Lemma 21.29.6. In Situation 21.29.1 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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