Math 530 Lecture Notes. Xi Chen

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1 Math 530 Lecture Notes Xi Chen 632 Central Academic Building, University of Alberta, Edmonton, Alberta T6G 2G1, CANADA address:

2 1991 Mathematics Subject Classification. Primary 14J28; Secondary 14E05 Key words and phrases. Algebraic Topology, Homology, Cohomology

3 CHAPTER 1 Homotopy, Fundamental Groups and Covering spaces 1.1. Homotopy Definition Let f : X Y and g : X Y be two continuous maps between topological spaces X and Y. We say that f and g are homotopic if there exists a continuous map F : X [0, 1] Y such that F (x, 0) = f(x) and F (x, 1) = g(x) for all x X. If we denote the set of continuous maps between X and Y by C(X, Y ), homotopy is clearly an equivalence relation on C(X, Y ). That is, it is trivial that f hom f and f hom h if f hom g and g hom h. The following is also obvious: Proposition If f 1 hom f 2 and g 1 hom g 2, then g 1 f 1 hom g 2 f 2, where f 1, f 2 C(X, Y ) and g 1, g 2 C(Y, Z). Let X be a topological space. Example We claim that all continuous maps f : X R n are homotopic to each other: for f, g C(X, R n ), let F : X [0, 1] R n be given by F (x, t) = f(x)(1 t) + g(x)t. Example The argument for the above example applies to all convex sets in R n : for V a convex set in R n and f, g C(X, V ), let F : X [0, 1] V be given by F (x, t) = f(x)(1 t) + g(x)t. Example Furthermore, the above argument works for all starshaped sets in R n. We call a set V R n star-shaped if there exists a point p V such that the line segment pq V for every q V. For f C(X, V ), setting F (x, t) = f(x)(1 t)+tp shows that every f C(X, V ) is homotopic to the constant map g : X V sending g(x) = p for all x X. Hence the continuous maps in C(X, V ) are homotopic to each other. Depending on situation, we sometimes impose some further constraints on f and g for them to be homotopic. For example, for two continuous curves f : [0, 1] X and g : [0, 1] X with the same ending points f(0) = g(0) and f(1) = g(1), we say that f and g are homotopic with fixed ending points if there exists a continuous map F : [0, 1] [0, 1] X such that F (t, 0) = f(t), F (t, 1) = g(t), F (0, s) = f(0) = g(0) and F (1, s) = f(1) = g(1). 3

4 4 1. HOMOTOPY, FUNDAMENTAL GROUPS AND COVERING SPACES In case that X is a C k manifold, we say that two C k -curves f : [0, 1] X and g : [0, 1] X are C k -homotopic if there exists a C k map F : [0, 1] [0, 1] X such that F (t, 0) = f(t) and F (t, 1) = g(t). We say that f C(X, Y ) and g C(Y, X) are homotopic inverses to each other if f g hom 1 Y and g f hom 1 X, where 1 X and 1 Y are the identity maps on X and Y, respectively. Definition We say two topological spaces X and Y are homotopic if there exist f C(X, Y ) and g C(Y, X) that are homotopic inverses to each other. If X is homotopic to a point, we call X contractible. It follows directly from Example that Proposition Every star-shaped set X in R n is contractible. Proof. Let Y = {p} X and let f : X Y and g : Y X be the maps that f(x) = p and g(p) = p. Clearly, f g = 1 Y and g f hom 1 X since all continuous maps in C(X, X) are homotopic to each other. Example We claim that (R n ) is homotopic to S n 1. Let g : (R n ) (R n ) be the map ( x 1 x g(x 1, x 2,..., x n ) = x x , 2 x2 n x x , x2 n ) (1.1) x n... x x x2 n ( ) x1 = x, x 2 x,..., x n x So g contracts (R n ) to S n 1. It suffices to find a continuous function F : (R n ) [0, 1] (R n ) such that F (x, 0) is the identity map of (R n ) and F (x, 1) = g(x). We simply take (1.2) F (x, t) = (1 t)x + tx x. Example Similarly, the annulus {r < x < R} R n is also homotopic to S n 1 by almost the same argument. Example If we remove a linear subspace V from R n, R n \V is homotopic to S n m 1 R m, where m = dim V. Proposition S 1 is not contractible. Lemma Let f : R S 1 be the map sending R to its quotient S 1 = R/Z.

