

Limited, 3438 Bank Street, Fort, Bombay, India.


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Editorial Note

This series of mathematical pamphlets is issued in response to awidespread demand from university teachers and research students inIndia who want to acquire a knowledge of some of those branches ofmathematics which are not a part of the curricula for ordinary universitydegrees. While some of these pamphlets are based on lectures givenby members of the Tata Institute of Fundamental Research at summerschools organized by the Institute, in cooperation with University ofBombay and the University Grants Commission, it is not the intention

to restrict the series to such lectures. Pamphlets will be issued fromtime to time which are of interest to students.

K. Chandrasekharan


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PREFACE

This pamphlet contains the notes of lectures given at a SummerSchool on Algebraic Topology at the Tata Institute of FundamentalResearch in 1962. The audience consisted of teachers and studentsfrom Indian Universities who desired to have a general knowledge ofthe subject, without necessarily having the intention of specializing it.The speakers were M.S. Narasimhan, S. Ramanan, R. Sridharan and K.Varadarajan.

Chapter I introduces the elements of Set Theory. Chapter II deals

with abelian groups and proves the structure theorem for finitely gener-ated abelian groups. Chapter III covers some topological preliminaries.Singular homology groups are defined and their invariance under ho-motopy is proved in Chapter IV. The homology of simplical complexesis treated in Chapter V, where proofs of the Euler-Poincare formulaand Brouwers fixed point thoerem are given (assuming the invariancetheorem for simplical homology).


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Contents

1 Set-theoretic Preliminaries 1

1.1 Sets and maps . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 Equivalence relations . . . . . . . . . . . . . . . . . 3

2 Abelian Groups 5

2.1 Groups and homomorphisms . . . . . . . . . . . . . . . . 52.1.1 Finitely generated abelian groups . . . . . . . . . . 8

3 Set Topology 15

3.1 Topological spaces . . . . . . . . . . . . . . . . . . . . . . 153.1.1 Compact spaces . . . . . . . . . . . . . . . . . . . 183.1.2 Connected spaces . . . . . . . . . . . . . . . . . . . 193.1.3 Continuous maps . . . . . . . . . . . . . . . . . . . 193.1.4 Some applications to real-valued functions . . . . . 223.1.5 Product spaces . . . . . . . . . . . . . . . . . . . . 233.1.6 Quotient spaces . . . . . . . . . . . . . . . . . . . . 233.1.7 Homotopy of maps . . . . . . . . . . . . . . . . . . 24

4 Singular Homology 27

4.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.1.1 Singular homology groups . . . . . . . . . . . . . . 284.1.2 Effect of a continuous map on the homology groups 314.1.3 Homomorphisms induced by homotopic maps . . . 32

5 Simplicial Complexes 375.1 Simplicial decomposition . . . . . . . . . . . . . . . . . . . 37

5.1.1 Homology of a simplicial decomposition . . . . . . 385.1.2 The Euler-Poincare Characteristic . . . . . . . . . 395.1.3 Homology of the sphere . . . . . . . . . . . . . . . 405.1.4 Brouwers fixed point theorem . . . . . . . . . . . 41

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Chapter 1

Set-theoretic Preliminaries

1.1 Sets and maps

We shall adopt the point of view of naive set theory. Asetis a collectionof objects which are called elements or pointsof the set. The set of allrational integers (i.e. integers positive, negative and zero) is denotedby Z, the set of all non-negative integers by Z+, the set of all rationalnumbers by Q, the set of all real numbers by R, and the set of allcomplex numbers by C.

Ifx is an element of a set A, we write x A. Ifxis not an element

ofA, we write x /A. (Thus x R will mean that xis a real number).IfPis a property, the set of all objects with propertyPwill be denotedby{x| x satisfies P}. Thus{x| x Z, x


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1.1. Sets and maps 3

X X which associates to each x X the element x itself is called

the identity map ofX, denoted by IX. Let f: X Y be a map. It iseasy to see that fis one-one and onto if and only if there exists a mapg: Y X such that g f =IX and f g = IY. If such a map g exists,it is unique and we shall denote it by f1; f1 is called the inverse off. IfA is a subset ofX, the map j: A X, which associates to eacha A the same element a in X, is called the inclusionmap ofA in X.Iff: XY is any map, the map fj: A Y is called the restrictionoff to A and is often denoted f|A.

1.1.1 Equivalence relations

Definition 1.1 Any subset R X X is said to be a relation in X.We write x R y if (x, y) R. An equivalence relation in a setX is arelation R in Xsuch that the following conditions are satisfied.

(i) For every x X, (x, x) R.

(ii) If (x, y) R then (y, x) R.

(iii) If (x, y) R and (y, z) R then (x, z) R.

We say x is equivalent to y under R, if xRy i.e. (x, y) R. Theabove conditions simply require that

(i) every element is equivalent to itself ( reflexivity),

(ii) ifx is equivalent to y , then y is equivalent to x ( symmetry),

(iii) ifx is equivalent toy, andy is equivalent toz, thenx is equivalentto z (transitivity).

Example 1.2 The subsetR X Xconsisting of elements (x, x), xX, is an equivalence relation in X. This is called the identity relationin X.

Example 1.3 R= X Xis an equivalence relation in X, in which allelements are equivalent to one another.

Example 1.4 Let q Z. Consider the subset of Z Z consisting ofpairs (m, n) of integers such that m n is divisible by q. This is anequivalence relation under which two integers are equivalent if and onlyif they are congruent modulo q.


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4 Chapter 1. Set-theoretic Preliminaries

Example 1.5 Consider the subset {(x, y)| (x, y) R2, x y} ofR2.

This satisfies (i) and (iii) but not (ii) and is therefore notan equivalencerelation.

Example 1.6 Iff: XYis a map, the subset RfX Xconsistingof (x1, x2) such that f(x1) =f(x2) is an equivalence relation.

