Lemma 67.17.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is quasi-compact and decent. (For example if $f$ is representable, or quasi-separated, see Lemma 67.17.2.) Then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

Proof. In Morphisms of Spaces, Lemma 66.42.1 we proved that any quasi-compact morphism which satisfies the existence part of the valuative criterion is universally closed. To prove the other, assume that $f$ is universally closed. In the proof of Proposition 67.16.1 we have seen that it suffices to show, for any valuation ring $A$, and any morphism $\mathop{\mathrm{Spec}}(A) \to Y$, that the base change $f_ A : X_ A \to \mathop{\mathrm{Spec}}(A)$ satisfies the existence part of the valuative criterion. By definition the algebraic space $X_ A$ has property $(\gamma )$ and hence Proposition 67.16.1 applies to the morphism $f_ A$ and we win. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).