5 CHAPTER 2 Singular Homology 2.1. Definition of Singular Homological Groups An n-simplex is the convex hull of n+1 points p 0, p 1,..., p n in R n. We call it a nondegenerate n-simplex if p 0, p 1,..., p n do not lie on a hyperplane in R n. The standard n-simplex n is the convex hull of the points e 1, e 2,..., e n, 0, where 0 is the origin. That is, (2.1) n = {(t 0, t 1,..., t n 1 ) : 0 t 0, t 1,..., t n 1 1} R n. To definte homological groups, we embed n to R n+1 by sending (2.2) (t 0, t 1,..., t n 1 ) (t 0, t 1,..., t n 1, 1 t 0 t 1... t n 1 ). That is, we let (2.3) n = {(t 0, t 1,..., t n ) : t 0 + t t n = 1, t 0, t 1,..., t n 0} R n+1. Depending on the context, n refers to either the simplex (2.1) or its embedding (2.2) in R n+1. The k-th face of n given in (2.3) is the embedding (2.4) δ k : n 1 n defined by (2.5) δ k (t 0, t 1,..., t n 1 ) = (t 0, t 1,..., t k 1, 0, t k,..., t n 1 ). Let X be a topological space. We define the group of singular n-chains C n (X) of X to be the abelian group freely generated by all continuous maps f : n X. That is, (2.6) C n (X) = {a 1 f 1 + a 2 f a m f m : a 1, a 2,..., a m Z, f 1, f 2,..., f m continuous functions n X} The singular chain complex C(X) of X is the direct sum of C n (X), i.e., (2.7) C(X) = C n (X) n= where we let C n (X) = 0 for n < 0. We have a group homomorphism : C n (X) C n 1 (X), called boundary operator, defined by sending a continuous map f : n X to n (2.8) f = ( 1) k f δ k k=0 5

6 6 2. SINGULAR HOMOLOGY and then extending by linearity via (2.9) (a 1 f 1 + a 2 f a m f m ) = a 1 f 1 + a 2 f a m f m. We can think of as an operator : C(X) C(X) from C(X) to itself. Theorem Let be the boundary operator on C(X) defined as above. Then 2 = 0. Proof. We defined the operator δ k : C n (X) C n 1 (X) by δ k f = f δ k and let δ k f = 0 for k > n. Then (2.10) f = ( 1) k δ k f and (2.11) 2 f = for f C n (X). Note that (2.12) Thus, we have k=0 ( 1) k+l δ k δ l f k,l=0 (δ k δ l f)(t 0, t 1,..., t n 2 ) { f(t 0, t 1,..., t l 1, 0, t l,..., t k 1, 0, t k,..., t n 2 ) = f(t 0, t 1,..., t k 1, 0, t k,..., t l 2, 0, t l 1,..., t n 2 ) (2.13) δ k δ l f δ l 1 δ k f = 0 for all k < l and hence 2 = 0. if k l if k < l Clearly, 2 = 0 implies that the image of is contained in the kernel of. More specifically, (2.14) C n+1 (X) ker(c n (X) C n 1 (X)) for all n. The n-th singular homological group of X is defined to be (2.15) H n (X) = ker(c n(x) C n 1 (X)). C n+1 (X) We can also replace the coefficients a i Z in (2.6) by a i R for a fixed commutative ring R. Namely, we let C n (X) be the module over R freely generated by all continuous functions f : n X. As another way to put it, we can replace C n (X) and C(X) by C n (X) Z R and C(X) Z R, respectively. We call the corresponding homological groups (R-modules) (2.16) H n (X, R) = ker(c n(x) Z R C n 1 (X) Z R) (C n+1 (X) Z R) the singular homology of X over R. The homology given in (2.15) is the singular homology of X over Z and we sometimes write H n (X) = H n (X, Z) to emphansize the fact that these groups are defined over Z. It is very easy to figure out H 0 (X). That is,