Let now x X and R an equivalence relation in X. The set of allelements of X equivalent to x under R is called an equivalence classx. Consider the family of distinct equivalence classes of X under R.It is easily verified that they are pairwise disjoint and that their unionis X. We shall define the quotient X/R of X by R as the set whose

elements are these equivalence classes. The natural map : X X/Rwhich associates to each x X, the equivalence class x which containsx, is clearly onto. In example 1.4 above, the residue classes are the usualcongruence classes modulo q and we denote the quotient set by Z/(q).Finally, letf: XYbe a map and Rf the equivalence relation definedby f. We define a map qf: X/Rf Y by setting qf(x) = f(x). Bydefinition ofRf, qf is well defined. Clearly qf is one-one. Moreover wehave qf = f. We have therefore proved the

Theorem 1.7 Let f: X Y be a map. Then there exists an equiva-lence relationRf onXand a one-one onto map qf: X/Rff(X)such

that f = j qf where j is the inclusion f(X) Y and is thenatural map XX/Rf.


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Chapter 2

Abelian Groups

2.1 Groups and homomorphisms

Definition 2.1 An abelian group (G, ) is a non-empty set G togetherwith a mapping: GG G,(x, y) being denoted byx +y, satisfyingthe following conditions.

(a) x+ (y+z) = (x+y) +z, for every x, y ,z in G(associativity).

(b) There exists an element 0, called the zero element of G, whichsatisfies 0 +x= x+ 0 =x for every x G.

(c) For everyx G, there exists an elementxcalled the negative ofx such that x + (x) = (x) +x= 0.

(d) x+y= y +x for every x, y in G (commutativity).

Remark 2.2 (i) We often writeGfor a group instead of (G, ), whenit is clear from the context to which map we are referring.

(ii) The map is called the addition in G.

(iii) The zero element is unique. In fact, if there is an element 0 G

such that (b) is valid for every x G with 0 replaced by 0

wehave, in particular, 0 = 0 + 0 = 0.

(iv) The negative of any element is unique (proof is easy).

(v) A group consisting of only a finite number of elements will becalled a finite group. The number of elements in a finite group

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6 Chapter 2. Abelian Groups

is called its order. A group which is not finite will be called an

infinite group.

(vi) In view of (a), for x, y ,z G, the element x+ y + z G isdefined without parenthesis. More generally, for xi G, i =1, 2, . . . , n ,

1in xi can be defined without ambiguity.

(vii) In what follows, for convenience, we sometimes drop the wordabelian and refer to abelian groups simply as groups.

Example 2.3 The set Z(resp. Q, resp. R, resp. C) of integers (resp.rationals, resp. reals, resp. complex numbers) with the usual operationof addition is an abelian group.

Example 2.4 LetG = Z/(m) where m is any integer 0. Setk+ l=k+l. It is easy to check that this defines an operation which satisfiesour axioms. Gbecomes thus an abelian group and is finite ifm >0.

Example 2.5 The non-zero real numbers denoted by R (resp. thenon-zero rational numbers denoted by Q, resp. the non-zero complexnumbers denoted by C) become an abelian group under the usual op-eration of multiplication.

Definition 2.6 LetG, G be two groups. A homomorphism f: G G

is a map such that f(x+y) =f(x) +f(y) for every x, y G.

Letf: G G be a homomorphism. Then f(0) = 0. In fact, f(0) =f(0 + 0) =f(0) +f(0) Adding f(0) to both sides we get 0 =f(0). Iff: G G and g: G G are homomorphisms, then g f: G G isalso a homomorphism. For any group G, the identity map IG: G Gis a homomorphism.

Definition 2.7 A homomorphism f: G G is called an isomorphismif there exists a homomorphism g: G G such that fg = IG (theidentity map ofG) and g f=IG (identity map ofG).

It is easily seen that a homomorphism f: G G is an isomorphismif and only if it is both one-one and onto.

Remark 2.8 IfG, G are groups such that there exists an isomorphismof G onto G then G and G are said to be isomorphic. If G and G

are isomorphic, we sometimes write G G. In some sense, isomorphicgroups have indistinguishable structures.


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2.1. Groups and homomorphisms 7

Example 2.9 The natural map : Z Z/(m) is an onto homomor-

phism. Ifm = 0, it is not one-one and hence not an isomorphism.

Example 2.10 The map Z Z given by x 2x is a one-one homo-morphism. It is not onto and hence not an isomorphism.

Example 2.11 The map Q Q given by x 2x is an isomorphism.

LetGbe a group. A non-empty subsetHofGis called asubgroupofGif for every x, y H, we have x y H. In particular 0 Hand foranyx H, we have x H. It can be easily checked that H, with theaddition induced by that of G, is a group with 0 as the zero-element

and the negative ofx Hbeingx.Let Hbe a subgroup ofG. Then the inclusion map j: H G is aone-one homomorphism.

For any group G, Gitself and the subset {0} are subgroups.

Example 2.12 LetH1, H2 be two groups of a group G. ThenH1 H2is a subgroup ofG. Also, the set of elements of the form h1+h2, h1H1, h2 H2, is a subgroup, H1 + H2, of G. If H1 H2 = (0) andH1+H2 = G, we say that G is the direct sum ofH1 and H2 and writeG= H1 H2.

Example 2.13 Z is a subgroup of Q, Q a subgroup of R and R asubgroup ofC.

Example 2.14 For m 0, the subset mZ = {m n | n Z} is asubgroup ofZ.

Proposition 2.15 The only subgroups ofZ aremZ, m Z+.

Proof: Let Hbe a subgroup of Z. IfH = 0, m can be taken to be0. IfH= (0) letn H, n= 0. Sincen H, we may assume withoutloss of generality that n > 0. Let m be the smallest strictly positiveinteger contained inH. (This exists since every non-empty subset of the

set of natural numbers has a least element.) We now assertH = mZ.In fact, let h H. Then h= m q+r with 0 r < m. Since m H,we havem q Hand hencer = h m q H. Sincer H andr < mwe must have r = 0, which proves our assertion.

Let f: G G be a homomorphism of groups. Clearly, the imagef(G) ofGby f is a subgroup ofG.


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Definition 2.16 The subset {x | x G, f(x) = 0 G} is called the

kernel off, usually denoted ker f.

