Lemma 61.19.5. Let $X$ be a scheme.

1. For an abelian sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $R\epsilon _*(\epsilon ^{-1}\mathcal{F}) = \mathcal{F}$.

2. For $K \in D^+(X_{\acute{e}tale})$ the map $K \to R\epsilon _*\epsilon ^{-1}K$ is an isomorphism.

Proof. Let $\mathcal{I}$ be an injective abelian sheaf on $X_{\acute{e}tale}$. Recall that $R^ q\epsilon _*(\epsilon ^{-1}\mathcal{I})$ is the sheaf associated to $U \mapsto H^ q(U_{pro\text{-}\acute{e}tale}, \epsilon ^{-1}\mathcal{I})$, see Cohomology on Sites, Lemma 21.7.4. By Lemma 61.19.4 we see that this is zero for $q > 0$ and $U$ affine and étale over $X$. Since every object of $X_{\acute{e}tale}$ has a covering by affine objects, it follows that $R^ q\epsilon _*(\epsilon ^{-1}\mathcal{I}) = 0$ for $q > 0$.

Let $K \in D^+(X_{\acute{e}tale})$. Choose a bounded below complex $\mathcal{I}^\bullet$ of injective abelian sheaves on $X_{\acute{e}tale}$ representing $K$. Then $\epsilon ^{-1}K$ is represented by $\epsilon ^{-1}\mathcal{I}^\bullet$. By Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) we see that $R\epsilon _*\epsilon ^{-1}K$ is represented by $\epsilon _*\epsilon ^{-1}\mathcal{I}^\bullet$. By Lemma 61.19.2 we conclude that $R\epsilon _*\epsilon ^{-1}\mathcal{I}^\bullet = \mathcal{I}^\bullet$ and the proof of (2) is complete. Part (1) is a special case of (2). $\square$

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).