Lemma 67.4.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Let $U$ be a scheme and let $\varphi : U \to X$ be an étale morphism. The following are equivalent:

1. $x$ is in the image of $|U| \to |X|$, and setting $R = U \times _ X U$ the fibres of both

$|U| \longrightarrow |X| \quad \text{and}\quad |R| \longrightarrow |X|$

over $x$ are finite,

2. there exists a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ with $k$ a field in the equivalence class of $x$, and the fibre product $\mathop{\mathrm{Spec}}(k) \times _ X U$ is a finite nonempty scheme over $k$.

Proof. Assume (1). This clearly implies the first condition of Lemma 67.4.3 and hence we obtain a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ in the class of $x$. Taking the fibre product we see that $\mathop{\mathrm{Spec}}(k) \times _ X U \to \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $\mathop{\mathrm{Spec}}(k)$ with finitely many points, hence a finite nonempty scheme over $k$, i.e., (2) holds.

Assume (2). By assumption $x$ is in the image of $|U| \to |X|$. The finiteness of the fibre of $|U| \to |X|$ over $x$ is clear since this fibre is equal to $|\mathop{\mathrm{Spec}}(k) \times _ X U|$ by Properties of Spaces, Lemma 65.4.3. The finiteness of the fibre of $|R| \to |X|$ above $x$ is also clear since it is equal to the set underlying the scheme

$(\mathop{\mathrm{Spec}}(k) \times _ X U) \times _{\mathop{\mathrm{Spec}}(k)} (\mathop{\mathrm{Spec}}(k) \times _ X U)$

which is finite over $k$. Thus (1) holds. $\square$

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