It is easily seen that the kernel of any homomorphism f: G G isa subgroup ofG.

LetG be an abelian group and Ha subgroup. The relation RH=Rdefined in G by x R y if and only if x y H is easily seen to bean equivalence relation. The quotient set G/R is usually denoted byG/H. We now define the structure of a group on G/H as follows. Forx,y G/H, set x+ y = x +y. It is easily seen that the element thusdefined depends only on the classes x and y and not on the particularrepresentativesx and y. It can be checked that G/H becomes a group

under this operation as addition. It is called the quotient group ofG byH. The natural map: G G/H is clearly a homomorphism which isonto.

Let f: G G be a homomorphism. Then the equivalence relationRf defined by f (Example 1.6, Chapter 1) is easily seen to be Rkerf.The induced map qf: G/Rf G

(defined by qf(x) =f(x)) is triviallychecked to be a homomorphism. fthen admits the following decompo-sition into homomorphisms

G G/Rf

qff(G)

jG

where is the natural map, qfis an isomorphism and j is the inclusion

off(G) in G.

Remark 2.17 (i) The above result is sometimes called the Funda-mental theorem of homomorphisms.

(ii) Letf: G G be a homomorphism of abelian groups. LetH bea subgroup ofG and H a subgroup ofG such that f(H) H.There exists then a homomorphism f: G/H G/H such thatf = f.

2.1.1 Finitely generated abelian groups

Definition 2.18 A torsion element of an abelian groupA is an elementa G such that a+ + a (n times), denoted by na, is 0 for someintegern >0.

Remark 2.19 The element 0 is always a torsion element of an abeliangroup. A non-zero torsion element may not exist in an abelian group.


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For instance, the group Q of rational numbers does not possess any non-

zero torsion element. On the other hand, every element ofZ/(n) (n >0)is a torsion element.

Ifa, b are torsion elements, then a b is also a torsion element. Infact, if ma = 0, nb = 0 with m = 0, n = 0 then m n(a b) = 0,and m n= 0. The torsion elements of an abelian group A thus form asubgroup, called the torsion subgroup ofA denoted by T(A).

Definition 2.20 An abelian group A for which T(A) = (0) is calledtorsion-free. A group A for which T(A) =A is called a torsion group.

Proposition 2.21 For any abelian group A, A/T(A) is torsion-free.In fact, let a A/T(A) be such thatna= 0 for somen Z+, n= 0.

Then we have na T(A). Therefore there exists an m Z+, m = 0such that m(na) = 0. Since m n = 0, it follows that a T(A) i.e.a= 0.

Definition 2.22 LetA be an abelian group and B a subgroup ofA. AsubsetSofB is said to be a set of generators for B if every element ofB can be written as

1in miai, mi Z, ai S. An abelian group

Ais said to be finitely generated if there exists a finite set of generatorsfor A.

Remark 2.23 Any quotient group of a finitely generated abelian groupA is finitely generated. In fact, if a1, . . . , an is a set of generators forA, their images under the natural map form a set of generators for thequotient.

Remark 2.24 Every finitely generated torsion group is finite. In fact,let a1, . . . , an generate a torsion group A. If mi is the least positiveinteger such that miai = 0, 1 i n then clearly every element ofAcan be written as

1in liai, with 0 li mi. Hence A is finite of

order 1in mi.Example 2.25 (Z, +) is generated by 1 (or 1).

Example 2.26 (Z/(n), +) is generated by 1 (remark above).

Example 2.27 (Q, +) is not finitely generated.


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Theorem 2.28 Every subgroup of a finitely generated abelian group is

finitely generated.

Proof: The proof is by induction on the number n of generators ofthe given group. The theorem being trivially true forn = 0, we shallassume the theorem for all groups which can be generated by less thann elements. Let now A be a group generated by e1, . . . , en and let Bbe a subgroup of A. The set {k | k Z, such that there exists b Bof the form

1in1 kiei + ken} is obviously a subgroup of Z and is

hence of the form l Z for some l Z (see Proposition 2.15). Letb0 B such that b0 =

1in1 kiei +len. The intersection of the

subgroup generated by e1, . . . , en1 with B, which we denote by B is

finitely generated, by induction hypothesis. Let f1, . . . , f q be a set ofgenerators forB . We assert that f1, . . . , f q, b0 generate B. In fact, letb =

1in miei B. By our choice of l, mn = p l for some p Z.Hencebpb0 =

1in1(mi pki)ei B

. Thusbpb0=

1iqtififor ti Z or, what is the same b = pb0+

1iqtifi. This proves our

assertion.

Remark 2.29 The above proof actually yields the following strongerassertion. If an abelian group is generated bynelements, every subgroupcan also be generated by n elements.

Definition 2.30 A seta1, . . . , anof elements of an abelian group is saidto be free if

1in miai = 0, mi Z, implies mi = 0 for every i with

1 i n. A subset S of an abelian group is said to be free if everyfinite subset ofS is free.

Definition 2.31 An abelian group A which admits of a free set S ofgenerators is said to be free. S is then called a base for A.

Remark 2.32 By a maximal free setSofAwe mean a free setSwhichis not a proper subset of any other free set. It is clear that a base of afree abelian group is a maximal free set.

Remark 2.33 It is easily seen that a finitely generated free abeliangroup admits of a finite base.

Remark 2.34 Let S be any set. Consider the set F(S) of all formallinear combinations

sSnss, ns Z, where ns= 0 for all but a finite


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number ofs S. We define addition in this set by

nss+ mss= (ns+ms)s.

Under this operation F(S) becomes an abelian group. Every s Swill be identified with the element ntt where nt = 0 for t = s andns = 1. With this identification, S clearly is a base for F(S). F(S)will be called the free abelian group generated by S. Let S, S be twosets and f: S S any map. It is easily seen that there exists a uniquehomomorphism F(f): F(S) F(S) such that F(f)/S= f.

Remark 2.35 There exist abelian groups which are not free. For in-

stance, the group Z/(n) is not free ifn = 0.

