Lemma 66.18.12. Let $S$ be a scheme. Let

$\xymatrix{ X' \ar[r] \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

be a cartesian square of algebraic spaces over $S$. Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. If $g$ is étale, then

1. $f'_{small, *}(\mathcal{F}|_{X'}) = (f_{small, *}\mathcal{F})|_{Y'}$ in $\mathop{\mathit{Sh}}\nolimits (Y'_{\acute{e}tale})$1, and

2. if $\mathcal{F}$ is an abelian sheaf, then $R^ if'_{small, *}(\mathcal{F}|_{X'}) = (R^ if_{small, *}\mathcal{F})|_{Y'}$.

Proof. Consider the following diagram of functors

$\xymatrix{ X'_{spaces, {\acute{e}tale}} \ar[r]_ j & X_{spaces, {\acute{e}tale}} \\ Y'_{spaces, {\acute{e}tale}} \ar[r]^ j \ar[u]^{V' \mapsto V' \times _{Y'} X'} & Y_{spaces, {\acute{e}tale}} \ar[u]_{V \mapsto V \times _ Y X} }$

The horizontal arrows are localizations and the vertical arrows induce morphisms of sites. Hence the last statement of Sites, Lemma 7.28.1 gives (1). To see (2) apply (1) to an injective resolution of $\mathcal{F}$ and use that restriction is exact and preserves injectives (see Cohomology on Sites, Lemma 21.7.1). $\square$

[1] Also $(f')_{small}^{-1}(\mathcal{G}|_{Y'}) = (f_{small}^{-1}\mathcal{G})|_{X'}$ because of commutativity of the diagram and (66.18.11.1)

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).