Lemma 24.5.1. Let $\mathcal{C}$ be a site with fibre products. Let $X$ be an object of $\mathcal{C}$. Let $K$ be a hypercovering of $X$. Let $\mathcal{F}$ be a sheaf of abelian groups on $\mathcal{C}$. Then $\check{H}^0(K, \mathcal{F}) = \mathcal{F}(X)$.

Proof. We have

$\check{H}^0(K, \mathcal{F}) = \mathop{\mathrm{Ker}}(\mathcal{F}(K_0) \longrightarrow \mathcal{F}(K_1))$

Write $K_0 = \{ U_ i \to X\}$. It is a covering in the site $\mathcal{C}$. As well, we have that $K_1 \to K_0 \times K_0$ is a covering in $\text{SR}(\mathcal{C}, X)$. Hence we may write $K_1 = \amalg _{i_0, i_1 \in I} \{ V_{i_0i_1j} \to X\}$ so that the morphism $K_1 \to K_0 \times K_0$ is given by coverings $\{ V_{i_0i_1j} \to U_{i_0} \times _ X U_{i_1}\}$ of the site $\mathcal{C}$. Thus we can further identify

$\check{H}^0(K, \mathcal{F}) = \mathop{\mathrm{Ker}}( \prod \nolimits _ i \mathcal{F}(U_ i) \longrightarrow \prod \nolimits _{i_0i_1 j} \mathcal{F}(V_{i_0i_1j}) )$

with obvious map. The sheaf property of $\mathcal{F}$ implies that $\check{H}^0(K, \mathcal{F}) = H^0(X, \mathcal{F})$. $\square$

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