Remark 2.36 The group Z is free with 1 as base.

Remark 2.37 Every free abelian group is torsion-free.

Theorem 2.38 If an abelian group has a base consisting ofn elements,then every other base also consists ofn elements.

Proof: This theorem is a consequence of the following lemmas andRemark 2.32 above.

Lemma 2.39 LetAbe a finitely generated abelian group andSa max-imal free set in A. If B is the subgroup generated by S, then A/B is

finite group.

Proof: Since A is finitely generated, so is A/B. We shall now provethat A/B is a torsion group. Let a A/B with a A as its represen-tative. SinceS is a maximal free set, {a}

Sis not free and there is a

nontrivial relation na+ misi = 0, n, mi Z, si S. Since S is free,n= 0 and we have na B . This proves that na= 0. By Remark 2.24following Proposition 2, A/B is finite.

Lemma 2.40 LetSbe a maximal free set of a finitely generated abeliangroup A and B the subgroup generated byS. For every primep whichdoes not divide the order of A/B, the group A/pA is finite if and onlyifS is finite. Moreover, ifS consists ofn elements, the order ofA/pAispn.


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Proof: Let: A A/B,: A A/pA be the natural maps and let q

denote the order ofA/B. Then, for anya A,qa B . Letl, m Z suchthat lp+mq= 1. Then mqa= a pla. That is (a) =(mqa) (B)and hence(S) generatesA/pA. Moreover,

mi(si) =

ni(si), si

S implies that mi ni(p). For

(mi ni)(si) = 0 implies

(mini)si = pa for some a A. As

(mi ni)si B, (pa) = p(a) = 0

and sincep does not divideq, (a) = 0, or what is the same, a =

lj sj .Since S is a free set, this proves that mi ni = pli. In other words,every element in A/pA can be written uniquely in the form

ni(si),

with 0 ni< p, si S. This proves that A/pAis infinite ifSis infiniteand if S is finite, consisting of n elements, then A/pA consists of pn

elements.

Lemma 2.41 LetSbe a maximal free set withn elements in a finitelygenerated abelian group. Then any other maximal free setS also hasnelements.

Proof: We denote by B, B the subgroups of A generated by S, S

respectively. Choose a prime p not dividing the orders of A/B andA/B. By Lemma 2.40, A/pA has pn elements which implies that S isfinite and that it has n elements.

Definition 2.42 The number ofn of the above theorem will be called

the rank of free abelian group.

Theorem 2.43 A finitely generated torsion-free abelian group is free(i.e. it has a base consisting of a finite number of elements).

Proof: We proceed by induction on the number n of generators. Ifn= 0 there is nothing to prove. Let us assume the theorem to be truefor all groups which can be generated by n 1 elements. Let nowA be atorsion-free group with generatorsa1, . . . an. LetB denote the subgroupgenerated by a1, . . . an1.Case i. Assume that there exists an integer m= 0 such that manB .The map : A A defined by (a) = ma is a homomorphism. It is

one-one sinceA is torsion-free. Moreover,(A) B by our assumption.ThusA is isomorphic to(A) and hence by the remark following Theo-rem 2.28, can be generated byn1 elements. The induction assumptionnow yields that A is free.

Case ii. Assume thatman B implies m= 0. B is torsion-free and isgenerated by n 1 elements. By induction assumption, B is free. Let


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b1, . . . , bk be a base for B . We assert that b1, . . . , bk, an is a base for A.

In fact, evidently they form a set of generators. If libi+lan = 0 thenlan B and this by our assumption implies that l = 0. Hence libi = 0which yields li = 0.

Remark 2.44 The above theorem is false if the group is not finitelygenerated. For instance, (Q, +) is torsion-free, but not free.

Definition 2.45 LetA be a finitely generated abelian group. The rankofAis by definition the rank of the (torsion-free and hence) free groupA/T(A).

If a1, . . . , an is a base ofA/T(A) with a1, . . . , an as representatives

in A, then a1, . . . , an clearly form a free set. Moreover, if a A thena = miai, mi Z. This means that ka kmiai = 0 for somek Z, k = 0. Thus (a1, . . . , an) is a maximal free set. In view of theabove lemmas, this means that the rank of a finitely generated abeliangroup is the same as the number of elements in any maximal free set.

Remark 2.46 The above considerations show that the subgroup Bgen-erated by a1, . . . , an is a free subgroup ofA such that every element ofA can be uniquely written as b+x, b B, x T(A). In other words,A is the direct sum of a free group and a finite group.

Proposition 2.47 LetA be a finitely generated abelian group andB bea subgroup. Then

rankA = rankB + rankA/B.

Proof: Since B is a subgroup of a finitely generated group and A/Ba quotient , it is clear, in view of our earlier results, that B and A/Bhave finite ranks. Let rank B = l and rank A/B = l. Then thereexist maximal free sets b1, . . . , bl in B and a1, . . . , al in A/B. Letai A represent ai A/B, 1 i l

. We claim that the set S ={a1, . . . , al , b1, . . . , bl} is a maximal free set. Suppose that

1ilniai+

1jlmj bj = 0. We then have 1jlniai = 0 and this means thatni = 0. This means that

1jlmj bj = 0, which implies mj = 0, 1

jl. This shows thatSis a free set. On the other hand, ifa A, thereexists an integer n = 0 such that na =

1jlniai. This means that

na

1ilniai B . Hence there exists an integerm= 0 such thatm(na

1ilniai) =

1jlmj bj , which proves that S is maximal.

By what we have said above, it now follows that rank A = l +l.


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Chapter 3

Set Topology

3.1 Topological spaces

A topological space is a set Xtogether with a collection Tof subsets ofX (called open sets) with the following properties.

(i) The empty set and Xare in T.

(ii) Any finiteintersection of sets in Tis again in T.

(iii) Anyunion of sets in Tis again in T.

Example 3.1 Let X be a set. Let T consist only ofX and . This iscalled the trivial topology on X.

Example 3.2 Let X be a set. Let T consist of all subsets ofX. Thistopology is called the discrete topology on X.

