Lemma 67.30.7. Let $S$ be a scheme. Let $f : X \to Y$ be a flat, quasi-compact, surjective morphism of algebraic spaces over $S$. A subset $T \subset |Y|$ is open (resp. closed) if and only $f^{-1}(|T|)$ is open (resp. closed) in $|X|$. In other words $f$ is submersive, and in fact universally submersive.

**Proof.**
Choose affine schemes $V_ i$ and étale morphisms $V_ i \to Y$ such that $V = \coprod V_ i \to Y$ is surjective, see Properties of Spaces, Lemma 66.6.1. For each $i$ the algebraic space $V_ i \times _ Y X$ is quasi-compact. Hence we can find an affine scheme $U_ i$ and a surjective étale morphism $U_ i \to V_ i \times _ Y X$, see Properties of Spaces, Lemma 66.6.3. Then the composition $U_ i \to V_ i \times _ Y X \to V_ i$ is a surjective, flat morphism of affines. Of course then $U = \coprod U_ i \to X$ is surjective and étale and $U \to V \times _ Y X$ is surjective. Moreover, the morphism $U \to V$ is the disjoint union of the morphisms $U_ i \to V_ i$. Hence $U \to V$ is surjective, quasi-compact and flat. Consider the diagram

By definition of the topology on $|Y|$ the set $T$ is closed (resp. open) if and only if $g^{-1}(T) \subset |V|$ is closed (resp. open). The same holds for $f^{-1}(T)$ and its inverse image in $|U|$. Since $U \to V$ is quasi-compact, surjective, and flat we win by Morphisms, Lemma 29.25.12. $\square$

## 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Your email address will not be published.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.

## Comments (2)

Comment #8842 by ZL on

Comment #9244 by Stacks project on

There are also: