The Stacks project

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 $\xi = 0$. $\square$