7 2.2. SOME BASIC NOTIONS IN HOMOLOGICAL ALGEBRA 7 Proposition Let X be a topological space and let X = α A X α with X α the connected components of X. Then (2.17) H n (X) = α A H n (X α ) for all n and (2.18) H 0 (X) = α A Z if all X α are path-connected. Proof. Clearly, C n (X) = α A C n (X α ) and hence (2.17) follows. To show (2.18), it suffices to show that H 0 (X) = Z for X path-connected. For every continuous map f : 1 X, f = f(0, 1) f(1, 0). Since X is path-connected, there is a continuous map f : 1 X such that f(0, 1) = p and f(1, 0) = q for every pair of points p, q X. Therefore, C 1 (X) is generated by p q for all p, q X. It follows that H 0 (X) = Z. Proposition Let X be the topological space consisting of one point. Then { Z if n = 0 (2.19) H n (X) = 0 if n 0 Proof. Obviously, C n (X) = Z is generated by the constant function f n : n X for n 0. Then (2.19) follows easily from the fact that { 0 if n is odd or n = 0 (2.20) f n = f n 1 if n is even and n > Some Basic Notions in Homological Algebra For simplicity, we work with abelian groups. One can replace them by R-modules. We call a sequence of homomorphisms of abelian groups (2.21) A f B g C exact at B if im(f) = ker(g), a left exact sequence if it is exact at B and f is injective, a right exact sequence if it is exact at B and g is surjective and a short exact or simply an exact sequence if it is both left and right exact. Usually, we write (2.22) 0 A f B g C for a left exact sequence, (2.23) A f B g C 0

8 8 2. SINGULAR HOMOLOGY for a right exact sequence and (2.24) 0 A f B g C 0 for a short exact sequence. A long exact sequence is simply a sequence (2.25) A 0 A 1 A 2... A n A n+1 of homomorphisms of abelian groups that is exact at all A 1, A 2,..., A n. A chain complex (2.26) K = n Z K n is an abelian group together with a boundary operator sending K n K n 1 and satisfying 2 = 0. Usually, we write it as (K, ). The homological groups of a chain complex (K, ) are defined to be (2.27) H n (K ) = ker(k n K n 1 ) K n+1. Clearly, H n (K ) = 0 if and only if the sequence (2.28)... K n+1 Kn Kn 1... is exact at K n. In other words, H n (K ) measures how (2.28) fails to be exact. Let (K, ) and (L, ) be two chain complexes. We call f : K L a map of chain complexes if it is a group homomorphism sending K n to L n and commuting with the boundary operators, i.e., f = f. More precisely, we have the commutative diagram (2.29) K n+1 K n K n 1... f f L n+1 L n L n 1... f The kernel ker(f) and cokernel coker(f) are also chain complexes in a natural way. That is, we can define boundary operators (2.30) : ker(k n f Ln ) ker(k n 1 f Ln 1 ) and (2.31) : coker(k n f Ln ) coker(k n 1 f Ln 1 )

9 2.2. SOME BASIC NOTIONS IN HOMOLOGICAL ALGEBRA 9 such that the commutative diagram (2.29) can be extended to (2.32) J n+1 J n J n 1... K n+1 K n K n 1... f f L n+1 L n L n 1... f... M n+1 M n M n where the vertical columns are exact. A short exact sequence of chain complexes is a sequence (2.33) 0 K f L g M 0 of maps of chain complexes that is exact as a sequence of abelian groups. A map f : K L of chain complexes induces a natural homomorphism (2.34) f : H n (K ) H n (L ) of homological groups of K and L sending ker(k n Kn 1 ) to ker(l n L n 1 ). A short exact sequence (2.33) of chain complexes leads to a long exact sequence on homology... H n+1 (K ) f H n+1 (L ) g H n+1 (M ) (2.35) H n (K ) f H n (L ) g H n (M ) H n 1 (K ) f H n 1 (L ) g H n 1 (M )... We say two maps f : K L and g : K L of chain complexes are homotopic if there are homomorphisms P n : K n L n+1 such that (2.36) P n + P n 1 = f g on K n for all n. Or equivalently, if we replace P n by ( 1) n P n, (2.36) becomes (2.37) P n P n 1 = ( 1) n (f g). It is obvious that f and g induce the same map on homology if they are homotopic.

