0.1 Universal Coefficient Theorem for Homology

Transcription

1 0.1 Universal Coefficient Theorem for Homology Tensor Products Let A, B be abelian groups. Define the abelian group A B = a b a A, b B / (0.1.1) where is generated by the relations (a + a ) b = a b + a b and a (b + b ) = a b + a b. The zero element of A B is 0 b = a 0 = 0 0 = 0 A B since, e.g., 0 b = (0+0) b = 0 b+0 b so 0 b = 0 A B. Similarly, the inverse of an element a b is (a b) = ( a) b = a ( b) since, e.g., 0 A B = 0 b = (a + ( a)) b = a b + ( a) b. Lemma The tensor product satisfies the following universal property which asserts that if ϕ : A B C is any bilinear map, then there exists a unique map ϕ : A B C such that ϕ = ϕ i, where i : A B A B is the natural map (a, b) a b. A B i ϕ A B! C Proof. Indeed, ϕ : A B C can be defined by a b ϕ(a, b). Proposition The tensor product satisfies the following properties: (1) A B = B A via the isomorphism a b b a. (2) ( i A i) B = i (A i B) via the isomorphism (a i ) i b (a i b) i. (3) A (B C) = (A B) C via the isomorphism a (b c) (a b) c. (4) Z A = A via the isomorphism n a na. (5) Z/nZ A = A/nA via the isomorphism l a la. Proof. These are easy to prove by using the above universal property. We sketch a few. (1) The map ϕ : A B B A defined by (a, b) b a is clearly bilinear and therefore induces a homomorphism ϕ : A B B A with a b b a. Similarly, there is the reverse map ψ : B A A B defined by (b, a) a b which induces a homomorphism ψ : B A A B with b a a b. Clearly, ϕ ψ = id B A and ψ ϕ = id A B and A B = B A. (4) The map ϕ : Z A A defined by (n, a) na is a bilinear map and therefore induces a homomorphism ϕ : Z A A with n a na. Now suppose ϕ(n a) = 0. Then na = 0 and n a = 1 (na) = 1 0 = 0 Z A. Thus ϕ is injective. Moreover, if a A, then ϕ(1 a) = a and ϕ is surjective as well. 1

2 (5) The map ϕ : Z/nZ A A/nA defined by (l, a) la is a bilinear map and therefore induces a homomorphism ϕ : Z/nZ A A/nA with l a la. Now suppose ϕ(l a) = la = 0. Then la = k i=1 na i and l a = 1 (la) = 1 ( k i=1 na i) = k i=1 (n a i) = 0 Z/nZ A, so ϕ is injective. Now let a A/nA. Then ϕ(1 a) = a and ϕ is surjective as well. More generally, if R is a ring and A and B are R-modules, a tensor product A R B can be defined as follows: (1) if R is commutative, define the R-module A R B := A B/, where is the relation generated by ra b = a rb = r(a b). (2) if R is not commutative, we need A a right R-module and B a left R-module and the relation is ar b = a rb. In this case A R B is only an abelian group. In both cases, A R B is not necessarily isomorphic to A B. Example Let R = Q[ 2] = a + b 2 a, b Q}. Now R R R = R which is a 2-dimensional Q-vector space. However, R R as a Z-module is a 4-dimesnional Q-vector space. Lemma If G is an abelian group, then the functor G is right exact, that is, if A i B j C 0 is exact, then A G i 1 G B G j 1 G C G 0 is exact. Proof. Let c g C G. Since j is onto, there exists, b B such that j(b) = c. Then (j 1 G )(b g) = c g and j 1 G is onto. Since j i = 0, we have (j 1 G ) (i 1 G ) = (j i) 1 G = 0 and thus, Image(i 1 G ) ker(j 1 G ). It remains to show that ker(j 1 G ) Image(i 1 G ). It is enough to show that ψ : B G/Image(i 1 G ) = C G, where ψ is the map induced by j 1 G. Construct an inverse of ψ, induced from the homomorphism ϕ : C G B G/Image(i 1 G ) defined by (c, g) b g, where j(b) = c. We must show that ϕ is a well-defined bilinear map and that the induced map ϕ satisfies ϕ ψ = id and ψ ϕ = id. If j(b) = j(b ) = c, then b b ker j = Image i, so b b = i(a) for some a A. Thus, b g b g = (b b ) g = i(a) g Image(i 1 G ). So ϕ is well defined. Now ϕ((c + c, g)) = d g where j(d) = c + c. Since j is surjective, choose b, b B such that j(b) = c and j(b ) = c. Then d (b + b ) ker j = Image i and so there exists a A such that i(a) = d (b + b ). Thus, ϕ((c + c, g)) = d g = (b + b ) g = 2

