Chain homotopies Tensor product of chain complexes Let R be a commutative ring with unity, and let C and D be chain complexes of R-modules (e.g. of abelian groups ≡ Z-modules). The tensor product of C and D is defined on its components by (cid:77) (C ⊗D) = C ⊗D n i j i+j=n Note that the tensor product on the right is the tensor product of R-modules. The boundary map of C⊗D is inherited from the boundary maps of C and D. Specifically, given c ∈ C and d ∈ D with i+j = n, we can define i j ∂(c⊗d) = c⊗(∂d)+(−1)n(∂c)⊗d Interval chain complex The interval chain complex I is defined by ··· → 0 → 0 → R → R2 where the boundary map R → R2 is defined by r (cid:55)→ (r,−r) for all r ∈ R. It is an interval object in the category of chain complexes of R-modules. Note in particular that if C is a chain complex then (C ⊗R) = (C ⊗R)⊕(C ⊗R2) ∼= C ⊕(C ⊗R2) n+1 n n+1 n n+1 for all n ∈ N, and (C ⊗R) = C ⊗R2. 0 0 Chain homotopies Given chain complexes C,D and chain maps f,g : C → D, a chain homotopy from f to g is a sequence p : C → D n n n+1 of R-module homomorphisms such that, for all n ∈ N, we have ∂p +p ∂ = f −g n+1 n n+1 n+1 and ∂p = f −g . 0 0 0 Proposition 1. Chain homotopies p from f to g correspond with chain maps h : C⊗I → D for which h (c⊗(1,0)) = f (c) and h (c⊗(0,1)) = g (c) n n n n for all n ∈ N and c ∈ C . n Proof. First suppose that p is a chain homotopy from f to g. Recall that (C ⊗R) = C ⊗R2 and (C ⊗R) = (C ⊗R)⊕(C ⊗R2) for all n ∈ N 0 0 n+1 n n+1 Define maps h : (C ⊗I) → D for all n ∈ N as follows: n n n • Let h (c⊗(1,0)) = f (c) and h (c⊗(0,1)) = g (c) for all n ∈ N and c ∈ C ; then extend n n n n n linearly. • Let h (c⊗1) = p (c) for all n ∈ N and c ∈ C ; then extend linearly. n+1 n n All we need to do is prove that h defines a chain map C ⊗I → D. Fix n ∈ N. If n > 0 and c ∈ C , then we have n h (∂(c⊗1)) = h (c⊗(1,−1)−(∂c)⊗1) by definition of ∂ on C ⊗I n n = h (c⊗(1,0))−h (c⊗(0,1))−h ((∂c)⊗1) by R-linearity of h n n n = f (c)−g (c)−p (∂c) by definition of h n n n−1 = ∂p (c) since p is a chain homotopy n = ∂h (c⊗1) by definition of h n+1 and if c ∈ C (for arbitrary n ∈ N, we have n+1 h (∂(c⊗(1,0))) = h (c⊗∂(1,0)+(∂c)⊗(1,0)) by definition of ∂ on C ⊗I n n = f (∂c) since ∂(1,0) = 0 and by definition of h n = ∂f (c) since f is a chain map n+1 = ∂h (c⊗(1,0)) by definition of h n+1 This proves that h ∂ = ∂h for all n ∈ N, so that h defines a chain map C ⊗ I → D. The n n remaining conditions hold by definition. Conversely, suppose we have a chain map h : C ⊗ I → D for which h (c ⊗ (1,0)) = f and n n h (c⊗(0,1)) = g (c) for all n ∈ N. For each n ∈ N, define p : C → D by n n n n n+1 p (c) = h (c⊗1) n n+1 for all c ∈ C . n 2 Then, given c ∈ C , we have n+1 (∂p +p ∂)(c) = ∂h (c⊗1)+h ((∂c)⊗1) by definition of p n+1 n n+2 n+1 = h ∂(c⊗1)+h ((∂c)⊗1) since h is chain map n+1 n+1 = h (∂(c⊗1)+(∂c)⊗1) by R-linearity of h n+1 = h (c⊗(1,−1)−(∂c)⊗1+(∂c⊗1)) by definition of ∂ on C ⊗I n+1 = h (c⊗(1,0))−h (c⊗(0,1)) by R-linearity of h n+1 n+1 = f (c)−g (c) by definition of h n+1 n+1 and likewise, given c ∈ C , we have 0 ∂p (c) = ∂h (c⊗1) = h (∂(c⊗1)) = h (c⊗(1,−1)) = f (c)−g (c) 0 1 0 0 0 0 So p is a chain homotopy. 3