10 10 2. SINGULAR HOMOLOGY 2.3. Homotopy Invariance of Homology Push-forward. We are going to establish the homotopic invariance of singular homology. First of all, a continuous map f : X Y induces a map of chain complexes (C (X), ) and (C (Y ), ). That is, the diagram (2.38)... C n+1 (X) C n (X) C n 1 (X)... f f f... C n+1 (Y ) C n (Y ) C n 1 (Y )... commutes for f : C n (X) C n (Y ) defined by f (g) = f g for a continuous map g : n X. So it induces a map on homological groups f : H n (X) H n (Y ), which is called the push-forward of f on homology. It is obvious that Proposition Let f : X Y and g : Y Z be continuous maps between topological spaces X, Y and Z. Then (g f) = g f for f : H n (X) H n (Y ) and g : H n (Y ) H n (Z) the push-forwards of f and g, respectively Homotopy invariance of singular homology. We first show that two homotopic maps f : X Y and g : X Y induce the same map on homology. Theorem Let f : X Y and g : X Y be two continuous maps between topological spaces X and Y and let f : H n (X) H n (Y ) and g : H n (X) H n (Y ) be the induced maps on homology, respectively. If f and g are homotopic, then f g. Combining the above theorem with Proposition 2.3.1, we arrive at our main result of this section Corollary Let X and Y be two homotopic topological spaces. Then H n (X) = H n (Y ) for all n. We will spend the rest of the section to prove Theorem Here is a sketch of our proof. It suffices to show that f : C(X) C(Y ) and g : C(X) C(Y ) are two homotopic maps of chain complexes. Namely, we are trying to find P n : C n (X) C n+1 (Y ) such that (2.39) (P n (σ)) P n 1 ( (σ)) = ( 1) n (f σ g σ) for all σ C n (X) and n. A natural way to construct P n is via the continuous map F : X I Y satisfying F (x, 0) = f and F (x, 1) = g, where I = 1 = [0, 1]. For every continuous map σ : n X, we have a continuous map F (σ 1 I ) : n I Y, where 1 I is the identity map on I and σ 1 = σ 1 I : n I X I is the map sending (p, t) to (σ(p), t).

11 Roughly, we let 2.3. HOMOTOPY INVARIANCE OF HOMOLOGY 11 (2.40) P n (σ) = F (σ 1) and extend it by linearity to all σ C n (X). The problem is that F (σ 1) is a continuous function on n I, n I is not a simplex and hence F (σ 1) is not an element of C n+1 (Y ). The remedy of this situation involves a technique called subdivision of convex polytopes. Roughly, we will find an (n + 1)-chain ξ n C n+1 ( n I) and define P n to be (2.41) P n (σ) = F (σ 1) ξ n such that (2.39) holds, where F : C(X I) C(Y ) is the push-forward of F on the chain complexes Oriented simplices. Let X be a convex set in R N. We let L n (X) denote the abelian group freely generated by the maps (2.42) σ p0 p 1...p n (t 0, t 1,..., t n ) = t 0 p 0 + t 1 p t n p n from n to X for some p 0, p 1,..., p n X. We call σ p0 p 1...p n the oriented n-simplex spanned by p 0, p 1,..., p n. Note that the order of p 0, p 1,..., p n is essential. Obviously, L n (X) is a subgroup of C n (X) and L n (X) L n 1 (X) since n (2.43) σ p0 p 1...p n = ( 1) k σ p0 p 1...ˆp k...p n. k=0 Let p 0, p 1,..., p n be n + 1 points in R n that do not lie on a hyperplane. Or equivalently, the determinant p 0 1 p 1 1 (2.44) det.. p n 1 (n+1) (n+1) does not vanish, where p k is the row vector representing the point p k. We let ξ p0 p 1...p n be the n-chain in L n (R n ) given by (2.45) ξ p0 p 1...p n = λ p0 p 1...p n σ p0 p 1...p n where λ p0 p 1...p n = 1 if the determinant (2.44) is negative, λ p0 p 1...p n = 1 if the determinant (2.44) is positive and 0 if it vanishes. For simplicity, we also write ξ σ = ξ p0 p 1...p n and λ σ = λ p0 p 1...p n for σ = σ p0 p 1...p n Convex polytope and subdivision. Definition A convex polytope T R n is the convex hull of a finite set S of points in R n. The dimension dim T of T is the dimension of the linear subspace spanned by S. We call T R n non-degenerate if dim T = n, or equivalently, the interior of T in R n is not empty. We can find {p 1, p 2,..., p m } S such that T is the convex hull of p 1, p 2,..., p m S

