Susan Marie Cooper

Tuesday, October 31, 2000

Exercise B16 - Tony’s Notes, Exercise 14A, page 27, part 2:

Chain homotopy is an equivalence relation on the chain maps from A_{•} to B_{•} in Ch_{R}.

Let f_{•} and g_{•} be chain maps from A_{•} to B_{•}. Define f_{•} ≈ g_{•} ⇔ f_{•} and g_{•} are chain homotopic. We will show that ≈ is an equivalence relation.

Proof. First we verify that ≈ is reflexive. Let f_{•} : A_{•} → B_{•} be a chain map. Then we have the following commutative diagram.

⋯ → A_{n+1} →^{d_{n+1}} A_{n} →^{d_{n}} A_{n−1} → ⋯
↓ f_{n+1}  ↓ f_{n} ↓ f_{n−1}
⋯ → B_{n+1} →^{d′_{n+1}} B_{n} →^{d′_{n}} B_{n−1} → ⋯

We want to show that f_{•} − f_{•} is null homotopic. Clearly f_{n} − f_{n} is the zero map for all n. Define s_{n}, for each n, to be the R-module homomorphism s_{n} : A_{n} → B_{n+1} such that s_{n}(a_{n}) = 0. Then

(d′_{n+1}s_{n} + s_{n−1}d_{n})(a_{n}) = d′_{n+1}s_{n}(a_{n}) + s_{n−1}d_{n}(a_{n})
= d′_{n+1}(0) + s_{n−1}(d_{n}(a_{n}))
= 0
= (f_{n} − f_{n})(a_{n}).

Thus, by definition, f_{•} − f_{•} is null homotopic, and hence, f_{•} and f_{•} are chain homotopic. Therefore, f_{•} ≈ f_{•} and ≈ is reflexive.

Secondly, we show that ≈ is symmetric. Suppose then that g_{•} : A_{•} → B_{•} is another chain map. Suppose further that f_{•} ≈ g_{•}. Then f_{•}, g_{•} are chain homotopic. Thus, by definition, f_{•} − g_{•} is null homotopic. So there exists R-module homomorphisms s_{n} : A_{n} → B_{n+1}, for all n, such that f_{n} − g_{n} = d′_{n+1}s_{n} + s_{n−1}d_{n}. We want to show that g_{•} − f_{•} is null homotopic. Define, for all n, the R-module homomorphisms α_{n} : A_{n} → B_{n+1} such that α_{n}(a_{n}) = −s_{n}(a_{n}). Then, since f_{n} − g_{n} = d′_{n+1}s_{n} + s_{n−1}d_{n}, we have that g_{n} − f_{n} = d′_{n+1}(−s_{n}) + (−s_{n−1})d_{n}. Thus, by definition, g_{•} − f_{•} is null homotopic, and hence, g_{•} and f_{•} are chain homotopic. Therefore, g_{•} ≈ f_{•} and ≈ is symmetric.

Finally, we show that ≈ is transitive. Thus, we now let h_{•} : A_{•} → B_{•} be another chain map. Suppose that f_{•} ≈ g_{•} and g_{•} ≈ h_{•}. Then f_{•}, g_{•} and g_{•}, h_{•} are two sets of chain homotopic chain maps. Thus, f_{•} − g_{•} and g_{•} − h_{•} are null homotopic. Hence, there exist R-module homomorphisms s_{n}, α_{n} : A_{n} → B_{n+1}, for all n, such that f_{n}−g_{n} = d′_{n+1}s_{n}+s_{n−1}d_{n} and g_{n}−h_{n} = d′_{n+1}α_{n}+α_{n−1}d_{n}. We want to show that f_{•}−h_{•} is null homotopic. For all n, define the R-module homorphisms β_{n} : A_{n} → B_{n+1} by β_{n}(a_{n}) = (s_{n} + α_{n})(a_{n}) = s_{n}(a_{n}) + α_{n}(a_{n}). Then, since f_{n} − g_{n} = d′_{n+1}s_{n} + s_{n−1}d_{n} and g_{n} −h_{n} = d′_{n+1}α_{n} +α_{n−1}d_{n}, we have f_{n} −h_{n} = d′_{n+1}(s_{n} +α_{n})+(s_{n−1} +α_{n−1})d_{n}. Thus, by definition, f_{•} − h_{•} is null homotopic, and hence, f_{•} and h_{•} are chain homotopic. Thus, f_{•} ≈ h_{•} and ≈ is transitive. This completes the proof.