Lemma 14.23.4. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. The canonical map $N(U_ n) \to U_ n$ gives rise to a morphism of complexes $N(U) \to s(U)$.
Proof. This is clear because the differential on $s(U)_ n = U_ n$ is $\sum (-1)^ i d^ n_ i$ and the maps $d^ n_ i$, $i < n$ are zero on $N(U_ n)$, whereas the restriction of $(-1)^ nd^ n_ n$ is the boundary map of $N(U)$ by definition. $\square$
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