Lemma 82.8.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. For $K \in D^+(X_{\acute{e}tale})$ with torsion cohomology sheaves the maps

$\pi _ X^{-1}K \longrightarrow R\epsilon _{X, *}a_ X^{-1}K \quad \text{and}\quad K \longrightarrow Ra_{X, *}a_ X^{-1}K$

are isomorphisms with $a_ X : \mathop{\mathit{Sh}}\nolimits ((\textit{Spaces}/X)_{ph}) \to \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$ as above.

Proof. We only prove the second statement; the first is easier and proved in exactly the same manner. There is a reduction to the case where $K$ is given by a single torsion abelian sheaf. Namely, represent $K$ by a bounded below complex $\mathcal{F}^\bullet$ of torsion abelian sheaves. This is possible by Cohomology on Sites, Lemma 21.19.8. By the case of a sheaf we see that $\mathcal{F}^ n = a_{X, *} a_ X^{-1} \mathcal{F}^ n$ and that the sheaves $R^ qa_{X, *}a_ X^{-1}\mathcal{F}^ n$ are zero for $q > 0$. By Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) applied to $a_ X^{-1}\mathcal{F}^\bullet$ and the functor $a_{X, *}$ we conclude. From now on assume $K = \mathcal{F}$ where $\mathcal{F}$ is a torsion abelian sheaf.

By Lemma 82.8.1 we have $a_{X, *}a_ X^{-1}\mathcal{F} = \mathcal{F}$. Thus it suffices to show that $R^ qa_{X, *}a_ X^{-1}\mathcal{F} = 0$ for $q > 0$. For this we can use $a_ X = \epsilon _ X \circ \pi _ X$ and the Leray spectral sequence (Cohomology on Sites, Lemma 21.14.7). By Lemma 82.8.1 we have $R^ i\epsilon _{X, *}(a_ X^{-1}\mathcal{F}) = 0$ for $i > 0$. We have $\epsilon _{X, *}a_ X^{-1}\mathcal{F} = \pi _ X^{-1}\mathcal{F}$ and by Lemma 82.5.5 we have $R^ j\pi _{X, *}(\pi _ X^{-1}\mathcal{F}) = 0$ for $j > 0$. This concludes the proof. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).