Lemma 35.9.1. Let $S$ be a scheme. Let

1. $\tau \in \{ Zariski, \linebreak[0] fppf, \linebreak[0] {\acute{e}tale}, \linebreak[0] smooth, \linebreak[0] syntomic\}$ and $\mathcal{C} = (\mathit{Sch}/S)_\tau$, or

2. let $\tau = {\acute{e}tale}$ and $\mathcal{C} = S_{\acute{e}tale}$, or

3. let $\tau = Zariski$ and $\mathcal{C} = S_{Zar}$.

Let $\mathcal{F}$ be an abelian sheaf on $\mathcal{C}$. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ be affine. Let $\mathcal{U} = \{ U_ i \to U\} _{i = 1, \ldots , n}$ be a standard affine $\tau$-covering in $\mathcal{C}$. Then

1. $\mathcal{V} = \{ \coprod _{i = 1, \ldots , n} U_ i \to U\}$ is a $\tau$-covering of $U$,

2. $\mathcal{U}$ is a refinement of $\mathcal{V}$, and

3. the induced map on Čech complexes (Cohomology on Sites, Equation (21.8.2.1))

$\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$

is an isomorphism of complexes.

Proof. This follows because

$(\coprod \nolimits _{i_0 = 1, \ldots , n} U_{i_0}) \times _ U \ldots \times _ U (\coprod \nolimits _{i_ p = 1, \ldots , n} U_{i_ p}) = \coprod \nolimits _{i_0, \ldots , i_ p \in \{ 1, \ldots , n\} } U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$

and the fact that $\mathcal{F}(\coprod _ a V_ a) = \prod _ a \mathcal{F}(V_ a)$ since disjoint unions are $\tau$-coverings. $\square$

Comment #2151 by Katha on

$\mathcal{U} = \{U_i \to U\}_{i=1,\ldots n}$ ?

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