Lemma 14.27.1. Let $\mathcal{A}$ be an additive category. Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{A}$. If $a$, $b$ are homotopic, then $s(a), s(b) : s(U) \to s(V)$ are homotopic maps of chain complexes. If $\mathcal{A}$ is abelian, then also $N(a), N(b) : N(U) \to N(V)$ are homotopic maps of chain complexes.
Proof. We may choose a sequence $a = a_0, a_1, \ldots , a_ n = b$ of morphisms from $U$ to $V$ such that for each $i = 1, \ldots , n$ either there is a homotopy from $a_ i$ to $a_{i - 1}$ or there is a homotopy from $a_{i - 1}$ to $a_ i$. The calculation above shows that in this case either $s(a_ i)$ is homotopic to $s(a_{i - 1})$ as a map of chain complexes or $s(a_{i - 1})$ is homotopic to $s(a_ i)$ as a map of chain complexes. Of course, these things are equivalent and moreover being homotopic is an equivalence relation on the set of maps of chain complexes, see Homology, Section 12.13. This proves that $s(a)$ and $s(b)$ are homotopic as maps of chain complexes.
Next, we turn to $N(a)$ and $N(b)$. It follows from Lemma 14.23.6 that $N(a)$, $N(b)$ are compositions
\[ N(U) \to s(U) \to s(V) \to N(V) \]
where we use $s(a)$, $s(b)$ in the middle. Hence the assertion follows from Homology, Lemma 12.13.1. $\square$
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