The Stacks project

Lemma 66.26.1. Let $S$ be a scheme. Let $f : X \to Y$ be an ├ętale morphism of algebraic spaces over $S$. Then $f^{-1}\mathcal{O}_ Y = \mathcal{O}_ X$, and $f^*\mathcal{G} = f_{small}^{-1}\mathcal{G}$ for any sheaf of $\mathcal{O}_ Y$-modules $\mathcal{G}$. In particular, $f^* : \textit{Mod}(\mathcal{O}_ Y) \to \textit{Mod}(\mathcal{O}_ X)$ is exact.

Proof. By the description of inverse image in Lemma 66.18.11 and the definition of the structure sheaves it is clear that $f_{small}^{-1}\mathcal{O}_ Y = \mathcal{O}_ X$. Since the pullback

\[ f^*\mathcal{G} = f_{small}^{-1}\mathcal{G} \otimes _{f_{small}^{-1}\mathcal{O}_ Y} \mathcal{O}_ X \]

by definition we conclude that $f^*\mathcal{G} = f_{small}^{-1}\mathcal{G}$. The exactness is clear because $f_{small}^{-1}$ is exact, as $f_{small}$ is a morphism of topoi. $\square$

Comments (2)

Comment #7746 by Mingchen on

Typo: f^* maps Mod(O_Y) to Mod(O_X), not the other direction.

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