Homology theory

Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - 2/15/2011 Tensor products and chain complexes Math 757 with coefficients in G The Tor functor Homology theory Lecture 21 - 2/16/2011

Lecture 22 - 2/18/2011 Universal coefficient theorem

February 21, 2011 Homology theory Review of tensor products Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - 2/15/2011 Tensor products and chain complexes Definition 132 (Bilinear function) Homology with coefficients in G The Tor functor A, B, C abelian groups Lecture 21 - 2/16/2011 ϕ : A × B → C Lecture 22 - 2/18/2011 Universal coefficient is bilinear if theorem ϕ(a1 + a2, b) = ϕ(a1, b) + ϕ(a2, b) and ϕ(a, b1 + b2) = ϕ(a, b1) + ϕ(a, b2) Homology theory Definition 133 ( of abelian groups)

Lecture 19 - A, B abelian groups. The tensor product of A and B is the group 2/14/2011 Review of tensor Z[A × B] products A ⊗ B = Lecture 20 - * + 2/15/2011 (a, b1 + b2) − (a, b1) − (a, b2) Tensor products and chain (a + a , b) − (a , b) − (a , b) complexes 1 2 1 2 Homology with coefficients in G The Tor functor The class of (a, b) in A ⊗ B is written a ⊗ b. Lecture 21 - 2/16/2011 Note: Lecture 22 - 2/18/2011 • {(a, b)} generates Z[A × B] Universal coefficient • so {a ⊗ b} generates A ⊗ B theorem • but just as general element of Z[A × B] is of the form

k X ni (ai , bi ) i=1 • general element of A ⊗ B is of the form k X ni · ai ⊗ bi i=1 Homology theory

Lecture 19 - Lemma 134 (Universal property of the tensor product) 2/14/2011 Review of tensor products A, B abelian groups. Let Lecture 20 - 2/15/2011 Tensor products κ : A × B → A ⊗ B and chain complexes Homology with coefficients in G be the map κ(a, b) = a ⊗ b. The Tor functor Lecture 21 - • Then κ is bilinear 2/16/2011 • Lecture 22 - for any bilinear ϕ : A × B → C 2/18/2011 0 Universal • there exists a unique homomorphism ϕ : A ⊗ B → C coefficient theorem • such that ϕ = ϕ0 ◦ κ. ϕ A × B / C J < JJ JJ κ JJ 0 JJ$ ∃!ϕ A ⊗ B Homology theory

Lecture 19 - Lemma 135 (Tensor product of homomorphisms) 2/14/2011 Review of tensor products Given homomorphisms of abelian groups Lecture 20 - 2/15/2011 f : A → A0 Tensor products and chain complexes 0 Homology with g : B → B coefficients in G The Tor functor Lecture 21 - There is unique homomorphism 2/16/2011 0 0 Lecture 22 - f ⊗ g : A ⊗ B → A ⊗ B 2/18/2011 Universal coefficient theorem satisfying (f ⊗ g)(a ⊗ b) = f (a) ⊗ g(b)

Idea of proof. Apply universal property to the map ϕ : A × B → A0 ⊗ B0 given by ϕ(a, b) = f (a) ⊗ g(b) Homology theory Proposition 136 (Properties of the tensor product)

Lecture 19 - 1 2/14/2011 Review of tensor (f2 ◦ f1) ⊗ (g2 ◦ g1) = (f2 ⊗ g2) ◦ (f1 ⊗ g1) products Lecture 20 - 2/15/2011 That is, the following diagram commutes Tensor products and chain complexes f1⊗g1 f2⊗g2 Homology with A1 ⊗ B1 / A2 ⊗ B2 / A3 ⊗ B3 coefficients in G 3 The Tor functor (f ◦f )⊗(g ◦g ) Lecture 21 - 2 1 2 1 2/16/2011 Lecture 22 - 2 2/18/2011 ∼ Universal (⊕αAα) ⊗ B = ⊕α(Aα ⊗ B) coefficient theorem 3 A ⊗ B =∼ B ⊗ A via the map a ⊗ b 7→ b ⊗ a

4 Z ⊗ A =∼ A via the map 1 ⊗ a 7→ a

5 For sets S and T

Z[S] ⊗ Z[T ] =∼ Z[S × T ] via the map s ⊗ t 7→ (s, t) Homology theory Proposition 137 If A0 ⊂ A and B0 ⊂ B are abelian groups then Lecture 19 - 2/14/2011 Review of tensor A ⊗ B products 0 0 ∼ (A/A ) ⊗ (B/B ) = 0 0 Lecture 20 - iA0 ⊗ 1B (A ⊗ B) + 1A ⊗ iB0 (A ⊗ B ) 2/15/2011 Tensor products and chain complexes Homology with coefficients in G Proof. The Tor functor Lecture 21 - 2/16/2011 • Let 0 Lecture 22 - q1 : A → A/A 2/18/2011 Universal 0 coefficient q2 : B → B/B theorem be the quotient maps • Then by Lemma 135 we have

0 0 q1 ⊗ q2 : A ⊗ B → (A/A ) ⊗ (B/B )

0 0 0 • Claim 1: A ⊗ B ⊂ ker q1 ⊗ q2. If a ∈ A and b ∈ B then

0 0 q1 ⊗ q2(a ⊗ b) = q1(a ) ⊗ q2(b)

= 0 ⊗ q2(b) = 0 Homology theory Proof of Proposition 137 (continued).

Lecture 19 - 2/14/2011 Review of tensor 0 products q1 : A → A/A Lecture 20 - 0 2/15/2011 q2 : B → B/B Tensor products and chain complexes Homology with coefficients in G • Similarly A ⊗ B0 ⊂ ker q ⊗ q . The Tor functor 1 2 0 0 Lecture 21 - • Let K = i 0 ⊗ 1 (A ⊗ B) + 1 ⊗ i 0 (A ⊗ B ) 2/16/2011 A B A B Lecture 22 - • q1 ⊗ q2(K) = 0 so q1 ⊗ q2 induces homomorphism 2/18/2011 Universal coefficient 0 0 theorem Q :(A ⊗ B)/K → (A/A ) ⊗ (B/B )

• Define r :(A/A0) × (B/B0) → (A ⊗ B)/K by setting r([a], [b]) = [a ⊗ b] • r is well-defined and bilinear. Homology theory Proof of Proposition 137 (continued).

