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0.1 Universal Coefficient Theorem for Homology

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0.1 Universal Coefficient Theorem for Homology 0.1 Universal Coefficient Theorem for Homology 0.1.1 Tensor Products Let A; B be abelian groups. Define the abelian group A ⊗ B = ha ⊗ b j a 2 A; b 2 Bi= ∼ (0.1.1) where ∼ is generated by the relations (a + a0) ⊗ b = a ⊗ b + a0 ⊗ b and a ⊗ (b + b0) = 0 a ⊗ b + a ⊗ b . The zero element of A ⊗ B is 0 ⊗ b = a ⊗ 0 = 0 ⊗ 0 = 0A⊗B since, e.g., 0⊗b = (0+0)⊗b = 0⊗b+0⊗b so 0⊗b = 0A⊗B. Similarly, the inverse of an element a⊗b is −(a ⊗ b) = (−a) ⊗ b = a ⊗ (−b) since, e.g., 0A⊗B = 0 ⊗ b = (a + (−a)) ⊗ b = a ⊗ b + (−a) ⊗ b. Lemma 0.1.1. The tensor product satisfies the following universal property which asserts that if ' : A × B ! C is any bilinear map, then there exists a unique map ' : A ⊗ B ! C such that ' = ' ◦ i, where i : A × B ! A ⊗ B is the natural map (a; b) 7! a ⊗ b. i A × B / A ⊗ B ' 9! % C Proof. Indeed, ' : A ⊗ B ! C can be defined by a ⊗ b 7! '(a; b). Proposition 0.1.2. The tensor product satisfies the following properties: (1) A ⊗ B ∼= B ⊗ A via the isomorphism a ⊗ b 7! b ⊗ a. L ∼ L (2) ( i Ai) ⊗ B = i(Ai ⊗ B) via the isomorphism (ai)i ⊗ b 7! (ai ⊗ b)i. (3) A ⊗ (B ⊗ C) ∼= (A ⊗ B) ⊗ C via the isomorphism a ⊗ (b ⊗ c) 7! (a ⊗ b) ⊗ c. ∼ (4) Z ⊗ A = A via the isomorphism n ⊗ a 7! na. ∼ (5) Z=nZ ⊗ A = A=nA via the isomorphism l ⊗ a 7! la. Proof. These are easy to prove by using the above universal property. We sketch a few. (1) The map ' : A×B ! B ⊗A defined by (a; b) 7! b⊗a is clearly bilinear and therefore induces a homomorphism ' : A ⊗ B ! B ⊗ A with a ⊗ b 7! b ⊗ a. Similarly, there is the reverse map : B × A ! A ⊗ B defined by (b; a) 7! a ⊗ b which induces a homomorphism : B ⊗ A ! A ⊗ B with b ⊗ a 7! a ⊗ b. Clearly, ' ◦ = idB⊗A and ◦ ' = idA⊗B and A ⊗ B ∼= B ⊗ A. (4) The map ' : Z × A ! A defined by (n; a) 7! na is a bilinear map and therefore induces a homomorphism ' : Z ⊗ A ! A with n ⊗ a 7! na. Now suppose '(n ⊗ a) = 0. Then na = 0 and n ⊗ a = 1 ⊗ (na) = 1 ⊗ 0 = 0Z⊗A. Thus ' is injective. Moreover, if a 2 A, then '(1 ⊗ a) = a and ' is surjective as well. 1 (5) The map ' : Z=nZ×A ! A=nA defined by (l; a) 7! la is a bilinear map and therefore induces a homomorphism ' : Z=nZ ⊗ A ! A=nA with l ⊗ a 7! la. Now suppose '(l ⊗ a) = Pk Pk Pk la = 0. Then la = i=1 nai and l ⊗a = 1⊗(la) = 1⊗( i=1 nai) = i=1(n⊗ai) = 0Z=nZ⊗A, so ' is injective. Now let a 2 A=nA. Then '(1 ⊗ a) = a and ' 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. p p ∼ Example 0.1.3. Let R = Q[ 2] = fa + b 2 j a; b 2 Qg. 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 0.1.4. If G is an abelian group, then the functor − ⊗ G is right exact, that is, if j j⊗1 A −!i B −! C ! 0 is exact, then A ⊗ G −−−!i⊗1G B ⊗ G −−−!G C ⊗ G ! 0 is exact. Proof. Let c ⊗ g 2 C ⊗ G. Since j is onto, there exists, b 2 B such that j(b) = c. Then (j ⊗ 1G)(b ⊗ g) = c ⊗ g and j ⊗ 1G is onto. Since j◦i = 0, we have (j⊗1G)◦(i⊗1G) = (j◦i)⊗1G = 0 and thus, Image(i⊗1G) ⊆ ker(j⊗1G). It remains to show that ker(j ⊗ 1G) ⊆ Image(i ⊗ 1G). It is enough to show that =∼ : B ⊗ G=Image(i ⊗ 1G) −! C ⊗ G; where is the map induced by j ⊗ 1G. Construct an inverse of , induced from the homo- morphism ' : C × G ! B ⊗ G=Image(i ⊗ 1G) defined by (c; g) 7! b ⊗ g, where j(b) = c. We must show that ' is a well-defined bilinear map and that the induced