Example 3.3 Metric space. A metric space is a set Xtogether with afunction d (called the metric or the distance) from X X to the non-negative real numbers such that the following conditions are satisfied:

(a) d(x, y) = 0 if and only ifx = y, where x, y X;(b) d(x, y) =d(y, x) for x, y X (symmetry)

(c)d(x, z) d(x, y) + d(y, z), for everyx, y ,z X (triangle inequal-ity).

Let x X and r a positive real number. By the open ball aroundx of radius r we mean the set {y | y X, d(x, y) < r}. We define atopology on the metric space by defining the open sets to be all setswhich are unions of open balls (together with the empty set).

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16 Chapter 3. Set Topology

Example 3.4 LetR be the set of real numbers. Rhas a natural metric

defined by d(x, y) =|x y|. This metric defines a topology on R.

Example 3.5 LetRn be then-fold cartesian product ofR, that is, theset ofn-tuples of real numbers (x1, . . . , xn). R

n is a metric space with the

metric defined by, d(x, y) =x y= {(x1 y1)2 + + (xn yn)2}1

2

where x = (x1, . . . , xn) and y = (y1, . . . , yn). Thus Rn has a natural

topology. Whenever we refer to Rn as a topological space, we assumethat it is provided with this topology.

Example 3.6 Induced topology. LetXbe a topological space and A X. We define a topology on A by defining the collection of open sets to

be sets of the formU AwhereUruns through all open sets ofX. Thistopology is called the induced topology on A.

Let Xbe a topological space. A subset ofX is said to be closed ifits complement is open.

The axioms for open sets are equivalent to the following.

(i) The empty set and Xare closed sets.

(ii) The union of a finite number of closed sets is closed.

(iii) Any intersection of closed sets is closed.

Example 3.7 In R, the set {x| x R, a x b} a, b R, a b, isclosed.

Definition 3.8 Let Xbe a topological space and x X. By a neigh-bourhood ofx we mean an open set containing x.

Example 3.9 Let X = R. The set{x | x R, |x| < } is a neigh-bourhood of the origin 0, if > 0.

Definition 3.10 Let A be a subset of a topological space X. A point

x X is said to be a closure point of A if every neighbourhood of xcontains a point ofA. The set of closure points ofA is called the closureofA, and is denoted by A.

Remark 3.11 A is the intersection of all closed sets containing A. Inparticular, the closure of a closed set is itself.


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3.1.1 Compact spaces

Definition 3.23 A family{U}Iof subsets of a setXis said to be acovering ofX if

IU= X, i.e. if each point ofXbelongs to at least

one U. If, further, Xis a topological space and each U is an open setofX, we say that{U}I is an open covering ofX.

Definition 3.24 A topological space X is said to be compact if thefollowing condition is satisfied: if{U}I is any open covering of X,then some finite sub-collection of{U}Iis already a covering; that is,if{U}I is any open covering ofX, then there exist a finite numberof elements 1, . . . , n I such that 1in Ui =X.Definition 3.25 A subsetAof a topological space is said to be compactif it is compact in the induced topology.

Remark 3.26 (i) The spaceRis not compact.

(ii) Let X be a topological space. A subset A of X consisting of afinite number of points is compact.

(iii) A closed subset of a compact space is compact.

Proposition 3.27 A compact subset of a Hausdorff space is closed.

Proof: LetXbe a Hausdorff space andAa compact subset ofX. Wehave to show that X A is open. For this, it is sufficient to prove thatevery point x X A has a neighbourhood which does not intersectA. Let x XA. If y A, since X is Hausdorff, we can find aneighbourhood Vy of x and a neighbourhood Uy of y in X such thatVy Uy =. If,U

y = Uy A,then{U

y}yA is an open covering ofA and

sinceAis compact, we can find y1, . . . , yn Asuch that

1in Uyi

=A.Then V =

1in Vyi is a neighbourhood ofx which does not intersect

A.

Remark 3.28 Since the space R is Hausdorff, it follows that everycompact subset of R is closed. On the other hand, it is easy to seethat every compact subset of R is bounded. Conversely, every closedbounded set of R can be proved to be compact. This result is usuallyknown as the Heine-Borel Theorem.


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Remark 3.32 We can easily prove thatf: XYis continuous if and

only if the inverse image by fof every open (resp. closed) set in Y is anopen (resp. closed) set ofX.

Remark 3.33 Letf: XY, g: Y Zbe continuous maps. Then thecomposite (g f): XZ is continuous.

Remark 3.34 The identity map IX: XX is continuous.

Example 3.35 Let f: R R be a map (i.e., fis a real-valued func-tion of a real variable). Let x0 R. Then f is continuous at x0 if andonly if for every > 0, there exists a > 0 such that for each x with

|x x0|< we have |f(x) f(x0)| < .

Example 3.36 Let X be a discrete space and Y a topological space.Then any map f: XY is continuous.

Example 3.37 Let Xbe a topological space and Y a set with trivialtopology. Then any map f: XY is continuous.

Definition 3.38 LetXandYbe two topological spaces. A continuousmap f: XY is said to be a homeomorphism if there exists a contin-uous map g : Y Xsuch that g f =IX, f g =Iy. Two topological

spaces are said to be homeomorphic if there exists a homeomorphism fofX onto Y.

In a sense, homeomorphic spaces have indistinguishable structures.

Remark 3.39 A one-one continuous map fof a topological space ontoanother is not necessarily a homeomorphism. This is because the settheoretic inverse off is not in general continuous. For instance, letXbe any set with more than one element. LetTbe the discrete topology onXand T the trivial topology onX. Then the identity mapIX: (X, T)(X, T) is continuous, but is not a homeomorphism.

Remark 3.40 Properties of topological spaces like compactness, con-nectedness etc. are preserved under homeomorphisms. Such proper-ties are said to be topologically invariant.

Example 3.41 LetX=Y =R. The mapf: XYdefined byf(x) =x3 is a homeomorphism. The mapf(x) =xis also a homeomorphism.


