Lemma 73.4.7. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of $(\textit{Spaces}/S)_{\acute{e}tale}$. The inclusion functor $Y_{spaces, {\acute{e}tale}} \to (\textit{Spaces}/X)_{\acute{e}tale}$ is cocontinuous and induces a morphism of topoi

\[ i_ f : \mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}) \longrightarrow \mathop{\mathit{Sh}}\nolimits ((\textit{Spaces}/X)_{\acute{e}tale}) \]

For a sheaf $\mathcal{G}$ on $(\textit{Spaces}/X)_{\acute{e}tale}$ we have the formula $(i_ f^{-1}\mathcal{G})(U/Y) = \mathcal{G}(U/X)$. The functor $i_ f^{-1}$ also has a left adjoint $i_{f, !}$ which commutes with fibre products and equalizers.

**Proof.**
Denote the functor $u : Y_{spaces, {\acute{e}tale}} \to (\textit{Spaces}/X)_{\acute{e}tale}$. In other words, given an étale morphism $j : U \to Y$ corresponding to an object of $Y_{spaces, {\acute{e}tale}}$ we set $u(U \to T) = (f \circ j : U \to S)$. The category $Y_{spaces, {\acute{e}tale}}$ has fibre products and equalizers and $u$ commutes with them. It is immediate that $u$ cocontinuous. The functor $u$ is also continuous as $u$ transforms coverings to coverings and commutes with fibre products. Hence the Lemma follows from Sites, Lemmas 7.21.5 and 7.21.6.
$\square$

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