Lemma 67.13.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Then there exists a unique monomorphism $Z \to X$ of algebraic spaces over $S$ such that $Z$ is an algebraic space which satisfies the equivalent conditions of Lemma 67.13.3 and such that the image of $|Z| \to |X|$ is $\{ x\} $.

**Proof.**
Choose a scheme $U$ and a surjective étale morphism $U \to X$. Set $R = U \times _ X U$ so that $X = U/R$ is a presentation (see Spaces, Section 64.9). Set

The canonical morphism $U' \to U$ is a monomorphism. Let

Because $U' \to U$ is a monomorphism we see that the projections $s', t' : R' \to U'$ factor as a monomorphism followed by an étale morphism. Hence, as $U'$ is a disjoint union of spectra of fields, using Remark 67.4.1, and using Schemes, Lemma 26.23.11 we conclude that $R'$ is a disjoint union of spectra of fields and that the morphisms $s', t' : R' \to U'$ are étale. Hence $Z = U'/R'$ is an algebraic space by Spaces, Theorem 64.10.5. As $R'$ is the restriction of $R$ by $U' \to U$ we see $Z \to X$ is a monomorphism by Groupoids, Lemma 39.20.6. Since $Z \to X$ is a monomorphism we see that $|Z| \to |X|$ is injective, see Morphisms of Spaces, Lemma 66.10.9. By Properties of Spaces, Lemma 65.4.3 we see that

is surjective which implies (by our choice of $U'$) that $|Z| \to |X|$ has image $\{ x\} $. We conclude that $|Z|$ is a singleton. Finally, by construction $U'$ is locally Noetherian and reduced, i.e., we see that $Z$ satisfies the equivalent conditions of Lemma 67.13.3.

Let us prove uniqueness of $Z \to X$. Suppose that $Z' \to X$ is a second such monomorphism of algebraic spaces. Then the projections

are monomorphisms. The algebraic space in the middle is nonempty by Properties of Spaces, Lemma 65.4.3. Hence the two projections are isomorphisms by Lemma 67.13.4 and we win. $\square$

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