12 12 2. SINGULAR HOMOLOGY and no one point among p 1, p 2,..., p m is a convex combination of the rest. Then the center (barycenter) of T is given by (2.46) c(t ) = 1 m (p 1 + p p m ). An orientation of T is an ordering of {p 1, p 2,..., p m }. Let m = dim T. An (m 1)-dimensional face of T is the intersection T = T H of T and a hyperplane H R n such that dim T = m 1 and H is a supporting hyperplane of T, i.e., T lies on one side of H. Inductively, every l-dimensional face of T is also an l-dimensional face of T. That is, the faces of T consist of all (m 1)-dimensional faces T and all the faces of such T. For convenience, we call T the m-dimensional face of itself. The faces of an n-simplex conv(p 0, p 1,..., p n ) are conv(s) for all nonempty set S {p 0, p 1,..., p n }. An oriented n-simplex σ = σ p0 p 1...p n gives orientations to all its faces. We call σ pi1 p i2...p im an oriented face of σ for 0 i 1 < i 2 <... < i m n. Definition (Subdivison of a convex polytope). Let T be a convex polytope in R n of dimension dim T = m. A subdivision of T is a way to divide T into a union of m-simplices (2.47) T = a T a such that T a T b is either empty or a face of both T a and T b for all a and b. A subdivision of T is denoted by a collection {T a } of m-simplices with the above properties. Furthermore, an oriented subdivision of T is a collection {σ p0 p 1...p m } of oriented m-simplices with the above properties. For T R n a nondegenerate convex polytope and an oriented subdivision T of T, we associate an n-chain ξ T L n (T ) by (2.48) ξ T = σ T with ξ σ defined in (2.45). Let T = n I R n+1 and let i 0 and i 1 be the maps n n I sending p to (p, 0) and (p, 1), respectively; i 0 and i 1 can be regarded as oriented faces of T. The vertices of i 0 and i 1 are the vertices of T and we orient T in the way {a 0, a 1,..., a n, b 0, b 1,..., b n } such that (2.49) i 0 = σ a0 a 1...a n and i 1 = σ b0 b 1...b n. The other n-dimensional faces of T are δ k ( n 1 ) I for k = 0, 1,..., n, where δ k : n 1 n is the k-th face of n defined by (2.5). We construct the following pseudo barycentric subdivision, denoted by pbcs(t ), inductively as follows: If n = 0, we let pbcs(t ) = {σ}, where σ = 1 I is the identity map on I = 1. ξ σ