3 b g + b g = ϕ(c, g) + ϕ(c, g) and ϕ is linear in the first component. For the second component, ϕ(c, g + g ) = b (g + g ) = b g + b g = ϕ(c, g) + ϕ(c, g ). Thus, ϕ is bilinear. Now by the universal property of the tensor product, the bilinear map ϕ induces a homomorphism ϕ : C G B G/Image(i 1 G ) defined by c g ϕ(c, g) = b g, where j(b) = c. For c g C G, ψ ϕ(c g) = ψ(b g) = j(b) g = c g, so ψ ϕ = id C G. Similarly, for b g B G/Image(i 1 G ), ϕ ψ(b g) = ϕ(j(b) g) = ϕ(j(b), g) = b g. Thus ϕ ψ = id. Remark Tensoring with a free group is an exact functor Homology with Arbitrary Coefficients Let G be an abelian group and X a topological space. We define the homology of X with G-coefficients, denoted H (X; G), as the homology of the chain complex C i (X; G) = C i (X) G (0.1.2) consisting of finite formal sums i η i σ i (σ i : i X, η i G), and with boundary maps given by G i := i id G. Since i satisfies i i+1 = 0 it follows that i G i+1 G = 0, so (C (X; G), G ) forms indeed a chain complex. We can construct versions of the usual modified homology groups (relative, reduced, etc.) in the natural way. Define relative chains with G-coefficients by C i (X, A; G) := C i (X; G)/C i (A; G), and reduced homology with G-coefficients via the augmented chain complex G i+1 C i (X; G) G i 2 G C 1 (X; G) G 1 C 0 (X; G) ɛ G 0, where ɛ( i η iσ i ) = i η i G. Notice that H i (X) = H i (X; Z) by definition. By studying the chain complex with G-coefficients, it follows that G i = 0 H i (pt; G) = 0 i 0. Nothing (other than coefficients) needs to change in describing the relationships between relative homology and reduced homology of quotient spaces, so we can compute the homology of a sphere as before by induction and using the long exact sequence of the pair (D n, S n ) to be H i (S n G i = 0, n ; G) = 3

4 Finally, we can build cellular homology with G-coefficients in the same way, defining The cellular boundary maps are given by: C G i (X) = H i (X i, X i 1 ; G) = G # i cells. d G i ( α η α e i α) = α,β η α d αβ e i 1 β, where d αβ is as before the degree of a map αβ : S i 1 S i 1. This follows from the easy fact that if f : S k S k has degree m, then f : H k (S k ; G) G H k (S k ; G) G is the multiplication by m. As it is the case for integers, we get an isomorphism for all i. H CW i (X; G) H i (X; G) Example We compute H i (RP n ; Z 2 ) using the cellular homology with Z 2 -coefficients. Notice that over Z the cellular boundary maps are d i = 0 or d i = 2 depending on the parity of i, and therefore with Z 2 -coefficients all of boundary maps vanish. Therefore, H i (RP n Z 2 0 i n ; Z 2 ) = Example Fix n > 0 and let g : S n S n be a map of degree m. Define the CW complex X = S n g e n+1, where the (n + 1)-cell e n+1 is attached to S n via the map g. Let f be the quotient map f : X X/S n. Define Y = X/S n = S n+1. The homology of X can be easily computed by using the cellular chain complex: 0 d n+2 Z d n+1 m Z dn... d 1 0 d 1 Z d 0 0 Therefore, Moreover, as Y = S n+1, we have Z i = 0 H i (X; Z) = Z m i = n H i (Y ; Z) = Z i = 0, n + 1 It follows that f induces the trivial homomorphisms in homology with Z-coefficients (except in degree zero, where f is the identity). So it is natural to ask if f is homotopic to the 4

