Lemma 59.22.1 (03OZ)—The Stacks project

Lemma 59.22.1. Let $\tau \in \{ fppf, syntomic, smooth, {\acute{e}tale}, Zariski\}$ . Let $S$ be a scheme. Let $\mathcal{F}$ be an abelian sheaf on $(\mathit{Sch}/S)_\tau$ , or on $S_\tau$ in case $\tau = {\acute{e}tale}$ , and let $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a standard $\tau$ -covering of this site. Let $V = \coprod _{i \in I} U_ i$ . Then

$V$ is an affine scheme,
$\mathcal{V} = \{ V \to U\}$ is an fpqc covering and also a $\tau$ -covering unless $\tau = Zariski$ ,
the Čech complexes $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ and $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F})$ agree.

Proof. The definition of a standard $\tau$ -covering is given in Topologies, Definition 34.3.4, 34.4.5, 34.5.5, 34.6.5, and 34.7.5. By definition each of the schemes $U_ i$ is affine and $I$ is a finite set. Hence $V$ is an affine scheme. It is clear that $V \to U$ is flat and surjective, hence $\mathcal{V}$ is an fpqc covering, see Example 59.15.3. Excepting the Zariski case, the covering $\mathcal{V}$ is also a $\tau$ -covering, see Topologies, Definition 34.4.1, 34.5.1, 34.6.1, and 34.7.1.

Note that $\mathcal{U}$ is a refinement of $\mathcal{V}$ and hence there is a map of Čech complexes $\check{\mathcal{C}}^\bullet (\mathcal{V}, \mathcal{F}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ , see Cohomology on Sites, Equation (21.8.2.1). Next, we observe that if $T = \coprod _{j \in J} T_ j$ is a disjoint union of schemes in the site on which $\mathcal{F}$ is defined then the family of morphisms with fixed target $\{ T_ j \to T\} _{j \in J}$ is a Zariski covering, and so

59.22.1.1

$\begin{equation} \label{etale-cohomology-equation-sheaf-coprod} \mathcal{F}(T) = \mathcal{F}(\coprod \nolimits _{j \in J} T_ j) = \prod \nolimits _{j \in J} \mathcal{F}(T_ j) \end{equation}$

by the sheaf condition of $\mathcal{F}$ . This implies the map of Čech complexes above is an isomorphism in each degree because

$V \times _ U \ldots \times _ U V = \coprod \nolimits _{i_0, \ldots i_ p} U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$

as schemes. $\square$

Comments (2)

Comment #1472 by Xiaowen Hu on May 23, 2015 at 13:59

The product in $V \times_U \ldots \times_U V = \prod\nolimits_{i_0, \ldots i_p} U_{i_0} \times_U \ldots \times_U U_{i_p}$ should be coproduct.

Comment #1490 by Johan on June 07, 2015 at 12:40

Thanks very much. Fixed here.

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