Lemma 59.40.1. Let $f : X \to Y$ be a morphism of schemes. The morphism of ringed sites $(f_{small}, f_{small}^\sharp )$ associated to $f$ is a morphism of locally ringed sites, see Modules on Sites, Definition 18.40.9.

**Proof.**
Note that the assertion makes sense since we have seen that $(X_{\acute{e}tale}, \mathcal{O}_{X_{\acute{e}tale}})$ and $(Y_{\acute{e}tale}, \mathcal{O}_{Y_{\acute{e}tale}})$ are locally ringed sites, see Lemma 59.33.5. Moreover, we know that $X_{\acute{e}tale}$ has enough points, see Theorem 59.29.10 and Remarks 59.29.11. Hence it suffices to prove that $(f_{small}, f_{small}^\sharp )$ satisfies condition (3) of Modules on Sites, Lemma 18.40.8. To see this take a point $p$ of $X_{\acute{e}tale}$. By Lemma 59.29.12 $p$ corresponds to a geometric point $\overline{x}$ of $X$. By Lemma 59.36.2 the point $q = f_{small} \circ p$ corresponds to the geometric point $\overline{y} = f \circ \overline{x}$ of $Y$. Hence the assertion we have to prove is that the induced map of stalks

is a local ring map. Suppose that $a \in (\mathcal{O}_ Y)_{\overline{y}}$ is an element of the left hand side which maps to an element of the maximal ideal of the right hand side. Suppose that $a$ is the equivalence class of a triple $(V, \overline{v}, a)$ with $V \to Y$ étale, $\overline{v} : \overline{x} \to V$ over $Y$, and $a \in \mathcal{O}(V)$. It maps to the equivalence class of $(X \times _ Y V, \overline{x} \times \overline{v}, \text{pr}_ V^\sharp (a))$ in the local ring $(\mathcal{O}_ X)_{\overline{x}}$. But it is clear that being in the maximal ideal means that pulling back $\text{pr}_ V^\sharp (a)$ to an element of $\kappa (\overline{x})$ gives zero. Hence also pulling back $a$ to $\kappa (\overline{x})$ is zero. Which means that $a$ lies in the maximal ideal of $(\mathcal{O}_ Y)_{\overline{y}}$. $\square$

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