Lecture 19 - 2/14/2011 Review of tensor 0 products q1 : A → A/A Lecture 20 - 0 2/15/2011 q2 : B → B/B Tensor products and chain complexes Homology with coefficients in G • By universal propoerty get homomorphism The Tor functor

Lecture 21 - 0 0 2/16/2011 R :(A/A ) ⊗ (B/B ) → (A ⊗ B)/K Lecture 22 - 2/18/2011 Universal with coefficient theorem R([a] ⊗ [b]) = [a ⊗ b] • QR = 1 and RQ = 1 • Hence

0 0 ∼ A ⊗ B (A/A ) ⊗ (B/B ) = 0 0 iA0 ⊗ 1B (A ⊗ B) + 1A ⊗ iB0 (A ⊗ B ) Homology theory

Lecture 19 - Example 138 2/14/2011 Review of tensor products Lecture 20 - Z ⊗ Z 2/15/2011 ∼ Tensor products (Z/nZ) ⊗ (Z/mZ) = and chain nZ ⊗ Z + Z ⊗ mZ complexes Homology with coefficients in G ∼ Z The Tor functor = Lecture 21 - nZ + mZ 2/16/2011 ∼ Lecture 22 - = Z/ gcd(n, m)Z 2/18/2011 Universal coefficient theorem For example (Z/6Z) ⊗ (Z/6Z) =∼ Z/6Z

(Z/15Z) ⊗ (Z/12Z) =∼ Z/3Z

(Z/9Z) ⊗ (Z/8Z) =∼ Z/Z = 0 Homology theory Right exactness of tensor product Lecture 19 - 2/14/2011 Review of tensor Proposition 139 (Right exactness of tensor product) products Lecture 20 - If 2/15/2011 α β Tensor products and chain A −→ B −→ C → 0 complexes Homology with coefficients in G is exact then for all D The Tor functor Lecture 21 - α⊗1D β⊗1D 2/16/2011 A ⊗ D −−−→ B ⊗ D −−−→ C ⊗ D → 0

Lecture 22 - 2/18/2011 is exact. Universal coefficient theorem Proof of Proposition 139.

• Define B ⊗ D ϕ : C × D → α ⊗ 1D (A ⊗ D) by setting ϕ(β(b), d) = [b ⊗ d] • Claim: ϕ is well-defined Homology theory Proof of Proposition 139 (continued).

Lecture 19 - 2/14/2011 • Suppose β(b) = β(b0) Review of tensor products • Then b − b0 ∈ ker β thus there is a ∈ A with α(a) = b − b0 Lecture 20 - 2/15/2011 • Tensor products Thus and chain complexes 0 0 Homology with b ⊗ d − b ⊗ d = (b − b ) ⊗ d coefficients in G The Tor functor = α(a) ⊗ d Lecture 21 - 2/16/2011 = α ⊗ 1D (a ⊗ d) Lecture 22 - 2/18/2011 Universal • so ϕ is well-defined. coefficient theorem • Claim 2: ϕ is bilinear

ϕ(β(b) + β(b0), d) = ϕ(β(b + b0), d) = [(b + b0) ⊗ d] = [b ⊗ d + b0 ⊗ d] = [b ⊗ d] + [b0 ⊗ d] = ϕ(β(b), d) + ϕ(β(b0), d) Homology theory Proof of Proposition 139 (continued).

Lecture 19 - 2/14/2011 • Review of tensor products 0 0 Lecture 20 - ϕ(β(b), d + d ) = [b ⊗ (d + d )] 2/15/2011 0 Tensor products = [b ⊗ d + b ⊗ d ] and chain complexes 0 Homology with = ϕ(β(b), d) + ϕ(β(b), d ) coefficients in G The Tor functor Lecture 21 - • Universal property of tensor product gives 2/16/2011 Lecture 22 - B ⊗ D 2/18/2011 Φ: C ⊗ D → Universal coefficient α ⊗ 1D (A ⊗ D) theorem

• Claim 3: β ⊗ 1D (α ⊗ 1D (A ⊗ D)) = 0

β ⊗ 1D (α ⊗ 1D (a ⊗ d)) = (β ◦ α) ⊗ (1D ◦ 1D )(a ⊗ d) = βα(a) ⊗ d = 0 ⊗ d = 0 Homology theory

Lecture 19 - 2/14/2011 Review of tensor Proof of Proposition 139 (continued). products Lecture 20 - 2/15/2011 • So Tensor products and chain β ⊗ 1D : B ⊗ D → C ⊗ D complexes Homology with coefficients in G induces The Tor functor B ⊗ D Lecture 21 - Θ: → C ⊗ D 2/16/2011 α ⊗ 1D (A ⊗ D) Lecture 22 - 2/18/2011 • ΘΦ(β(b) ⊗ d) = Θ([b ⊗ d]) = β ⊗ d Universal coefficient theorem • ΦΘ([b ⊗ d]) = Φ(β(b) ⊗ d) = [b ⊗ d] • So Φ is an isomorphism establishing exactness at B ⊗ D.