map ' satisfies ' ◦ = id and ◦ ' = id. If j(b) = j(b0) = c, then b − b0 2 ker j = Image i, so b − b0 = i(a) for some a 2 A. 0 0 Thus, b ⊗ g − b ⊗ g = (b − b ) ⊗ g = i(a) ⊗ g 2 Image(i ⊗ 1G). So ' is well defined. Now '((c + c0; g)) = d ⊗ g where j(d) = c + c0. Since j is surjective, choose b; b0 2 B such that j(b) = c and j(b0) = c0. Then d − (b + b0) 2 ker j = Image i and so there exists a 2 A such that i(a) = d − (b + b0). Thus, '((c + c0; g)) = d ⊗ g = (b + b0) ⊗ g = 2 b ⊗ g + b0 ⊗ g = '(c; g) + '(c0; g) and ' is linear in the first component. For the second component, '(c; g + g0) = b ⊗ (g + g0) = b ⊗ g + b ⊗ g0 = '(c; g) + '(c; g0). Thus, ' is bilinear. Now by the universal property of the tensor product, the bilinear map ' induces a ho- momorphism ' : C ⊗ G ! B ⊗ G=Image(i ⊗ 1G) defined by c ⊗ g 7! '(c; g) = b ⊗ g, where j(b) = c. For c ⊗ g 2 C ⊗ G, ◦ '(c ⊗ g) = (b ⊗ g) = j(b) ⊗ g = c ⊗ g; so ◦ ' = idC⊗G. Similarly, for b ⊗ g 2 B ⊗ G=Image(i ⊗ 1G), ' ◦ (b ⊗ g) = '(j(b) ⊗ g) = '(j(b); g) = b ⊗ g. Thus ' ◦ = id. Remark 0.1.5. Tensoring with a free group is an exact functor. 0.1.2 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 Ci(X; G) = Ci(X) ⊗ G (0.1.2) P consisting of finite formal sums i ηi · σi (σi : ∆i ! X, ηi 2 G), and with boundary maps given by G @i := @i ⊗ idG: G G G Since @i satisfies @i ◦ @i+1 = 0 it follows that @i ◦ @i+1 = 0, so (C∗(X; G);@∗ ) forms in- deed 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 Ci(X; A; G) := Ci(X; G)=Ci(A; G), and reduced homology with G-coefficients via the aug- mented chain complex @G G G G i+1 @i @2 @1 ··· −! Ci(X; G) −! · · · −! C1(X; G) −! C0(X; G) −! G ! 0; P P where ( i ηiσi) = i ηi 2 G. Notice that Hi(X) = Hi(X; Z) by definition. By studying the chain complex with G-coefficients, it follows that ( G i = 0 Hi(pt; G) = 0 i 6= 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 (Dn;Sn) to be ( n G i = 0; n Hi(S ; G) = 0 otherwise: 3 Finally, we can build cellular homology with G-coefficients in the same way, defining G ∼ # i−cells Ci (X) = Hi(Xi;Xi−1; G) = G : The cellular boundary maps are given by: G X i X i−1 di ( ηαeα) = ηαdαβeβ ; α α,β i−1 i−1 where dαβ is as before the degree of a map ∆αβ : S ! S . This follows from the easy k k k k fact that if f : S ! S has degree m, then f∗ : Hk(S ; G) ' G ! Hk(S ; G) ' G is the multiplication by m. As it is the case for integers, we get an isomorphism CW Hi (X; G) ' Hi(X; G) for all i. n Example 0.1.6. We compute Hi(RP ; Z2) using the cellular homology with Z2-coefficients. Notice that over Z the cellular boundary maps are di = 0 or di = 2 depending on the parity of i, and therefore with Z2-coefficients all of boundary maps vanish. Therefore, ( n Z2 0 ≤ i ≤ n Hi(RP ; Z2) = 0 otherwise: Example 0.1.7. Fix n > 0 and let g : Sn ! Sn be a map of degree m. Define the CW complex n n+1 X = S [g e ; where the (n + 1)-cell en+1 is attached to Sn via the map g. Let f be the quotient map f : X ! X=Sn. Define Y = X=Sn = Sn+1. The homology of X can be easily computed by using the cellular chain complex: dn+2 dn+1 dn d1 d1 d0 0 / Z m / Z / ::: / 0 / Z / 0 Therefore, 8 i = 0 <>Z Hi(X; Z) = Zm i = n :>0 otherwise: Moreover, as Y = Sn+1, we have ( Z i = 0; n + 1 Hi(Y ; Z) = 0 otherwise: It follows that f induces the trivial homomorphisms in homology with Z-coefficients (except in degree zero, where f∗ is the identity).
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