13 2.3. HOMOTOPY INVARIANCE OF HOMOLOGY 13 Let T be an n-dimensional face of T. If T is i 0 or i 1, we let pbcs(t ) = {σ} with σ the identity map on n. If T = δ k ( n 1 ) I, we let pbcs(t ) be the subdivision obtained by induction. For every oriented n-simplex σ p0 p 1...p n pbcs(t ), we replace it by σ p0 p 1...p np with p = c(t ) the center of T. That is, we let (2.50) pbcs(t ) = {σ p0 p 1...p np} T σ p0 p 1...pn pbcs(t ) for T running over all n-dimensional faces of T. Every simplex σ pbcs(t ) is given in the form of σ = σ p0 p 1...p k p k+1...p n+1, where σ p0 p 1...p k is a face of i 0 or i 1 and p j is the center of a face T j = j 1 I of T for j = k + 1,..., n + 1 satisfying σ p0 p 1...p k T k+1...t n+1 = T. As in (2.48), we define ξ n L n+1 (T ) similarly by (2.51) ξ n = σ pbcs(t ) and we claim that ξ n is the (n + 1)-chain such that (2.39) holds if we define P n as in (2.41). Namely, we can prove Lemma For every continuous map F : X I Y and every σ C n (X), (2.52) (F (σ 1) ξ n ) F ( σ 1) ξ n 1 = ( 1) n (f σ g σ) where F (x, 0) = f(x) and F (x, 1) = g(x). Proof. It suffices to prove it for X = n, Y = n I, F = 1 T : T = n I T the identity map on n I and σ = 1 n : n X the identity map on n. Indeed, since F commutes with, the LHS of (2.52) becomes (2.53) ξ σ (F (σ 1) ξ n ) F ( σ 1) ξ n 1 = F ( ((σ 1) ξ n ) ( σ 1) ξ n 1 ) If (2.52) holds for (X, Y, F, σ) = ( n, n I, 1 T, 1 n ), then (2.54) ((σ 1) ξ n ) ( σ 1) ξ n 1 = ( 1) n (i 0 i 1 ) for σ = 1 n. For all σ C n (X), σ = σ 1 n, σ = σ 1 n (2.55) and hence ((σ 1) ξ n ) ( σ 1) ξ n 1 = (σ 1) ( ((1 n 1) ξ n ) ( 1 n 1) ξ n 1 ) = ( 1) n (σ 1) (i 0 i 1 ). Combining this with (2.53), we see that (2.52) follows. So it suffices to prove it for (X, Y, F, σ) = ( n, n I, 1 T, 1 n ). Indeed, it remains to verify (2.54) for σ = 1 n. That is, (2.56) ξ n ( 1 n 1) ξ n 1 = ( 1) n (i 0 i 1 )

14 14 2. SINGULAR HOMOLOGY or equivalently, (2.57) ξ n n ( 1) k (δ k 1) ξ n 1 = ( 1) n (i 0 i 1 ) k=0 where δ k 1 : n 1 I n I is the map sending (p, t) to (δ k (p), t). If an (n+1)-simplex σ pbcs(t ) has an n-dimensional face ε containing the center c(t ) of T, then there must exist another (n + 1)-simplex σ pbcs(t ) such that σ σ = ε. More precisely, there are oriented simplices in the form of σ = σ p0 p 1...p k...p n+1 and σ = σ p0 p 1...p k...p in pbcs(t ) such that n+1 ε = σ p0 p 1...ˆp k...p n+1. Note that p n+1 = c(t ). It is easy to check that (2.58) λ p0 p 1...p k...p n+1 = λ p0 p 1...p k...p n+1 and hence (2.59) ξ p0 p 1...p k...p n+1 δ k = ξ p0 p 1...p k...p n+1 δ k. Therefore, (2.60) ξ n = = σ p0 p 1...p n+1 pbcs(t ) σ p0 p 1...p n+1 pbcs(t ) ξ p0 p 1...p n+1 ( 1) n+1 λ p0 p 1...p n+1 σ p0 p 1...p n. Obviously, σ p0 p 1...p n is either one of i 0 and i 1 or a simplex in pbcs(t ) for a face T = δ k ( n 1 ) I of T. And it is easy to check that λ p0 p 1...p n+1 = 1 if σ p0 p 1...p n = i 0 and λ p0 p 1...p n+1 = 1 if σ p0 p 1...p n = i 1. Therefore, (2.61) ξ n ( 1 n 1) ξ n 1 = ( 1) n (i 0 i 1 ) n + ( 1) n+1 λ q0 q 1...q nqσ q0 q 1...q n k=0 n ( 1) k σ p0 p 1...pn pbcs( n 1 I) k=0 σ p0 p 1...pn pbcs( n 1 I) λ p0 p 1...p n σ q0 q 1...q n where q 0 = δ k (p 0 ), q 1 = δ k (p 1 ),..., q n = δ k (p n ) and q = c(t ). Here we use δ k for the map δ k 1 : n 1 I n I. It remains to verify that (2.62) ( 1) n+1 λ q0 q 1...q nq = ( 1) k λ p0 p 1...p n for all p 0, p 1,..., p n n 1 I, q j = δ k (p j ) for j = 0, 1,..., n and q = c(t ), which is quite straightforward.

