The Stacks project

Lemma 66.17.11. 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 66.17.2.) Then $f$ is universally closed if and only if the existence part of the valuative criterion holds.

Proof. In Morphisms of Spaces, Lemma 65.42.1 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 66.16.1 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 66.16.1 applies to the morphism $f_ A$ and we win. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03M6. Beware of the difference between the letter 'O' and the digit '0'.