Lemma 66.30.7. 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 65.6.1. 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 65.6.3. 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 $U = V \times _ Y X$. 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. $\square$

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