A geometric point of an algebraic space gives a point of its étale topos.

Lemma 66.19.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\overline{x}$ be a geometric point of $X$. Consider the functor

$u : X_{\acute{e}tale}\longrightarrow \textit{Sets}, \quad U \longmapsto |U_{\overline{x}}|$

Then $u$ defines a point $p$ of the site $X_{\acute{e}tale}$ (Sites, Definition 7.32.2) and its associated stalk functor $\mathcal{F} \mapsto \mathcal{F}_ p$ (Sites, Equation 7.32.1.1) is the functor $\mathcal{F} \mapsto \mathcal{F}_{\overline{x}}$ defined above.

Proof. In the proof of Lemma 66.19.5 we have seen that the scheme $U_{\overline{x}}$ is a disjoint union of schemes isomorphic to $\overline{x}$. Thus we can also think of $|U_{\overline{x}}|$ as the set of geometric points of $U$ lying over $\overline{x}$, i.e., as the collection of morphisms $\overline{u} : \overline{x} \to U$ fitting into the diagram of Definition 66.19.2. From this it follows that $u(X)$ is a singleton, and that $u(U \times _ V W) = u(U) \times _{u(V)} u(W)$ whenever $U \to V$ and $W \to V$ are morphisms in $X_{\acute{e}tale}$. And, given a covering $\{ U_ i \to U\} _{i \in I}$ in $X_{\acute{e}tale}$ we see that $\coprod u(U_ i) \to u(U)$ is surjective by Lemma 66.19.5. Hence Sites, Proposition 7.33.3 applies, so $p$ is a point of the site $X_{\acute{e}tale}$. Finally, the our functor $\mathcal{F} \mapsto \mathcal{F}_{\overline{s}}$ is given by exactly the same colimit as the functor $\mathcal{F} \mapsto \mathcal{F}_ p$ associated to $p$ in Sites, Equation 7.32.1.1 which proves the final assertion. $\square$

Comment #984 by on

Suggested slogan: Taking stalks at geometric points of algebraic spaces induces topos-theoretic points.

Comment #7432 by on

In the third row of the proof of Lemma 04K0, "...fitting into the diagram of Definition 0486." should be "...fitting into the diagram of Definition 04JV."

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