Lemma 25.5.1. 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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