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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