• Surjectivity of β gives surjectivity of β ⊗ 1D establishing exactness at C ⊗ D. Homology theory

Lecture 19 - 2/14/2011 Review of tensor Example 140 (Tensor product is not left exact) products Lecture 20 - 2/15/2011 • Consider the Tensor products and chain complexes ×5 Homology with 0 → Z −−→ Z → Z/5Z coefficients in G The Tor functor Lecture 21 - • Tensor with Z/5Z to get 2/16/2011 Lecture 22 - (×5)⊗1 2/18/2011 0 → Z ⊗ (Z/5Z) −−−−−→ Z ⊗ (Z/5Z) → Z/5Z ⊗ Z/5Z Universal coefficient theorem • which becomes

∼ 0 → Z/5Z −→0 Z/5Z −→= Z/5Z

• No longer exact at leftmost Z/5Z Homology theory Tensor products and chain Lecture 19 - 2/14/2011 Review of tensor complexes products Given a (C, ∂) Lecture 20 - 2/15/2011 Tensor products ∂n+1 ∂n ∂n−1 ∂n−2 and chain complexes ··· −−→ Cn −→ Cn−1 −−−→ Cn−2 −−−→· · · Homology with coefficients in G The Tor functor and an G we get a new chain complex (C ⊗ G, ∂ ⊗ 1G ). Lecture 21 - 2/16/2011 ∂n+1⊗1G ∂n⊗1G ∂n−1⊗1G ∂n−2⊗1G Lecture 22 - ··· −−−−−→ Cn ⊗ G −−−−→ Cn−1 ⊗ G −−−−−→ Cn−2 ⊗ G −−−−−→· · · 2/18/2011 Universal coefficient theorem Lemma 141 If C is a chain complex then C ⊗ G is a chain complex.

Proof.

(∂n ⊗ 1G ) ◦ (∂n−1 ⊗ 1G ) = (∂n ◦ ∂n−1) ⊗ 1G = 0 ⊗ 1G = 0 Homology theory

Lemma 142 Lecture 19 - 2/14/2011 Review of tensor If products f : C → D Lecture 20 - 2/15/2011 Tensor products is a chain map then and chain complexes Homology with coefficients in G f ⊗ 1G : C ⊗ G → D ⊗ G The Tor functor Lecture 21 - 2/16/2011 is a chain map.

Lecture 22 - 2/18/2011 Universal Proof. coefficient theorem f : C → D is a chain map so ∂f = f ∂

(f ⊗ 1G ) ◦ (∂ ⊗ 1G ) = (f ◦ ∂) ⊗ 1G

= (∂ ◦ f ) ⊗ 1G

= (∂ ⊗ 1G ) ◦ (f ⊗ 1G ) Homology theory Lemma 143 Lecture 19 - 2/14/2011 If Review of tensor products T : C → D Lecture 20 - 2/15/2011 is a chain homotopy between chain maps f , g : C → D then Tensor products and chain complexes Homology with T ⊗ 1G : C ⊗ G → D ⊗ G coefficients in G The Tor functor Lecture 21 - is a chain homotopy between chain maps f ⊗ 1G and g ⊗ 1G 2/16/2011

Lecture 22 - 2/18/2011 Proof. Universal coefficient theorem T : C → D is a chain homotopy so ∂T + T ∂ = g − f

(∂ ⊗ 1) ◦ (T ⊗ 1) + (T ⊗ 1) ◦ (∂ ⊗ 1) = (∂T ) ⊗ 1 + (T ∂) ⊗ 1 = (∂T + T ∂) ⊗ 1 = (g − f ) ⊗ 1 = g ⊗ 1 − f ⊗ 1 Homology theory Homology with coefficients in G Lecture 19 - 2/14/2011 Review of tensor products Definition 144 (Homology with coefficients in G) Lecture 20 - 2/15/2011 Given a chain complex C let Tensor products and chain complexes Homology with Hn(C; G) = Hn(C ⊗ G) coefficients in G The Tor functor Lecture 21 - If X is a space let 2/16/2011

Lecture 22 - 2/18/2011 Hn(X ; G) = Hn(C(X ) ⊗ G) Universal coefficient theorem

H˜n(X ; G) = Hn(C˜(X ) ⊗ G)

CW CW Hn (X ; G) = Hn(C (X ) ⊗ G) For (X , A) a topological pair let

Hn(X , A; G) = Hn(C(X , A) ⊗ G) Homology theory Example 145 (Homology of RP4)

Lecture 19 - 4 2/14/2011 Standard CW chain complex for RP is Review of tensor products ×2 ×0 ×2 ×0 ×0 Lecture 20 - · · · → 0 → Z −−→ Z −−→ Z −−→ Z −−→ Z −−→ 0 2/15/2011 Tensor products and chain CW 4 complexes Tensoring with Z2 we get C (RP ) ⊗ Z2 Homology with coefficients in G The Tor functor ×0 ×0 ×0 ×0 ×0 · · · → 0 → Z2 −−→ Z2 −−→ Z2 −−→ Z2 −−→ Z2 −−→ 0 Lecture 21 - 2/16/2011

Lecture 22 - So 2/18/2011  CW 4 Z2, 0 ≤ n ≤ 4 Universal H (RP ; Z ) =∼ coefficient n 2 theorem 0, otherwise Tensoring with Q we get C CW(RP4) ⊗ Q

· · · → 0 → Q −−→×2 Q −−→×0 Q −−→×2 Q −−→×0 Q −−→×0 0 ∼= ∼= So  Q, n = 0 H CW(RP4; Q) =∼ n 0, n 6= 0 Homology theory

Lecture 19 - 2/14/2011 Note that C ⊗ Z = C so Review of tensor products ∼ Lecture 20 - Hn(C; Z) = Hn(C) 2/15/2011 Tensor products and chain complexes Why consider homology with other coefficent groups? Homology with coefficients in G The Tor functor Lecture 21 - 1 If G is a right R-module then Hn(C; G) is a right R-module 2/16/2011 2 In particular if G is a field then H (C; G) is a vector space over Lecture 22 - n 2/18/2011 G. Universal coefficient theorem 3 Computation of Hn(C; Z2) are often easier.

