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The Stacks project

Lemma 21.30.6. In Situation 21.30.1 let f : X \to Y be in \mathcal{P} such that \{ X \to Y\} is a \tau -covering. Let \mathcal{F}' be in \mathcal{A}'_ Y. If n \geq 0 and

\theta \in \text{Equalizer}\left( \xymatrix{ H^{n + 1}_{\tau '}(X, \mathcal{F}') \ar@<1ex>[r] \ar@<-1ex>[r] & H^{n + 1}_{\tau '}(X \times _ Y X, \mathcal{F}') } \right)

then there exists a \tau '-covering \{ Y_ i \to Y\} such that \theta restricts to zero in H^{n + 1}_{\tau '}(Y_ i \times _ Y X, \mathcal{F}').

Proof. Observe that X \times _ Y X exists by (1). For Z in \mathcal{C}/Y denote \mathcal{F}'|_ Z the restriction of \mathcal{F}' to \mathcal{C}_{\tau '}/Z. Recall that H^{n + 1}_{\tau '}(X, \mathcal{F}') = H^{n + 1}(\mathcal{C}_{\tau '}/X, \mathcal{F}'|_ X), see Lemma 21.7.1. The lemma asserts that the image \overline{\theta } \in H^0(Y, R^{n + 1}f_{\tau ', *}\mathcal{F}'|_ X) of \theta is zero. Consider the cartesian diagram

\xymatrix{ X \times _ Y X \ar[d]_{\text{pr}_1} \ar[r]_{\text{pr}_2} & X \ar[d]^ f \\ X \ar[r]^ f & Y }

By trivial base change (Lemma 21.21.1) we have

f_{\tau '}^{-1}R^{n + 1}f_{\tau ', *}(\mathcal{F}'|_ X) = R^{n + 1}\text{pr}_{1, \tau ', *}(\mathcal{F}'|_{X \times _ Y X})

If \text{pr}_1^{-1}\theta = \text{pr}_2^{-1}\theta , then the section f_{\tau '}^{-1}\overline{\theta } of f_{\tau '}^{-1}R^{n + 1}f_{\tau ', *}(\mathcal{F}'|_ X) is zero, because it is clear that \text{pr}_1^{-1}\theta maps to the zero element in H^0(X, R^{n + 1}\text{pr}_{1, \tau ', *}(\mathcal{F}'|_{X \times _ Y X})). By (2) we have \mathcal{F}'|_ X in \mathcal{A}'_ X. Thus \mathcal{G}' = R^{n + 1}f_{\tau ', *}(\mathcal{F}'|_ X) is an object of \mathcal{A}'_ Y by (4). Thus \mathcal{G}' satisfies the sheaf property for \tau -coverings by (3). Since \{ X \to Y\} is a \tau -covering we conclude that restriction \mathcal{G}'(Y) \to \mathcal{G}'(X) is injective. It follows that \overline{\theta } is zero. \square


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