15 2.4. BARYCENTRIC SUBDIVISION Barycentric Subdivision Definition (Barycentric subdivision). Let T be a convex polytope of dimension dim T = m in R n. The barycentric subdivision of T, denoted by bcs(t ), is the oriented subdivision of T constructed inductively as follows: If T = {p} is a point, bcs(t ) = {σ p }. For every (m 1)-dimensional face T of T and every oriented (m 1)-simplex σ p0 p 1...p m 1 bcs(t ), we replace it by σ p0 p 1...p m 1 p, where p = c(t ) is the center of T. That is, we let (2.63) bcs(t ) = {σ p0 p 1...p m 1 p} T σ p0 p 1...p m 1 bcs(t ) for T running over all (m 1)-dimensional faces of T. By the construction of bcs(t ), it is easy to see that every oriented m- simplex σ bcs(t ) is spanned by a sequence p 0, p 1,..., p m with the property that p j is the center of a j-dimensional face T j of T for j = 0, 1,..., m with the property that T 0 T 1... T m. For a nondegenerate convex polytope T R n, we associate an n-chain α T L n (T ) to T by (2.64) α T = ξ σ. σ bcs(t ) as in (2.48). For every continuous map g : n X, we let (2.65) β g = g α n = λ σ (g σ) σ bcs( n) and we can define β g for all g C n (X) by extension of linearity. This defines an operator β : C(X) C(X) by sending g to β g. It is not hard to see, though a little tedious to check, that β commutes with the boundary operator. That is, Lemma Let X be a topological space and β : C(X) C(X) be the operator defined as above via barycentric subdivision. Then β = β. Proof. By the same argument as in the proof of Lemma 2.3.6, we see that it suffices show that β(1 n ) = β( 1 n ) for X = n. That is, (2.66) α n = n ( 1) k δ k α n 1. k=0 Again, by the same argument for Lemma 2.3.6, we see that all (n 1)- dimensional faces of σ bcs( n ) inside σ cancel out each other. Thus,

16 16 2. SINGULAR HOMOLOGY we have (2.67) α n = ξ q0 q 1...q n 1 q = σ q0 q 1...q n 1 q bcs( n) σ q0 q 1...q n 1 q bcs( n) ( 1) n λ q0 q 1...q n 1 qσ q0 q 1...q n 1 where q = c( n ). Obviously, σ q0 q 1...q n 1 is a simplex in bcs(δ k ( n 1 )). To show (2.66), it is enough to verify that (see also (2.62)) (2.68) ( 1) n λ q0 q 1...q n 1 q = ( 1) k λ p0 p 1...p n 1 for all p 0, p 1,..., p n 1 n 1, q j = δ k (p j ) and q = c( n ), which is quite straightforward. So β defines a map of complexes C(X) C(X). The main result of this section is that it is more or less homotopic to the identity map C(X) C(X). Theorem Let X be a topological space and β : C(X) C(X) be the operator defined above via barycentric subdivision. Then there exists a homomorphism P n : C n (X) C n+1 (X) for each n such that (2.69) P n (σ) P n ( σ) = β(σ) σ for all σ C n (X). As a consequence, we see that (2.70) β m (σ) = β β... β(σ) = σ }{{} m for all σ H n (X) and m 0. This fact is important to us later on. To construct P n, we again find ξ n L n+1 ( n ) and let P n (σ) = σ ξ n, similar to (2.41). We just have to find ξ n such that n (2.71) ξ n ( 1) k δ k ξ n 1 = α n 1 n. k=0 We let e 1, e 2,..., e n, e n+1 be the n + 1 vertices of n, which are the standard basis of R n+1 when we embed n to R n+1 by (2.2). For every permutation τ of {1, 2,..., n + 1} and every 1 k n + 1 satisfying that τ(1) < τ(2) <... < τ(k), we define an oriented (n + 1)- simplex (2.72) γ τ,k = σ eτ(1) e τ(2)...e τ(k) q k...q n+1 where q j is the center of the simplex σ eτ(1) e τ(2)...e τ(j) for k j n + 1. Finally, we let (2.73) ξ n = ( 1) k 1 sgn(τ)γ τ,k τ k

