Lemma 65.19.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

1. The functor $f_{small}^{-1} : \textit{Ab}(Y_{\acute{e}tale}) \to \textit{Ab}(X_{\acute{e}tale})$ is exact.

2. The functor $f_{small}^{-1} : \mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$ is exact, i.e., it commutes with finite limits and colimits, see Categories, Definition 4.23.1.

3. For any étale morphism $V \to Y$ of algebraic spaces we have $f_{small}^{-1}h_ V = h_{X \times _ Y V}$.

4. Let $\overline{x} \to X$ be a geometric point. Let $\mathcal{G}$ be a sheaf on $Y_{\acute{e}tale}$. Then there is a canonical identification

$(f_{small}^{-1}\mathcal{G})_{\overline{x}} = \mathcal{G}_{\overline{y}}.$

where $\overline{y} = f \circ \overline{x}$.

Proof. Recall that $f_{small}$ is defined via $f_{spaces, small}$ in Lemma 65.18.8. Parts (1), (2) and (3) are general consequences of the fact that $f_{spaces, {\acute{e}tale}} : X_{spaces, {\acute{e}tale}} \to Y_{spaces, {\acute{e}tale}}$ is a morphism of sites, see Sites, Definition 7.14.1 for (2), Modules on Sites, Lemma 18.31.2 for (1), and Sites, Lemma 7.13.5 for (3).

Proof of (4). This statement is a special case of Sites, Lemma 7.34.2 via Lemma 65.19.7. We also provide a direct proof. Note that by Lemma 65.19.8. taking stalks commutes with sheafification. Let $\mathcal{G}'$ be the sheaf on $Y_{spaces, {\acute{e}tale}}$ whose restriction to $Y_{\acute{e}tale}$ is $\mathcal{G}$. Recall that $f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}'$ is the sheaf associated to the presheaf

$U \longrightarrow \mathop{\mathrm{colim}}\nolimits _{U \to X \times _ Y V} \mathcal{G}'(V),$

see Sites, Sections 7.13 and 7.5. Thus we have

\begin{align*} (f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}')_{\overline{x}} & = \mathop{\mathrm{colim}}\nolimits _{(U, \overline{u})} f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}'(U) \\ & = \mathop{\mathrm{colim}}\nolimits _{(U, \overline{u})} \mathop{\mathrm{colim}}\nolimits _{a : U \to X \times _ Y V} \mathcal{G}'(V) \\ & = \mathop{\mathrm{colim}}\nolimits _{(V, \overline{v})} \mathcal{G}'(V) \\ & = \mathcal{G}'_{\overline{y}} \end{align*}

in the third equality the pair $(U, \overline{u})$ and the map $a : U \to X \times _ Y V$ corresponds to the pair $(V, a \circ \overline{u})$. Since the stalk of $\mathcal{G}'$ (resp. $f_{spaces, {\acute{e}tale}}^{-1}\mathcal{G}'$) agrees with the stalk of $\mathcal{G}$ (resp. $f_{small}^{-1}\mathcal{G}$), see Equation (65.19.6.1) the result follows. $\square$

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