What advantages does the coefficient group Z have? 1 Homology groups with coefficients in Z are the strongest topological invariants Homology theory

Lecture 19 - Homology with coefficients in Z are “universal” since they determine 2/14/2011 Review of tensor homology with coefficients in G for all abelian groups G. products Lecture 20 - 2/15/2011 Theorem 146 (Universal coefficient theorem) Tensor products and chain complexes For C a chain complex there is a split short exact sequence Homology with coefficients in G The Tor functor 0 → Hn(C) ⊗ G → Hn(C; G) → Tor(Hn−1(C), G) → 0. Lecture 21 - 2/16/2011

Lecture 22 - This sequence is natural in that if C → D is a chain map then we get 2/18/2011 commutative Universal coefficient theorem 0 / Hn(C) ⊗ G / Hn(C; G) / Tor(Hn−1(C), G) / 0

     0 / Hn(D) ⊗ G / Hn(D; G) / Tor(Hn−1(D), G) / 0 Homology theory The Tor functor Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - 2/15/2011 Definition 147 (Free ) Tensor products and chain complexes A free resolution of the abelian group A is an exact sequence Homology with coefficients in G The Tor functor · · · → F2 → F1 → F0 → A → 0 Lecture 21 - 2/16/2011 Lecture 22 - where each Fi is free abelian. 2/18/2011 Universal coefficient theorem Example 148 (Two free resolutions of Z5)

×5 · · · → 0 → 0 → Z −−→ Z → Z5 → 0

×1 ×0 ×1 ×0 ×5 ··· −−→ Z −−→ Z −−→ Z −−→ Z −−→ Z → Z5 → 0 Homology theory Theorem 149 (Existence of free resolutions)

Lecture 19 - 2/14/2011 Every abelian group A has a free resolution Review of tensor products Lecture 20 - · · · → 0 → R → F → A → 0 2/15/2011 Tensor products and chain where F and R are free abelian. complexes Homology with coefficients in G The Tor functor Theorem 150 Lecture 21 - 2/16/2011 Every subgroup of a is a free abelian group. Lecture 22 - 2/18/2011 Universal coefficient Proof of Theorem 149. theorem We have a surjection p Z[A] −→ A given by p(a) = a. Let F = Z[A] and R = ker p.

Then p · · · → 0 → R → F −→ A → 0 is a free resolution of A. Homology theory

Lecture 19 - 2/14/2011 Definition 151 (The Tor functor) Review of tensor products Let A and B be abelian groups and Lecture 20 - 2/15/2011 Tensor products ∂3 ∂2 ∂1 and chain ··· −→ F2 −→ F1 −→ F0 → A → 0 complexes Homology with coefficients in G The Tor functor be a free resolution of A. Lecture 21 - 2/16/2011 Let ∂0 : F0 → 0 be the 0 map so we have a chain complex (F , ∂) Lecture 22 - 2/18/2011 ∂3 ∂2 ∂1 ∂0 Universal ··· −→ F −→ F −→ F −→ 0 coefficient 2 1 0 theorem Tensoring with B we get the chain complex F ⊗ B

∂3⊗1B ∂2⊗1B ∂1⊗1B ∂0⊗1B ··· −−−−→ F2 ⊗ B −−−−→ F1 ⊗ B −−−−→ F0 ⊗ B −−−−→ 0

Torn(A, B) = Hn(F ⊗ B) Homology theory

Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - 2/15/2011 Tensor products and chain complexes Homology with coefficients in G The Tor functor Theorem 152 (Tor is well-defined) Lecture 21 - 2/16/2011 Torn(A, B) is independent of the free resolution F of A. Lecture 22 - 2/18/2011 Universal coefficient theorem Homology theory Lemma 153 (Free abelian groups are projective)

Lecture 19 - Suppose we have commutative 2/14/2011 Review of tensor products F Lecture 20 - C ∃ψ CC 0 2/15/2011 ϕ CC Tensor products CC and chain ~  C! complexes 0 00 Homology with M / M / M coefficients in G i j The Tor functor Lecture 21 - 2/16/2011 1 If F is a free abelian group Lecture 22 - 2/18/2011 0 i j 00 Universal 2 M −→ M −→ M is exact coefficient theorem There is a homomorphism ψ making the diagram commute.

Proof.

• Let {eα} be a free basis for F

• jϕ(eα) = 0(eα) = 0 so ϕ(eα) ∈ ker j = Im i 0 0 0 • Thus there is mα ∈ M such that i(mα) = ϕ(eα) 0 • Let ψ(eα) = mα Homology theory We will prove Theorem 152 using the following lemma Lemma 154 Lecture 19 - 0 2/14/2011 If F and F are two free resolutions of A then there is a chain map Review of tensor 0 products f : F → F which is a chain homotopy equivalence. Lecture 20 - 2/15/2011 Tensor products Proof. and chain complexes Homology with coefficients in G • We have The Tor functor Lecture 21 - ∂2 ∂1 ε 0 2/16/2011 ··· / F1 / F0 / A / 0 Lecture 22 - 2/18/2011 1 Universal A coefficient 0 0 theorem ∂ ∂ 0 2 0 1 0 ε  0 ··· / F1 / F0 / A / 0

• Apply Lemma 153 (free ab. grps. are proj.) with ϕ = 1Aε 0 0 • Get ψ : F0 → F0 with ε ψ = 1Aε • Let f0 = ψ 0 • Now apply Lemma 153 with ϕ = f0∂1 to get f1 : F1 → F1 • Continue inductively to get chain map f : F → F 0 Homology theory Proof of Lemma 154 (continued).