17 2.5. PRELIMINARIES IN HOMOLOGICAL ALGEBRA 17 where sgn(τ) is the sign of the permutation τ, i.e., it is 1 if τ is an even permutation and 1 if τ is an odd permutation. It is not hard, though quite tedious, to check that (2.71) holds Preliminaries in Homological Algebra Direct sums and products. Let {A i : i I} be a set of abelian groups indexed by a set I. Then the direct sum of A i is (2.74) A i = {(a i ) i I : a i A i, a i 0 only for finitely many i I}, i I while the direct product of A i is (2.75) A i = {(a i ) i I : a i A i }. i I Clearly, the direct sum of A i is a subgroup of their direct product. They are the same if I is finite but different if I is infinite and A i is nontrivial. In addition, it is easy to see that ) (2.76) Hom Z ( i I A i, Z = i I Hom Z (A i, Z). Let {A i : i I} and {B j : j J} be two sets of abelian groups. We have natural maps ( ) ( ) (2.77) A i Z B j A i Z B j i I i J and (2.78) ( ) ( ) A i Z B j i I i J (i,j) I J (i,j) I J A i Z B j sending (a i, b j ) to (a i b j ). It is easy to see that (2.77) is an isomorphism. However, (2.78) is usually not a surjection. Proposition If I and J are infinite, each A i is nontrivial and torsion free and B j = Z, then the map (2.78) is not surjective. Proof. Obviously, we just have to show this for I = J = Z +. For each A i, we choose r i 0 A i. Let v (i,j) I J A i Z B j be given by v ij = r i 1 if i = j and 0 otherwise. Let v j = (v 1j, v 2j,...). Then v 1, v 2,... are linearly independent over Z. That is, the map (2.79) Z j=1 i=1 A i

18 18 2. SINGULAR HOMOLOGY sending (a j ) to a j v j is injective. If v lies in the image of (2.78), then m (2.80) v = x k y k where x k i I A i and y k j J B j. Let y k = (y k1, y k2,...). Then (2.81) v j = for all j. That is, (2.82) v j k=1 m y kj x k k=1 m Zx k k=1 for all j. This contradicts the fact that v 1, v 2,... are linearly independent over Z Direct limits. Let I be a partially ordered the set. A directed system {A i : i I} is a collection of abelian groups together with homomorphisms f ij : A i A j for i j satisfying f ik = f jk f ij for i j k. The direct limit (2.83) A = lim A i i I is an abelian group together with homomorphisms g i : A i A satisfying f i = g j f ij for i j and the universal property that if there are another A and homomorphisms g i : A i A with the same property, then there is a homomorphism h : A A with the commutative diagram g i (2.84) A i A g i h A Example Let A be an abelian group, I be the set of all finite subsets i of A and A i be the subgroup of A generated by the elements in i. In other words, A i are all finitely generated subgroups of A. Obviously, {A i } forms a directed system with the direct limit A. One can think of lim as a functor from the category of directed systems to that of abelian groups. This is a very nice functor: it is exact. That is, for directed systems {A i }, {B i } and {C i }, the direct limit (2.85) 0 lim A i lim B i lim C i 0 i I i I i I of the exact sequences (2.86) 0 A i B i C i 0

19 2.6. UNIVERSAL COEFFICIENTS THEOREM FOR HOMOLOGY 19 is still exact. Therefore, the direct limit of the kernels is the kernel of the direct limits (2.87) lim ker(a i B i ) = ker(lim A i lim B i ) i I i I i I and the direct limit of the images is the image of the direct limits (2.88) lim im(a i B i ) = im(lim A i lim B i ). i I i I i I Consequently, for a directed system of chain complexes {K,i }, the direct limit of holomology is the homology of the direct limit: (2.89) lim H n (K,i ) = H n (lim K,i ) i I i I for all n Universal coefficients theorem for homology

20

21 Bibliography [F] Bill Fulton, Algebraic topology, a first course, GTM 153. [M] William Massey, Singular homology theory, GTM 70. [B-T] Raoul Bott and Loring W. Tu, Differential Forms in Algebraic Topology, GTM