Lemma 25.4.4. Let $\mathcal{C}$ be a site. Let $K$ be a simplicial presheaf. Let $\mathcal{G}$ be a presheaf. Let $K \to \mathcal{G}$ be an augmentation of $K$ towards $\mathcal{G}$. Assume that

The morphism of presheaves $K_0 \to \mathcal{G}$ becomes a surjection after sheafification.

The morphism

\[ (d^1_0, d^1_1) : K_1 \longrightarrow K_0 \times _\mathcal {G} K_0 \]

becomes a surjection after sheafification.

For every $n \geq 1$ the morphism

\[ K_{n + 1} \longrightarrow (\text{cosk}_ n \text{sk}_ n K)_{n + 1} \]

turns into a surjection after sheafification.

Then $H_ i(K) = 0$ for $i > 0$ and $H_0(K) = \mathbf{Z}_\mathcal {G}^\# $.

**Proof.**
Denote $K^ n = \text{cosk}_ n \text{sk}_ n K$ for $n \geq 1$. Define $K^0$ as the simplicial object with terms $(K^0)_ n$ equal to the $(n + 1)$-fold fibred product $K_0 \times _\mathcal {G} \ldots \times _\mathcal {G} K_0$, see Simplicial, Example 14.3.5. We have morphisms

\[ K \longrightarrow \ldots \to K^ n \to K^{n - 1} \to \ldots \to K^1 \to K^0. \]

The morphisms $K \to K^ i$, $K^ j \to K^ i$ for $j \geq i \geq 1$ come from the universal properties of the $\text{cosk}_ n$ functors. The morphism $K^1 \to K^0$ is the canonical morphism from Simplicial, Remark 14.20.4. We also recall that $K^0 \to \text{cosk}_1 \text{sk}_1 K^0$ is an isomorphism, see Simplicial, Lemma 14.20.3.

By Lemma 25.4.2 we see that $H_ i(K^0) = 0$ for $i > 0$ and $H_0(K^0) = \mathbf{Z}_\mathcal {G}^\# $.

Pick $n \geq 1$. Consider the morphism $K^ n \to K^{n - 1}$. It is an isomorphism on terms of degree $< n$. Note that $K^ n \to \text{cosk}_ n \text{sk}_ n K^ n$ and $K^{n - 1} \to \text{cosk}_ n \text{sk}_ n K^{n - 1}$ are isomorphisms. Note that $(K^ n)_ n = K_ n$ and that $(K^{n - 1})_ n = (\text{cosk}_{n - 1} \text{sk}_{n - 1} K)_ n$. Hence by assumption, we have that $(K^ n)_ n \to (K^{n - 1})_ n$ is a morphism of presheaves which becomes surjective after sheafification. By Lemma 25.4.3 we conclude that $H_ i(K^ n) = H_ i(K^{n - 1})$. Combined with the above this proves the lemma.
$\square$

## Comments (0)