5 constant map. As we will see below, by considering Z m -coefficients we can show that this is not the case. Let us now consider H (X; Z m ) where m is, as above, the degree of the map g. We return to the cellular chain complex level and observe that we have 0 d n+2 Z m d n+1 m Z m d n... d 1 0 d 1 Z m d 0 0 Multiplication by m is now the zero map, so we get Z m i = 0, n, n + 1 H i (X; Z m ) = Also, as already discussesd, H i (Y ; Z m ) = Z m i = 0, n + 1 We next consider the induced homomorphism f : H n+1 (X; Z m ) H n+1 (Y ; Z m ). The claim is that this map is injective, thus non-trivial, so f cannot be homotopic to the constant map. As noted before, we have an isomorphism H n+1 (Y ; Z m ) H n+1 (X, S n ; Z m ). This leads us to consider the long exact sequence of the pair (X, S n ) in dimension n + 1. We have H n+1 (S n ; Z m ) H n+1 (X; Z m ) f H n+1 (X, S n ; Z m ) But, H n+1 (S n ; Z m ) = 0 and so f is injective on H n+1 (X; Z m ). Since H n+1 (X; Z m ) = Z m 0 and H n+1 (X, S n ; Z m ) H n+1 (Y ; Z m ) it follows that f is not trivial on H n+1 (X; Z m ), which proves our claim The Tor functor and the Universal Coefficient Theorem In this section, we explain how to compute H (X; G) in terms of H (X; Z) and G. More generally, given a chain complex C : C n n Cn 1 n 1 1 C 0 0 of free abelian groups and G an abelian group, we aim to compute H (C ; G) := H (C G) in terms of H (C ; Z) and G. The answer is provided by the following result: Theorem (Universal Coefficient Theorem) For each n, there are natural short exact sequences: 0 H n (C ) G H n (C ; G) Tor(H n 1 (C ), G) 0. (0.1.3) Naturality here means that if C C is a chain map, then there is an induced map of short exact sequences with commuting squares. Moreover, these short exact sequences split, but not naturally. 5

6 In particular, if C = C (X, A) is the relative singular chain complex of a pair (X, A), then there are natural short exact sequences 0 H n (X, A) G H n (X, A; G) Tor(H n 1 (X, A), G) 0. (0.1.4) Naturality is with respect to maps of pairs (X, A) f (Y, B). The exact sequence (0.1.4) splits, but not naturally. Indeed, if we assume that A = B =, then we have splittings H n (X; G) = ( H n (X) G ) Tor(H n 1 (X), G), H n (Y ; G) = ( H n (Y ) G ) Tor(H n 1 (Y ), G). If these splittings were natural, and f induces the trivial map f = 0 on H ( ; Z) then f induces the trivial map on H ( ; G), for any coefficient group G. But this is in contradiction with Example Let us next explain the Tor functor appearing in the statement of the universal coefficient theorem. Definition A free resolution of an abelian group H is an exact sequence: with each F n a free abelian group. f 2 f 1 f 0 F 2 F1 F0 H 0, Given an abelian group G, from a free resolution F of H, we obtain a modified chain complex: F G : F 2 G F 1 G F 0 G 0. We define Moreover, the following holds: Tor n (H, G) := H n (F G). (0.1.5) Lemma For any two free resolutions F and F of H there are canonical isomorphisms H n (F G) = H n (F G) for all n. Thus, Tor n (H, G) is independent of the free resolution F of H used for its definition. Proposition For any abelian group H, we have that and Tor n (H, G) = 0 if n > 1, (0.1.6) Tor 0 (H, G) = H G. (0.1.7) Proof. Indeed, given an abelian group H, take F 0 to be the free abelian group on a set of f 0 generators of H to get F 0 H 0. Let F 1 := ker(f 0 ), and note that F 1 is a free (and abelian) group, as it is a subgroup of a free abelian group F 0. Let F 1 F 0 be the inclusion map. Then 0 F 1 F 0 H 0 is a free resolution of H. Thus, Tor n (H, G) = 0 if n > 1. Moreover, it follows readily that Tor 0 (H, G) = H G. 6

7 Definition In what follows, we adopt the notation: Tor(H, G) := Tor 1 (H, G). Proposition The Tor functor satisfies the following properties: (1) Tor(A, B) = Tor(B, A). (2) Tor( i A i, B) = i Tor(A i, B). (3) Tor(A, B) = 0 if either A or B is free or torsion-free. (4) Tor(A, B) = Tor(Torsion(A), B), where Torsion(A) is the torsion subgroup of A. (5) Tor(Z/nZ, A) = ker(a n A). (6) For a short exact sequence: 0 B C D 0 of abelian groups, there is a natural exact sequence: 0 Tor(A, B) Tor(A, C) Tor(A, D) A B A C A D 0. Proof. (2) Choose a free resolution for i A i as the direct sum of free resolutions for the A i s. (5) The exact sequence 0 Z n Z Z/nZ 0 is a free resolution of Z/nZ. Tensoring with A and dropping the right-most term yields the complex Z A n 1 A Z A 0, which by property (4) of the tensor product is A n A 0. Thus, Tor(Z/nZ, A) = ker(a n A). (3) If A is free, we can choose the free resolution: F 1 = 0 F 0 = A A 0 which implies that Tor(A, B) = 0. On the other hand, if B is free, tensoring the exact sequence 0 F 1 F 0 A 0 with B = Z s gives a direct sum of copies of 0 F 1 F 0 A 0. Hence, it is an exact sequence and so H 1 of this complex is 0. For the torsion free case, see below. (6) Let 0 F 1 F 0 A 0 be a free resolution of A, and tensor it with the short exact sequence 0 B C D 0 to get a commutative diagram: F 1 B F 1 C F 1 D 0 0 F 0 B F 0 C F 0 D