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3.1. Topological spaces 21

Example 3.42 The spaceR and the open subset{x| 1< x 0, are not homeomorphic. The parabola and the hyperbola{(x, y)| (x, y) R2, x2y2 = 1} are not homeomorphic. (The parabolais connected whereas the hyperbola is not.)

Example 3.44 Let X be the complement of the origin in R2 (withthe induced topology). Let Y be the cylinder in R3 defined by Y ={(x,y ,z)| (x,y ,z) R3, x2+y2 = 1, z >0}with the induced topology.X and Yare homeomorphic (Use polar coordinates in R2 (0)).

Proposition 3.45 LetXbe a compact space, Ya topological space andf: X Y a continuous map. Then f(X) is compact. (That is, acontinuous image of a compact space is compact.)

Proof: Let {U}Ibe a family of open sets off(X) such that

IU=f(X). Since clearly f: Xf(X) is continuous, {f1(U)} forms anopen covering ofX. SinceXis compact, there exist1, . . . , n Isuchthat

1in f

1(Ui) =X. Then it is clear that

1in Ui =f(X).

Corollary 3.46 LetXbe compact, Y Hausdorff andf: XY contin-uous. Thenf(X) is closed inY.

Corollary 3.47 Let X be a compact space and Y a Hausdorff space.Letf: XY be a one-one continuous onto map. Thenf is a homeo-

morphism.Contrast the above with the Remark 3.39 following the definition of

homeomorphism.

Proposition 3.48 LetXbe a connected topological space, Y a topolog-ical space andf: XY a continuous map. Thenf(X) is connected.


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Proof: Iff(X) =A were not connected, there would exist non-empty

open setsU1, U2 off(X) such thatf(X) =U1 U2, U1 U2 = . Sinceclearly f: X f(X) is continuous, f1(Ui), i = 1, 2 are open (non-empty) sets ofXwithX=f1(U1) f1(U2) andf1(U1) f1(U2) =. This contradicts the connectedness ofX.

3.1.4 Some applications to real-valued functions

LetXbe a set. A map fromXto the real numbers is called a real-valuedfunction on X.

Proposition 3.49 Let X be a compact topological space and f a con-

tinuous real-valued function onX. Then we have the following two prop-erties.

1. The function f is bounded on X i.e. there exists a positive realnumber a such that |f(x)| a for everyx X.

2. The functionfattains its least upper bound (denoted by sup) andits greatest lower bound (denoted by inf ) on X, i.e. if M =supxXf(x)andm = infxXf(x), then there exist elementsx1, x2inX such thatf(x1) =M, f(x2) =m.

Proof: The result follows from the fact that f(X) is compact andthat a compact subset ofRis bounded.

Since f(X) is compact it is a closed and bounded set. Hence thegreatest lower bound and the least upper bound off(X) belong tof(X).

Proposition 3.50 LetXbe a connected topological space andfa con-tinuous real-valued function onX. Thenfattains every value betweenany two of its values, i.e. if f(x1) = a1 and f(x2) = a2, a2 a1 anda any real number such that a2 a a1, then there exists an elementy X such thatf(y) =a.

Proof: Since f is continuous, f(X) is connected and hence is an in-terval. Since a1 and a2 belong to f(X), a f(X).

Remark 3.51 If X is an interval in R, the above property reducesto a well-known property of continuous real-valued functions of a realvariable.


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3.1.5 Product spaces

LetX andYbe two topological spaces. Consider the cartesian productX Y. We define a topology on X Y, called the product topology, asfollows: an open set in X Ywill be, by definition, a union of sets ofthe form U V where U is an open set in X and V an open set in Y.(Verify that these sets satisfy the axioms for open sets).

Ifp1: X Y Xand p2: X Y Y are theprojections, defined byp1(x, y) =x and p2(x, y) =y respectively, thenp1and p2 are continuousmaps, by the definition of topology on X Y.

In a similar way, the product of a finite number of topological spacescan be defined.

Remark 3.52 The metric topology on Rn is the product topology onthe n-fold product of the topological space R.

Problem 3.1 The product of two topological spaces is compact if eachcomponent is compact.

3.1.6 Quotient spaces

Let Xbe a topological space and R an equivalence relation in X. LetY =X/R be the quotient set and : XY the natural map. We puton Y the following topology: a set in Y is open if and only if its inverse

image by is open in X. This topology on Y is called the quotienttopology on X/R. X/R endowed with this topology will be referred toas thequotient space. The map: XX/Ris continuous, be definition.

Let X and Ybe two topological spaces and f: XY a continuousmap. Let Rfdenote the equivalence relation defined by F. Then it iseasy to check that the map qf: X/R Y is continuous and f admitsthe following decomposition into continuous maps:

X X/Rf

qff(X)

jY.

Remark 3.53 qf is one-one, onto and continuous, but not necessarilya homeomorphism ontof(X).

Example 3.54 Let X = R be the group of real numbers and Z thesubgroup of integers. Let R be the equivalence relation in R definedby the subgroup Z. The quotient space is homeomorphic to the circle{(x, y)| (x, y) R2, x2 + y2 = 1}, with the topology induced from thatofR2.


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Remark 3.62 A contractible space is arcwise connected. In fact, let

F: XIXbe a homotopy ofIXinto a constant map, i.e. F(x, 0) =xand F(x, 1) = x0 for every x X. Clearly, for everyx X, the path: IX defined by (t) =F(x, t) joins x and x0. HenceX is arcwiseconnected. Problem 3.3

Remark 3.63 It is clear that a space Xis contractible if and only if itis of the same homotopy type as a space consisting of a single point.