Lecture 19 - • Now we will show that the chain homotopy type of f is unique. 2/14/2011 Review of tensor 0 products • That is if g : F → F is another chain map extending Lecture 20 - 1A : A → A then there is a chain homotopy T between f and g 2/15/2011 Tensor products • and chain Let τ = g − f complexes 0 Homology with • Assume inductively that ∂ Tn + Tn−1∂n = τn coefficients in G n+1 The Tor functor • We have (non-commutating diagram) Lecture 21 - 2/16/2011

Lecture 22 - ∂n+1 ∂n 2/18/2011 Fn+1 / Fn / Fn−1 Universal | coefficient Tn | Tn−1 {{ theorem τn+1 || τn {{ || {{ ∂0  } |∂0  } { 0 n+2 0 | n+1 0 { Fn+2 / Fn+1 / Fn

0 • Consider the map (τn+1 − Tn∂n+1): Fn+1 → Fn+1 0 0 0 ∂n+1(τn+1 − Tn∂n+1) = ∂n+1τn+1 − ∂n+1Tn∂n+1 0 = ∂n+1τn+1 − (τn − Tn−1∂n)∂n+1 0 = ∂n+1τn+1 − τn∂n+1 + Tn−1∂n∂n+1 = 0 Homology theory

Lecture 19 - Proof of Lemma 154 (continued). 2/14/2011 Review of tensor products • Apply Lemma 153 (free ab. grps. are proj.) with Lecture 20 - 2/15/2011 ϕ = τn+1 − Tn∂n+1 Tensor products and chain 0 complexes • Let Tn+1 = ψ : Fn+1 → Fn+2 Homology with 0 coefficients in G • Then ∂ Tn+1 = ϕ = τn+1 − Tn∂n+1 The Tor functor n+2 Lecture 21 - • Hence 2/16/2011 0 ∂ Tn+1 + Tn∂n+1 = τn+1 = gn+1 − fn+1 Lecture 22 - n+2 2/18/2011 Universal • We may start the induction in degree −2 where all groups are 0. coefficient theorem • Thus the chain homotopy class of f : F → F 0 is unique. • Similarly we get f 0 : F 0 → F

• 1F : F → F extends 1A : A → A 0 • so f ◦ f is chain homotopic to 1F • That is, f is a chain homotopy equivalence. Homology theory

Now we finish proof that Torn(A, B) is independent of free resolution

Lecture 19 - of A. 2/14/2011 Review of tensor products Proof of Theorem 152 (Tor is well-defined). Lecture 20 - 2/15/2011 0 Tensor products • Let F and F be two free resolutions of A and chain complexes • By Lemma 154 we have a chain homotopy equivalence Homology with coefficients in G The Tor functor f : F → F 0 Lecture 21 - 2/16/2011 0 0 0 Lecture 22 - • Hence there is a chain map f : F → F such that f ◦ f is chain 2/18/2011 Universal homotopic to 1F coefficient theorem 0 • Tensoring with B we get (f ⊗ 1B ) ◦ (f ⊗ 1B ) is chain homotopic to 1F ⊗ B • Thus 0 (f ⊗ 1B )∗ : Hn(F ⊗ B) → Hn(F ⊗ B) is an isomorphism ∼ 0 • Torn(A, B) = Hn(F ⊗ B) = Hn(F ⊗ B) is well-defined Homology theory • As we saw in Lemma 149 every abelian group A has a free Lecture 19 - resolution 2/14/2011 ∂2 ∂1 ε Review of tensor · · · → 0 −→ F1 −→ F0 −→ A → 0 products Lecture 20 - • So we have the chain complexs F 2/15/2011 Tensor products and chain ∂ ∂ ∂ complexes · · · → 0 −→2 F −→1 F −→0 0 Homology with 1 0 coefficients in G The Tor functor • and F ⊗ B Lecture 21 - 2/16/2011

Lecture 22 - ∂2⊗1B ∂1⊗1B ∂0⊗1B 2/18/2011 · · · → 0 −−−−→ F1 ⊗ B −−−−→ F0 ⊗ B −−−−→ 0 Universal coefficient theorem •  F0⊗B , n = 0  Im(∂1⊗1B )  Torn(A, B) = Hn(F ⊗ B) = ker(∂1 ⊗ 1B ), n = 1   0, n 6= 0, 1

• We can say more about Tor0(A, B). Homology theory

Lecture 19 - • We have the exact sequence 2/14/2011 Review of tensor products F −→∂1 F −→ε A → 0 Lecture 20 - 1 0 2/15/2011 Tensor products and chain • Which remains exact after tensoring with B so we get exact complexes Homology with coefficients in G ∂1⊗1B ε⊗1B The Tor functor F1 ⊗ B −−−−→ F0 ⊗ B −−−→ A ⊗ B → 0 Lecture 21 - 2/16/2011 • Hence Lecture 22 - F ⊗ B 2/18/2011 ∼ 0 ∼ Universal Tor0(A, B) = = A ⊗ B coefficient Im(∂1 ⊗ 1B ) theorem • Since Tor1(A, B) is the only new object we define Definition 155

Tor(A, B) = Tor1(A, B) Homology theory

Lecture 19 - • Note that if we have exact 2/14/2011 Review of tensor products ∂1 ε 0 → F1 −→ F0 −→ A → 0 Lecture 20 - 2/15/2011 Tensor products and chain • Then we have exact complexes Homology with coefficients in G ∂1⊗1B ε⊗1B The Tor functor F1 ⊗ B −−−−→ F0 ⊗ B −−−→ A ⊗ B → 0 Lecture 21 - 2/16/2011 • and hence exact Lecture 22 - 2/18/2011 ∂1⊗1B ε⊗1B Universal 0 → ker(∂ ⊗ 1 ) → F ⊗ B −−−−→ F ⊗ B −−−→ A ⊗ B → 0 coefficient 1 B 1 0 theorem • and hence exact

∂1⊗1B ε⊗1B 0 → Tor(A, B) → F1 ⊗ B −−−−→ F0 ⊗ B −−−→ A ⊗ B → 0

• In particular if Tor(A, B) = 0 then tensoring with B preserves exactness. Homology theory Example 156 Lecture 19 - 2/14/2011 Review of tensor Let’s compute Tor(Z60, Z42) products Lecture 20 - • Free resolution F of Z60 2/15/2011 Tensor products and chain ×60 complexes · · · → 0 → Z −−→ Z → Z60 → 0 Homology with coefficients in G The Tor functor • F ⊗ Z42 is Lecture 21 - 2/16/2011 (×60)⊗1 Lecture 22 - 0 → Z ⊗ Z42 −−−−−→ Z ⊗ Z42 → 0 2/18/2011 Universal coefficient theorem • Simplifying ×60 0 → Z42 −−→ Z42 → 0 • Hence Z Z Tor(Z , Z ) =∼ ker(×60) =∼ =∼ =∼ Z 60 42 42Z + 60Z gcd(42, 60)Z 6 Homology theory Proposition 157 (Properties of Tor)

