Lemma 68.13.5. 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 68.13.3 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 65.9). Set
\[ U' = \coprod \nolimits _{u \in U\text{ lying over }x} \mathop{\mathrm{Spec}}(\kappa (u)). \]
The canonical morphism $U' \to U$ is a monomorphism. Let
\[ R' = U' \times _ X U' = R \times _{(U \times _ S U)} (U' \times _ S U'). \]
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 68.4.1, 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 65.10.5. 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 67.10.9. By Properties of Spaces, Lemma 66.4.3 we see that
\[ |U'| = |Z \times _ X U'| \to |Z| \times _{|X|} |U'| \]
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 68.13.3.
Let us prove uniqueness of $Z \to X$. Suppose that $Z' \to X$ is a second such monomorphism of algebraic spaces. Then the projections
\[ Z' \longleftarrow Z' \times _ X Z \longrightarrow Z \]
are monomorphisms. The algebraic space in the middle is nonempty by Properties of Spaces, Lemma 66.4.3. Hence the two projections are isomorphisms by Lemma 68.13.4 and we win. $\square$
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