Example 3.64 Rn is contractible. In fact the map F: Rn I Rn

given byF((x1, x2, . . . , xn), t) = ((1 t)x1, (1 t)x2, . . . , (1 t)xn), xiR, t Iis a homotopy between the identify map and the constant map

0. Similarly it can be proved that the closed (resp. open) unit ball inRn is contractible.


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Chapter 4

Singular Homology

4.1 Notation

LetR denote the set whose elements are sequences x= (x0, x1, x2, . . . ,xj , . . .) of real numbers with xj = 0 except for a finite number ofj . For R, x R, we write x for the element (x0, x1, . . .) R.For x = (x1, x2, . . .), y = (y1, y2, . . .) R

, we define x+y = (x1+y1, x2+ y2, . . .) R

. R is provided with the following metric: ifx = (x0, x1, x2, . . .) and y = (y0, y1, y2, . . .) are any two elements of

R, d(x, y) = {

j=0(xj yj )

2}1

2 , where by square root we mean the

positive square root.

j=o(xj yj )2

, is a finite sum and the verificationthat d(x, y) is a metric on R is left to the reader. Let n be anyinteger 0. Consider the map Rn+1 R given by (x0, . . . , xn) (x0, . . . , xn, 0, 0, . . .). This is one-one and we shall from now on identifyRn+1 with its image in R. For each j Z+, let ej be the element(x0, x1, . . .) given by xj = 1 and xi = 0 fori =j .

Definition 4.1 By the standard Euclidean n-simplex we mean the sub-set ofRn+1 consisting of points (x0, x1, . . . , xn) withxj 0 and

0jleq n xj =

1 with the induced topology.

The standard Euclideann-simplex is denoted by n.

Trivially, the elements ej , j = 0, 1, . . . , n lie in n and any ele-ment x of n can uniquely written as x =

0jn xj ej , xj 0 and

0jn xj = 1. The elements {ej}j=0,1,...,n are known as the vertices ofn.

For instance, 0 is a point, 1 is a segment, 2 is a triangle and 3a tetrahedron.

27


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28 Chapter 4. Singular Homology

4.1.1 Singular homology groups

Let q be any integer 1. Consider the maps iq: q1 q, i =0, 1, . . . , q , defined as follows:

iq(x0, . . . , xq1) = (x0, . . . , xi1, 0, xi, . . . , xq1)

for every (x0, . . . , xq1) q1. Actually the maps iq can be described

as follows. The vertices ej , j = 0, 1, 2, . . . , q 1, of q1 are mappedinto vertices of q by the following assignment:

iq(ej ) =ej for j < i

iq(ej ) =ej+1 for j i.

With this vertex assignment the maps are extended linearly, that is tosay the point x =

0jq1 xj ej of q1 is mapped into the point

0ji xj ej+

ijq1 xj ej+1 of q byiq. These maps

iq are trivially

continuous.

Lemma 4.2 Forq 2, iq jq1=

jq

i1q1, 0 j < i q. In fact, for

0 j < i q, we have

iq jq1(x0, . . . , xq2) = (x0, . . . , xj1, 0, . . . , xi2, 0, . . . , xq2)

= iq i1q1(x0, . . . , xq2).

Definition 4.3 For any integer q 0, a singular q-simplex in a topo-logical space Xis a continuous map f: q X.

For anyq 1 and for any singular q-simplexf: q Xthe singular(q1)-simplexfiq: q1 Xis called thei-th face offand is denoted

byf(i), for i = 0, 1, . . . , q .

Lemma 4.4 {f(i)}(j) = {f(j)}(i1) for any singular q-simplex f of Xand0 j < i q, withq 2.

This is an immediate consequence of the above lemma.

Example 4.5 For any a X, the constant map aq: q X given by

aq(y) =a for every y q is a singular q-simplex. Clearly, a(i)q = aq1

for 0 i q, ifq 1.


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0j


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Hence Cn()() Zn(Y). Similarly we can show that if Bn(X),

we have Cn()() Bn(Y). Hence Cn() induces a homomorphism ofZn(X)/Bn(X) =Hn(X) intoZn(Y)/Bn(Y) =Hn(Y), which we denoteby, Hn().

Remark 4.13 If I: X X is the identity map, Hn(I): Hn(X) Hn(X) is the identity homomorphism.

Remark 4.14 If: XX and : X X are continuous maps, thehomomorphismHn(

): Hn(X) Hn(X) induced by : XX

is the same as Hn() Hn().

Remark 4.15 Let : X X

be a homeomorphism From (i) and (ii)it can be easily seen that Hn(): Hn(X) Hn(X) is an isomorphism

of groups.

Remark 4.16 Let: XYbe a constant map. SinceHn((X)) = (0)for n = 0, Hn() = 0 for n = 0.

4.1.3 Homomorphisms induced by homotopic maps

Our main aim in this section is to show that if and are homotopic,the induced homomorphisms Hn() and Hn(

) are the same for everyintegern. To this end, we prove first the following

Lemma 4.17 If, are two continuous maps ofX into Y, which arehomotopic, then there exists a sequenceq: Cq(X) Cq+1(Y) of homo-morphisms such that

q+1 q+q1 q = Cq() Cq().

Proof: Let q I denote the cylinder over q. As usual, the ver-tices of q will be denoted by e0, . . . , eq. Let the vertices of q (0) bedenoted byle0, . . . , l eq and the vertices of q (1) by ue0, . . . , u eq. Foreach i = 0, 1, 2, . . . , q , we denote the linear map of q+1 in q Iwhich takes ej into lej for j i and ej into uej1 for j > i by(le0, le1, . . . , l ei, uei, uei+1, . . . , u eq). Clearly it is a singular (q+ 1)-simplex of q I.

Consider the singular (q+ 1)-chain q of q Igiven by

q =

0iq

(1)i(le0, le1, . . . , l ei, uei, uei+1, . . . , u eq).