Lecture 19 - 2/14/2011 ∼ Review of tensor 1 Tor(A, B) = Tor(B, A) products ∼ Lecture 20 - 2 Tor(⊕αAα, B) = ⊕α Tor(Aα, B) 2/15/2011 Tensor products 3 Tor(A, B) = 0 if A or B is free or free. and chain complexes ∼ Homology with 4 Tor(A, B) = Tor(Ator, B) where Ator is the of coefficients in G The Tor functor A.  ×n  Lecture 21 - ∼ 2/16/2011 5 Tor(Zn, A) = ker A −−→ A

Lecture 22 - 2/18/2011 6 The short exact sequence Universal coefficient theorem 0 → B → C → D → 0

yeilds a natural exact sequence

0 → Tor(A, B) → Tor(A, C) → Tor(A, D) → A ⊗ B → A ⊗ C → A ⊗ D → 0 Homology theory Proof of Proposition 157.

Lecture 19 - ∼ 2/14/2011 2 Tor(⊕αAα, B) = ⊕α Tor(Aα, B) Review of tensor products • Let Fα be a free resolution of Aα Lecture 20 - • Then ⊕αFα is a free resolution of Aα 2/15/2011 • Tensor products and chain complexes ∼ Homology with Tor(⊕αAα, B) = H1((⊕αFα) ⊗ B) coefficients in G The Tor functor =∼ H1(⊕α(Fα ⊗ B)) Lecture 21 - ∼ 2/16/2011 = ⊕αH1(Fα ⊗ B)

Lecture 22 - =∼ ⊕α Tor(Aα, B) 2/18/2011 Universal coefficient theorem ∼  ×n  5 Tor(Zn, A) = ker A −−→ A

• Use the free resolution of Zn

· · · → 0 → Z −−→×n Z → A → 0

• Tensoring with A simplifies to

(×n) · · · → 0 → A −−−→ A → 0 Homology theory

Lecture 19 - 2/14/2011 Proof of Proposition 157 (continued). Review of tensor products Lecture 20 - 3 Tor(A, B) = 0 if A or B is free (we will address torsion free 2/15/2011 Tensor products later) and chain complexes • Suppose A is free. Homology with coefficients in G • Use the free resolution of A The Tor functor Lecture 21 - · · · → 0 → 0 → A → A → 0 2/16/2011

Lecture 22 - 2/18/2011 • Tor(A, B) = ker(0 → 0) = 0 Universal coefficient • Suppose B is Z theorem • Then tensoring an exact free resolution 0 → F1 → F0 → A → 0 with B remains exact • Suppose B =∼ ⊕αZ • Then tensoring an exact free resolution 0 → F1 → F0 → A → 0 with B is a direct sum of exact sequences which is exact. Homology theory Proof of Proposition 157 (continued).

Lecture 19 - 2/14/2011 6 The short exact sequence Review of tensor products Lecture 20 - 0 → B → C → D → 0 2/15/2011 Tensor products and chain complexes yeilds a natural exact sequence Homology with coefficients in G The Tor functor Lecture 21 - 0 → Tor(A, B) → Tor(A, C) → Tor(A, D) 2/16/2011 → A ⊗ B → A ⊗ C → A ⊗ D → 0 Lecture 22 - 2/18/2011 Universal coefficient theorem • Choose a free resolution F of the form 0 → F1 → F0 → A → 0 • All the terms of F are free so tensoring 0 → B → C → D → 0 with Fn remains exact. • Get short exact sequence of chain complexes

0 → (F ⊗ B) → (F ⊗ C) → (F ⊗ D) → 0

• Apply Snake Lemma to get natural exact sequence above. Homology theory Proof of Proposition 157 (continued).

Lecture 19 - ∼ 2/14/2011 1 Tor(A, B) = Tor(B, A) Review of tensor products • Consider the six term exact sequence from part 6 coming from Lecture 20 - the short exact 2/15/2011 Tensor products 0 → F1 → F0 → B → 0 and chain complexes • Homology with coefficients in G The Tor functor Lecture 21 - 0 → Tor(A, F1) → Tor(A, F0) → Tor(A, B) 2/16/2011 → A ⊗ F1 → A ⊗ F0 → A ⊗ B → 0 Lecture 22 - 2/18/2011 Universal • F0 and F1 are free so by part 3 Tor(A, F1) =∼ Tor(A, F0) =∼ 0 coefficient theorem •

0 → Tor(A, B) → A ⊗ F1 → A ⊗ F0 → A ⊗ B → 0 •

0 / Tor(A, B) / A ⊗ F1 / A ⊗ F0 / A ⊗ B / 0

∼= ∼= ∼=    0 / Tor(B, A) / F1 ⊗ A / F0 ⊗ A / B ⊗ A / 0 Homology theory Proof of Proposition 157 (continued).

