Lemma 14.31.8. Let $f : X \to Y$ be a homomorphism of simplicial abelian groups which is termwise surjective and induces a quasi-isomorphism on associated chain complexes. Then $f$ is a trivial Kan fibration of simplicial sets.

**Proof.**
Consider a commutative solid diagram

as in Definition 14.30.1. The map $a$ corresponds to $x_0, \ldots , x_ n \in X_{n - 1}$ satisfying $d_ i x_ j = d_{j - 1} x_ i$ for $i < j$. The map $b$ corresponds to an element $y \in Y_ n$ such that $d_ iy = f(x_ i)$. Our task is to produce an $x \in X_ n$ such that $d_ ix = x_ i$ and $f(x) = y$.

Since $f$ is termwise surjective we can find $x \in X_ n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_ i$ by $x_ i - d_ ix$. Then we see that we may assume $y = 0$. In particular $f(x_ i) = 0$. In other words, we can replace $X$ by $\mathop{\mathrm{Ker}}(f) \subset X$ and $Y$ by $0$. This works, because by Homology, Lemma 12.13.6 the homology of the chain complex associated to $\mathop{\mathrm{Ker}}(f)$ is zero and hence $\mathop{\mathrm{Ker}}(f) \to 0$ induces a quasi-isomorphism on associated chain complexes.

Since $X$ is a Kan complex (Lemma 14.31.6) we can find $x \in X_ n$ with $d_ i x = x_ i$ for $i = 0, \ldots , n - 1$. After replacing $x_ i$ by $x_ i - d_ ix$ for $i = 0, \ldots , n$ we may assume that $x_0 = x_1 = \ldots = x_{n - 1} = 0$. In this case we see that $d_ i x_ n = 0$ for $i = 0, \ldots , n - 1$. Thus $x_ n \in N(X)_{n - 1}$ and lies in the kernel of the differential $N(X)_{n - 1} \to N(X)_{n - 2}$. Here $N(X)$ is the normalized chain complex associated to $X$, see Section 14.23. Since $N(X)$ is quasi-isomorphic to $s(X)$ (Lemma 14.23.9) and thus acyclic we find $x \in N(X_ n)$ whose differential is $x_ n$. This $x$ answers the question posed by the lemma and we are done. $\square$

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