The Stacks project

Lemma 64.18.10. Let $S$ be a scheme, and let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is étale. In this case there is a functor

\[ j : X_{\acute{e}tale}\to Y_{\acute{e}tale}, \quad (\varphi : U \to X) \mapsto (f \circ \varphi : U \to Y) \]

which is cocontinuous. The morphism of topoi $f_{small}$ is the morphism of topoi associated to $j$, see Sites, Lemma 7.21.1. Moreover, $j$ is continuous as well, hence Sites, Lemma 7.21.5 applies. In particular $f_{small}^{-1}\mathcal{G}(U) = \mathcal{G}(jU)$ for all sheaves $\mathcal{G}$ on $Y_{\acute{e}tale}$.

Proof. Note that by our very definition of an étale morphism of algebraic spaces (Definition 64.16.2) it is indeed the case that the rule given defines a functor $j$ as indicated. It is clear that $j$ is cocontinuous and continuous, simply because a covering $\{ U_ i \to U\} $ of $j(\varphi : U \to X)$ in $Y_{\acute{e}tale}$ is the same thing as a covering of $(\varphi : U \to X)$ in $X_{\acute{e}tale}$. It remains to show that $j$ induces the same morphism of topoi as $f_{small}$. To see this we consider the diagram

\[ \xymatrix{ X_{\acute{e}tale}\ar[r] \ar[d]^ j & X_{spaces, {\acute{e}tale}} \ar@/_/[d]_{j_{spaces}} \\ Y_{\acute{e}tale}\ar[r] & Y_{spaces, {\acute{e}tale}} \ar@/_/[u]_{v : V \mapsto X \times _ Y V} } \]

of categories. Here the functor $j_{spaces}$ is the obvious extension of $j$ to the category $X_{spaces, {\acute{e}tale}}$. Thus the inner square is commutative. In fact $j_{spaces}$ can be identified with the localization functor $j_ X : Y_{spaces, {\acute{e}tale}}/X \to Y_{spaces, {\acute{e}tale}}$ discussed in Sites, Section 7.25. Hence, by Sites, Lemma 7.27.2 the cocontinuous functor $j_{spaces}$ and the functor $v$ of the diagram induce the same morphism of topoi. By Sites, Lemma 7.21.2 the commutativity of the inner square (consisting of cocontinuous functors between sites) gives a commutative diagram of associated morphisms of topoi. Hence, by the construction of $f_{small}$ in Lemma 64.18.7 we win. $\square$


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