Lecture 19 - 1 Tor(A, B) =∼ Tor(B, A) (continued) 2/14/2011 Review of tensor • α β products 0 / Tor(A, B) / A ⊗ F1 / A ⊗ F0 / A ⊗ B / 0 Lecture 20 - 2/15/2011 γ τ ∼= µ ∼= ∼= Tensor products  0  β0   and chain α complexes 0 / Tor(B, A) / F1 ⊗ A / F0 ⊗ A / B ⊗ A / 0 Homology with coefficients in G The Tor functor • We will define a homorphism γ : Tor(A, B) → Tor(B, A) Lecture 21 - preserving commutativity 2/16/2011 • Let x ∈ Tor(A, B) Lecture 22 - 0 2/18/2011 • Claim: τα(x) ∈ Im α Universal 0 coefficient • By commutativity β τα(x) = µβα(x) = µ(0) = 0 theorem • So τα(x) ∈ ker β0 = Im α0 • By injectivity of α0 there is a unique x 0 ∈ Tor(B, A) with α0(x 0) = τα(x) • Set γ(x) = x 0. • γ takes 0 to 0 and sums to sums so it is a homomorphism. • 0 / 0 / Tor(A, B) / A ⊗ F1 / A ⊗ F0

∼= ∼= γ τ ∼= µ ∼=      0 / 0 / Tor(B, A) / F1 ⊗ A / F0 ⊗ A Homology theory

Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - Proof of Proposition 157 (continued). 2/15/2011 Tensor products and chain ∼ complexes 1 Tor(A, B) = Tor(B, A) (continued) Homology with coefficients in G • Add some trivial groups and homomorphisms to get The Tor functor Lecture 21 - 2/16/2011 0 / 0 / Tor(A, B) / A ⊗ F1 / A ⊗ F0

Lecture 22 - ∼= ∼= γ τ ∼= µ ∼= 2/18/2011    Universal   coefficient 0 / 0 / Tor(B, A) / F1 ⊗ A / F0 ⊗ A theorem • Now apply Five Lemma to show γ : Tor(A, B) → Tor(B, A) is an isomorphism. Homology theory Proof of Proposition 157 (continued).

Lecture 19 - 3 Tor(A, B) =∼ 0 if A or B is torsion free. 2/14/2011 Review of tensor • Applying part 1 assume B is torsion free. products • Let Lecture 20 - ∂1 2/15/2011 0 → F1 −→ F0 → A Tensor products and chain complexes be a free resolution of A. Homology with coefficients in G • Get exact The Tor functor ∂1⊗1B Lecture 21 - 0 → Tor(A, B) → F1 ⊗ B −−−−→ F0 ⊗ B 2/16/2011

Lecture 22 - • Claim ∂1 ⊗ 1B is injective. 2/18/2011 P Universal • Suppose fi ⊗ bi ∈ ker ∂1 ⊗ 1B coefficient i theorem P • Then then i (∂1fi ) ⊗ bi = 0 in F0 ⊗ B. • Hence in Z[F0 × B]

X X 0 1 2 0 1 0 2 (∂1fi , bi ) = (fj , bj + bj ) − (fj , bj ) − (fj , bj ) i j X 1 2 0 1 0 2 2 + (fk + fk , bk ) − (fk , bk ) − (fk , bk ) k

• Let B0 ⊂ B be the subgroup generated by the finite set 0 1 2 {bi , bk , bj , bj } Homology theory Proof of Proposition 157 (continued).

Lecture 19 - ∼ 2/14/2011 3 Tor(A, B) = 0 if A or B is torsion free. (continued) Review of tensor products • Hence in Z[F0 × B0] Lecture 20 - 2/15/2011 X X 0 1 2 0 1 0 2 Tensor products (∂1fi , bi ) = (fj , bj + bj ) − (fj , bj ) − (fj , bj ) and chain complexes i j Homology with X 1 2 0 1 0 2 2 coefficients in G + (f + f , b ) − (f , b ) − (f , b ) The Tor functor k k k k k k k k Lecture 21 - 2/16/2011 • Lecture 22 - Therefore in F0 ⊗ B0 2/18/2011 Universal X coefficient (∂1fi ) ⊗ bi = 0 theorem i

• Let B0 ⊂ B is torsion free and finitely generated so free abelian. • Hence ∂ ⊗1 1 B0 F1 ⊗ B0 −−−−→ F0 ⊗ B0 is injective. • Thus in F ⊗ B 1 0 X fi ⊗ bi = 0 i Homology theory

Lecture 19 - 2/14/2011 Review of tensor Proof of Proposition 157 (continued). products Lecture 20 - 2/15/2011 3• Tor(A, B) =∼ 0 if A or B is torsion free. (continued) Tensor products and chain • Hence in Z[F × B ] complexes 1 0 Homology with coefficients in G The Tor functor X X 3 4 5 3 4 3 5 (fi , bi ) = (fn , bn + bn) − (fn , bn) − (fn , bn) Lecture 21 - i n 2/16/2011 X 4 5 3 4 3 4 3 Lecture 22 - + (fm + fk , bm) − (fm, bm) − (fm, bm) 2/18/2011 Universal k coefficient theorem • This equality holds in Z[F1 × B] • Thus in F ⊗ B 1 X fi ⊗ bi = 0 i

• It follows that Tor(A, B) = ker (∂1 ⊗ 1B ) = 0 Homology theory

Lecture 19 - 2/14/2011 Proof of Proposition 157 (continued). Review of tensor products ∼ Lecture 20 - 4 Tor(A, B) = Tor(Ator, B) where Ator is the torsion subgroup of 2/15/2011 Tensor products A. and chain complexes • We have the exact sequence 0 → Ator → A → (A/Ator) → 0 Homology with coefficients in G • Apply part 6 to get exact The Tor functor Lecture 21 - 2/16/2011 0 → Tor(B, Ator) → Tor(B, A) → Tor(B, A/Ator) Lecture 22 - 2/18/2011 → B ⊗ Ator → B ⊗ A → B ⊗ A/Ator → 0 Universal coefficient theorem • A/Ator is torsion free so by part 3 Tor(B, A/Ator) = 0 • Get exact 0 → Tor(B, Ator) → Tor(B, A) → 0

• Thus Tor(B, Ator) =∼ Tor(B, A) • and by part 1 Tor(Ator, B) =∼ Tor(A, B) Homology theory Universal coefficient theorem Lecture 19 - 2/14/2011 Review of tensor Theorem 158 (Universal coefficient theorem) products Lecture 20 - 2/15/2011 If C a chain complex with each Cn free abelian there is a split short Tensor products and chain exact sequence complexes Homology with coefficients in G 0 → Hn(C) ⊗ G → Hn(C; G) → Tor(Hn−1(C), G) → 0. The Tor functor Lecture 21 - 2/16/2011 In particular Lecture 22 - 2/18/2011 ∼  Universal Hn(C; G) = Hn(C) ⊗ G ⊕ Tor(Hn−1(C), G) coefficient theorem The short exact sequence is natural in that if C → D is a chain map then we get commutative

0 / Hn(C) ⊗ G / Hn(C; G) / Tor(Hn−1(C), G) / 0

     0 / Hn(D) ⊗ G / Hn(D; G) / Tor(Hn−1(D), G) / 0 Homology theory Proof of Theorem 158.

