Lemma 66.43.2 (Valuative criterion separatedness). Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

1. the morphism $f$ is quasi-separated, and

2. 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

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] \ar@{-->}[ru] & X \times _ Y X }$

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 66.42.1 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 64.5.8). 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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