Lemma 25.4.3. Let $\mathcal{C}$ be a site. Let $f : L \to K$ be a morphism of simplicial objects of $\textit{PSh}(\mathcal{C})$. Let $n \geq 0$ be an integer. Assume that

1. For $i < n$ the morphism $L_ i \to K_ i$ is an isomorphism.

2. The morphism $L_ n \to K_ n$ is surjective after sheafification.

3. The canonical map $L \to \text{cosk}_ n \text{sk}_ n L$ is an isomorphism.

4. The canonical map $K \to \text{cosk}_ n \text{sk}_ n K$ is an isomorphism.

Then $H_ i(f) : H_ i(L) \to H_ i(K)$ is an isomorphism.

Proof. This proof is exactly the same as the proof of Lemma 25.4.2 above. Namely, we first let $K_ n' \subset K_ n$ be the sub presheaf which is the image of the map $L_ n \to K_ n$. Assumption (2) means that the sheafification of $K_ n'$ is equal to the sheafification of $K_ n$. Moreover, since $L_ i = K_ i$ for all $i < n$ we see that get an $n$-truncated simplicial presheaf $U$ by taking $U_0 = L_0 = K_0, \ldots , U_{n - 1} = L_{n - 1} = K_{n - 1}, U_ n = K'_ n$. Denote $K' = \text{cosk}_ n U$, a simplicial presheaf. Because we can construct $K'_ m$ as a finite limit, and since sheafification is exact, we see that $(K'_ m)^\# = K_ m$. In other words, $(K')^\# = K^\#$. We conclude, by exactness of sheafification once more, that $H_ i(K) = H_ i(K')$. Thus it suffices to prove the lemma for the morphism $L \to K'$, in other words, we may assume that $L_ n \to K_ n$ is a surjective morphism of presheaves!

In this case, for any object $U$ of $\mathcal{C}$ we see that the morphism of simplicial sets

$L(U) \longrightarrow K(U)$

satisfies all the assumptions of Simplicial, Lemma 14.32.1. Hence it is a trivial Kan fibration. In particular it is a homotopy equivalence (Simplicial, Lemma 14.30.8). Thus

$\mathbf{Z}_ L(U) \longrightarrow \mathbf{Z}_ K(U)$

is a homotopy equivalence too. This for all $U$. The result follows. $\square$

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