8 Rows are exact since tensoring with a free group preserves exactness. Thus we get a short exact sequence of chain complexes. Recall now that for any short exact sequence of chain complexes 0 B C D 0, there is an associated long exact sequence of homology groups H n (B ) H n (C ) H n (D ) H n 1 (B )... So in our situation, with B = F B, C = F C and D = F D, we obtain the homology long exact sequence: 0 H 1 (F B) H 1 (F C) H 1 (F D) H 0 (F B) H 0 (F C) H 0 (F D) 0 Since H 1 (F B) = Tor(A, B) and H 0 (F B) = A B, the above long exact sequence reduces to: 0 Tor(A, B) Tor(A, C) Tor(A, D) A B A C A D 0. (1) Apply (6) to a free resolution 0 F 1 F 0 B 0 of B, and get a long exact sequence: 0 Tor(A, F 1 ) Tor(A, F 0 ) Tor(A, B) A F 1 A F 0 A B 0. Because F 1, F 0 are free, by (3) we have that Tor(A, F 1 ) = Tor(A, F 0 ) = 0, so the long exact sequence becomes: 0 Tor(A, B) A F 1 A F 0 A B 0. Also, by definition of Tor, we have a long exact sequence: So we get a diagram: 0 Tor(B, A) F 1 A F 0 A B A 0. 0 Tor(A, B) A F 1 A F 0 A B 0 φ 0 Tor(B, A) F 1 A F 0 A B A 0 with the arrow labeled φ defined as follows. The two squares on the right commute since is naturally commutative. Hence, there exists φ : Tor(A, B) Tor(B, A) which makes the left square commutative. Moreover, by the 5-lemma, we get that φ is an isomorphism. We can now prove the torsion free case of (3). Assume that B is torsion free. Let 0 F 1 F 0 A 0 be a free resolution of A. The claim about the vanishing of Tor(A, B) is equivalent to the injectivity of the map f id B : F 1 B F 0 B. Assume i x i b i ker(f id B ). So i f(x i) b i = 0 F 1 B. In other words, i f(x i) b i can be reduced to zero by a finite number of applications of the defining relations for tensor products. Only a finite number of elements of B, generating a finitely generated subgroup B 0 of B, are involved in 8 f

9 this process, so in fact i x i b i ker(f id B0 ). But B 0 is finitely generated and torsion free, hence free, so Tor(A, B 0 ) = 0. Thus i x i b i = 0, which proves the claim. The case when A is torsion free follows now by using (1) to reduce to the previous case. (4) Apply (6) to the short exact sequence: 0 Torsion(A) A A/Torsion(A) 0 to get: 0 Tor(G, Torsion(A)) Tor(G, A) Tor(G, A/Torsion(A)) Because A/Torsion(A) is torsion free, Tor(G, A/Torsion(A)) = 0 by (3), so: Tor(G, Torsion(A)) Tor(G, A) Now by (1), we get that Tor(A, G) Tor(Torsion(A), G). Remark It follows from (5) that Tor(Z/nZ, Z/mZ) = Z (n, m)z = Z/nZ Z/mZ, where (n, m) is the greatest common divisor of n and m. More generally, if A and B are finitely generated abelian groups, then Tor(A, B) = Torsion(A) Torsion(B) (0.1.8) where Torsion(A) and Torsion(B) are the torsion subgroups of A and B respectively. Let us conclude with some examples: Example Suppose G = Q, then Tor(H n 1 (X), Q) = 0, so H n (X; Q) H n (X) Q. It follows that the n-th Betti number of X is given by b n (X) := rkh n (X) = dim Q H n (X; Q). Example Suppose X = T 2, and G = Z/4. Recall that H 1 (T 2 ) = Z 2. So: H 0 (T 2 ; Z/4) = H 0 (T 2 ) Z/4 = Z/4 H 1 (T 2 ; Z/4) = ( H 1 (T 2 ) Z/4 ) Tor(H 0 (T 2 ), Z/4) = Z 2 Z/4 = (Z/4) 2 H 2 (T 2 ; Z/4) = ( H 2 (T 2 ) Z/4 ) Tor(H 1 (T 2 ), Z/4) = Z/4. Example Suppose X = K is the Klein bottle, and G = Z/4. Recall that H 1 (K) = Z Z/2, and H 2 (K) = 0, so: H 2 (K; Z/4) = ( H 2 (K) Z/4 ) Tor(H 1 (K), Z/4) = Tor(Z, Z/4) Tor(Z/2, Z/4) = 0 Z/2 = Z/2. 9