The Stacks project

Proof. Consider a commutative solid diagram

as in Definition 14.31.1. The map $a$ corresponds to $x_0, \ldots , \hat x_ k, \ldots , x_ n \in X_{n - 1}$ satisfying $d_ i x_ j = d_{j - 1} x_ i$ for $i < j$, $i, j \not= k$. The map $b$ corresponds to an element $y \in Y_ n$ such that $d_ iy = f(x_ i)$ for $i \not= k$. Our task is to produce an $x \in X_ n$ such that $d_ ix = x_ i$ for $i \not= k$ 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 $x_ i - d_ ix$ for $i \not= k$. 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 $0$. In this case the statement become Lemma 14.31.6. $\square$