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4.1. Notation 33

2

ue

ue

ue

le 0

le1

2

1

0

le

Figure 4.1:

The map iq: q1 q(defined in 2) gives a mapiq: q1 I

q I defined by iq(, t) = (iq, t), q1, t I. This in turn

induces a homomorphismC(iq): C(q1 I) C(q I). We denote

by(i)q the image of the q-chain q1 byCq(iq). We can verify that

() q+1q = (ue0, . . . , u eq) (le0, . . . , l eq)

0iq

(1)i(i)q ,

where (le0, . . . , l eq)[resp.(ue0, . . . , u eq)] denotes the singularq-simplex ofq I which sends ei to lei(resp. ei to uei) for 0 i qand which islinear. {The above formula makes precise the following geometric factwhich is intuitively clear : the boundary of the cylinder over the simplexq consists of the cylinder over the boundary of q and the upper andlower faces of the cylinder, with proper signs.}

Let now f: q X be any singular q-simplex. We have a map(f): q IY defined by (f) (, t) =F(f(), t)), q, t I,whereFis a homotopy between and . This induces a homomorphismC((f)): C(q I) C(Y). We setq(f) = Cq+1((f)) (q). (Here

q(f) is the cylinder constructed over the singular simplexfby meansof the homotopy F.) Using (*) one easily verifies that

q+1 q(f) +q+1 q(f) = (Cq() Cq()) (f);

Ifq is extended as a homomorphismCq(X) Cq+1(Y), it is clear that(q) has the required property.


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(f)l

f x I

I

Y

F

X

Figure 4.2:

Theorem 4.18 Let X and Y be two topological spaces. If : X Yand : X Y are homotopic maps, then the induced homomorphismsHn(): Hn(X) Hn(Y) andHn(): Hn(X) Hn(Y) are the same.

Proof: Let q: Cq(X) Cq+1(Y) be a sequence of homomorphismssuch thatq+1 q+ q1 q =Cq() Cq(). Let now Zq(X) bean element representing Hq(X). Then q= 0. Thus q+1 q=Cq(

) Cq() or, what is the same, Hq()() Hq()() = 0. Thisproves the theorem.

Theorem 4.19 IfXandYare topological spaces of the same homotopytype, they have isomorphic singular homology groups.

Proof: Let f: X Y , g: Y Xbe two continuous maps such that

g f is homotopic to IX and f g is homotopic to IY. Then we haveHn(g)) Hn(f) =Hn(g f) =Hn(IX) =IHn(X) and similarly Hn(f) Hn(g) =IHn(Y).

Corollary 4.20 IfX is a contractible topological space, thenHn(X) =0 forn= 0.


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4.1. Notation 35

Proof: In fact,IX and any constant map ofX into itself are homo-

topic, and the constant map induces 0 on the homology groups Hn(X)for n = 0 (Remark 4.16).

In particular, the corollary applies to closed and open balls ofRn.


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38 Chapter 5. Simplicial Complexes

Definition 5.3 A simplicial decomposition of a set X Rn is a finite

collection Tof subsets {s} ofX such that

(i)

s = X;

(ii) each s is a simplex;

(iii) ifs is in the collection, so is every face ofs;

(iv) ifs1 and s2 in the collection have a non-empty intersection, thens1 s2 is a common face of both s1 and s2.

The largest integernsuch that there exists ann-dimensional simplex

occurring in T is called the dimensionof the decomposition. Note thatthere can be more than one decomposition for the same set.

Remark 5.4 The spaceX(with the topology induced from Rn) is com-pact, since it is a finite union of compact sets.

Remark 5.5 Let v0, . . . , vN be the 0-simplices of T. For every sim-plex s = [vi0 , . . . , vin ] in T, consider the face = [ei0 , . . . , eir ] of thestandard Euclidean simplex N. Let |T| be the union of all faces ,for s T, provided with the topology induced from that of N. Wedefine a continuous map (x0, . . . , xN)

0iNxivi of N into R

n.It is easily seen that the restriction of this map to |T| is a continu-ous map of |T| onto X. In view of condition (iv) in the definition of asimplicial decomposition, is one-one and since |T| is compact, is ahomeomorphism.

5.1.1 Homology of a simplicial decomposition

Let X Rn and T a simplicial decomposition of X. Let r be thedimension ofT. We consider the set of all vertices of the simplices of thedecomposition. Order the set of all vertices in some way. If{vi}i=1,...,Nis the set of all vertices, every q-simplex in the decomposition is simply[vi0 , . . . , viq .] for some i0


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40 Chapter 5. Simplicial Complexes

Proof: : Cq(T) Cq1(T) is a homomorphism. The kernel is Zqand

the image is Bq1. By the Fundamental theorem of homomorphisms,we have Cq/Zq Bq1. By the proposition on ranks of abelian groups,we have

nq = (rank Zq) + (rank Bq1).

Moreover, since Hq(X) Zq/Bq, we have rank Hq(X) = rank Zq rank Bq. Thus nq rank Hq(X) = rankBq + rankBq1. Hence

(1)q(nq rank Hq(X)) = 0 or(X) =

(1)qnq.

Remark 5.10 The above formula is a generalization of the well-known

Euler formula connecting the number of vertices, edges and faces of aconvex polyhedron (V +F =E+ 2).

5.1.3 Homology of the sphere

Let us compute the homology of the sphere S2 = {x | x R3, x =1}. The sphere is homeomorphic to the surface of a tetrahedron. Thesurface of the tetrahedron 3 consists of four vertices e0, e1, e2, e3. The1-simplices are

1 = (e0, e1), 2= (e0, e2), 3 = (e0, e3), 4= (e1, e2), 5 = (e1, e3),

6 = (e2, e3), The 2-simplices are 1= (e1, e2, e3), 2= (e0, e2, e3),

3 = (e0, e1, e3), 4= (e0, e1, e2).

We have

21 = 6 5+4

22 = 6 3+2

23 = 5 3+1

24 = 4 2+1

It is easy to see that 2

4i=1 nii

= 0 if and only ifn1 =n2 =n3= n4; i.e. Z2consists of element of the formm(12+34). ThusZ2 is isomorphic to Z. Since there are no 3-simplices, B2 = (0). HenceH2 Z. It is also easy to see that H1= (0) and H0 Z. Similarly forthen-sphereSn, we haveHi(S

n) = (0) ifi = 0, n; H0(Sn) Hn(S

n)Z. (Sn ={x| x Rn+1, x= 1}, n 1)


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44 Bibliography

[14] Lefschetz, S.,Introduction to Topology, Chap. III and IV, Prince-

ton (1949).

TIFR pamphlet on Algebraic Topology

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