Lemma 75.4.1. The morphism $\epsilon $ of (75.4.0.1) is a flat morphism of ringed sites. In particular the functor $\epsilon ^* : \textit{Mod}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_{\acute{e}tale})$ is exact. Moreover, if $\epsilon ^*\mathcal{F} = 0$, then $\mathcal{F} = 0$.
Proof. The flatness of the morphism $\epsilon $ is Descent, Lemma 35.10.1. Here is another proof. We have to show that $\mathcal{O}_{\acute{e}tale}$ is a flat $\epsilon ^{-1}\mathcal{O}_ X$-module. To do this it suffices to check $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is flat for any geometric point $\overline{x}$ of $X$, see Modules on Sites, Lemma 18.39.3, Sites, Lemma 7.34.2, and Étale Cohomology, Remarks 59.29.11. By Étale Cohomology, Lemma 59.33.1 we see that $\mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x}$. Thus $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is faithfully flat by More on Algebra, Lemma 15.45.1.
The exactness of $\epsilon ^*$ follows from the flatness of $\epsilon $ by Modules on Sites, Lemma 18.31.2.
Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. If $\epsilon ^*\mathcal{F} = 0$, then with notation as above
\[ 0 = \epsilon ^*\mathcal{F}_{\overline{x}} = \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{{\acute{e}tale}, \overline{x}} \]
(Modules on Sites, Lemma 18.36.4) for all geometric points $\overline{x}$. By faithful flatness of $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ we conclude $\mathcal{F}_ x = 0$ for all $x \in X$. $\square$
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