Lemma 14.31.9. Let f : X \to Y be a map of simplicial abelian groups. If f is a homotopy equivalence of simplicial sets, then f induces a quasi-isomorphism of associated chain complexes.
Proof. In this proof we will write H_ n(Z) = H_ n(s(Z)) = H_ n(N(Z)) when Z is a simplicial abelian group, with s and N as in Section 14.23. Let \mathbf{Z}[X] denote the free abelian group on X viewed as a simplicial set and similarly for \mathbf{Z}[Y]. Consider the commutative diagram
of simplicial abelian groups. Since taking the free abelian group on a set is a functor, we see that the horizontal arrow is a homotopy equivalence of simplicial abelian groups, see Lemma 14.28.4. By Lemma 14.27.2 we see that H_ n(g) : H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y]) is bijective for all n \geq 0.
Let \xi \in H_ n(Y). By definition of N(Y) we can represent \xi by an element y \in N(Y_ n) whose boundary is zero. This means y \in Y_ n with d^ n_0(y) = \ldots = d^ n_{n - 1}(y) = 0 because y \in N(Y_ n) and d^ n_ n(y) = 0 because the boundary of y is zero. Denote 0_ n \in Y_ n the zero element. Then we see that
is an element with d^ n_0(\tilde y) = \ldots = d^ n_{n - 1}(\tilde y) = 0 and d^ n_ n(\tilde y) = 0. Thus \tilde y is in N(\mathbf{Z}[Y])_ n has boundary 0, i.e., \tilde y determines a class \tilde\xi \in H_ n(\mathbf{Z}[Y]) mapping to \xi . Because H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y]) is bijective we can lift \tilde\xi to a class in H_ n(\mathbf{Z}[X]). Looking at the commutative diagram above we see that \xi is in the image of H_ n(X) \to H_ n(Y).
Let \xi \in H_ n(X) be an element mapping to zero in H_ n(Y). Exactly as in the previous parapgraph we can represent \xi by an element x \in N(X_ n) whose boundary is zero, i.e., d^ n_0(x) = \ldots = d^ n_{n - 1}(x) = d^ n_ n(x) = 0. In particular, we see that [x] - [0_ n] is an element of N(\mathbf{Z}[X])_ n whose boundary is zero, whence defines a lift \tilde\xi \in H_ n(\mathbf{Z}[x]) of \xi . The fact that \xi maps to zero in H_ n(Y) means there exists a y \in N(Y_{n + 1}) whose boundary is f_ n(x). This means d^{n + 1}_0(y) = \ldots = d^{n + 1}_ n(y) = 0 and d^{n + 1}_{n + 1}(y) = f(x). However, this means exactly that z = [y] - [0_{n + 1}] is in N(\mathbf{Z}[y])_{n + 1} and
This proves that \tilde\xi maps to zero in H_ n(\mathbf{Z}[y]). As H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y]) is bijective we conclude \tilde\xi = 0 and hence \xi = 0. \square
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