Lemma 67.43.2 (Valuative criterion separatedness). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume
the morphism $f$ is quasi-separated, and
the morphism $f$ satisfies the uniqueness part of the valuative criterion.
Then $f$ is separated.
Proof. Assumption (1) means $\Delta _{X/Y}$ is quasi-compact. We claim the morphism $\Delta _{X/Y} : X \to X \times _ Y X$ satisfies the existence part of the valuative criterion. Let a solid commutative diagram
be given. The lower right arrow corresponds to a pair of morphisms $a, b : \mathop{\mathrm{Spec}}(A) \to X$ over $Y$. By assumption (2) we see that $a = b$. Hence using $a$ as the dotted arrow works. Hence Lemma 67.42.1 applies, and we see that $\Delta _{X/Y}$ is universally closed. Since always $\Delta _{X/Y}$ is locally of finite type and separated, we conclude from More on Morphisms, Lemma 37.44.1 that $\Delta _{X/Y}$ is a finite morphism (also, use the general principle of Spaces, Lemma 65.5.8). At this point $\Delta _{X/Y}$ is a representable, finite monomorphism, hence a closed immersion by Morphisms, Lemma 29.44.15. $\square$
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