Lecture 19 - 2/14/2011 • Let C be a chain complex with each Cn free abelain Review of tensor products • Let Zn = ker ∂n ⊂ Cn be the n-cycles Lecture 20 - 2/15/2011 • Let Bn = Im ∂n+1 ⊂ Cn be the n-boundaries Tensor products and chain complexes • View Z and B as chain complexes. Homology with coefficients in G • Z and B are subgroups of free abelian groups and hence free The Tor functor n n Lecture 21 - abelian. 2/16/2011 • Lecture 22 - Thus we have exact 2/18/2011 Universal coefficient 0 → Zn → Cn → Bn−1 → 0 theorem

• View above as a free resolution of Bn−1 • Get exact

0 → Tor(Bn−1, G) → Zn ⊗ G → Cn ⊗ G → Bn−1 ⊗ G → 0

• Bn−1 is free so Tor(Bn−1, G) = 0 Homology theory Proof of Theorem 158 (continued). Lecture 19 - 2/14/2011 Review of tensor • Bn−1 is free so Tor(Bn−1, G) = 0 products Lecture 20 - • Thus we have exact 2/15/2011 Tensor products and chain complexes 0 → Zn ⊗ G → Cn ⊗ G → Bn−1 ⊗ G → 0 Homology with coefficients in G The Tor functor • And hence exact sequence of chain complexes Lecture 21 - 2/16/2011

Lecture 22 - 0 → Z ⊗ G → C ⊗ G → B ⊗ G → 0 2/18/2011 Universal coefficient • Apply Snake Lemma and fact that ∂ = 0 for Z and B to get theorem exact

κ · · · → Bn ⊗ G −→ Zn ⊗ G → Hn(C ⊗ G) κ → Bn−1 ⊗ G −→ Zn−1 ⊗ G → · · ·

where κ is the connecting homomorphism Homology theory Proof of Theorem 158 (continued).

Lecture 19 - • What is κ? 2/14/2011 Review of tensor products • We have Lecture 20 - 2/15/2011 ∂n⊗1 Tensor products 0 / Zn ⊗ G / Cn ⊗ G / Bn−1 ⊗ G / 0 and chain complexes Homology with coefficients in G ∂n⊗1 The Tor functor   ∂n−1⊗1  Lecture 21 - 0 / Zn−1 ⊗ G / Cn−1 ⊗ G / Bn−2 ⊗ G / 0 2/16/2011

Lecture 22 - 2/18/2011 • Given bn−1 ⊗ g ∈ Bn−1 ⊗ G Universal coefficient theorem • There is cn ∈ Cn such that ∂ncn = bn−1

• Maps to bn−1 ⊗ g now viewed as a element of Cn−1 ⊗ G

• Which is the image of the element bn−1 ⊗ g ∈ Zn−1 ⊗ G

• Thus the connecting homomorphism is κ = in−1 ⊗ 1 where

in−1 : Bn−1 → Zn−1

is inclusion Homology theory Proof of Theorem 158 (continued).

Lecture 19 - • From long exact sequence 2/14/2011 Review of tensor products in⊗1 Lecture 20 - · · · → Bn ⊗ G −−−→ Zn ⊗ G → Hn(C ⊗ G) 2/15/2011 Tensor products in−1⊗1 and chain complexes → Bn−1 ⊗ G −−−−→ Zn−1 ⊗ G → · · · Homology with coefficients in G The Tor functor • Get short exact sequence Lecture 21 - 2/16/2011 Zn ⊗ G Lecture 22 - 0 → → Hn(C ⊗ G) → ker(in−1 ⊗ 1) → 0 2/18/2011 Im(in ⊗ 1) Universal coefficient theorem • By the definition of Hn−1(C) we have a free resolution

in−1 0 → Bn−1 −−→ Zn−1 → Hn−1(C) → 0

• Tensoring with G we get exact

in⊗1 0 → Tor(Hn(C), G) → Bn ⊗ G −−−→ Zn ⊗ G

→ Hn(C) ⊗ G → 0 Homology theory Proof of Theorem 158 (continued).

Lecture 19 - • Apply exactness of 2/14/2011 Review of tensor products in⊗1 Lecture 20 - 0 → Tor(Hn(C), G) → Bn ⊗ G −−−→ Zn ⊗ G 2/15/2011 Tensor products and chain → Hn(C) ⊗ G → 0 complexes Homology with coefficients in G • To short exact sequence The Tor functor Lecture 21 - 2/16/2011 Z ⊗ G 0 → n → H (C ⊗ G) → ker(i ⊗ 1) → 0 Lecture 22 - n n−1 2/18/2011 Im(in ⊗ 1) Universal coefficient theorem • We see that Zn ⊗ G ∼ = Hn(C) ⊗ G Im(in ⊗ 1) • and ∼ ker(in−1 ⊗ 1) = Tor(Hn(C), G) • Conclude with exactness of

0 → Hn(C) ⊗ G → Hn(C; G) → Tor(Hn−1(C), G) → 0. Homology theory

Lecture 19 - 2/14/2011 Review of tensor products Lecture 20 - 2/15/2011 Tensor products and chain complexes Homology with Exercise 159 (HW7) coefficients in G The Tor functor Lecture 21 - • 2/16/2011 Hatcher pg. 155-159 Problems 19, 24, 40 Lecture 22 - • Hatcher pg. 267 Problems 1, 2 2/18/2011 Universal coefficient theorem