Lemma 101.27.15. Let $f : \mathcal{X} \to \mathcal{Y}$ be flat and locally of finite presentation. Then $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is open.

**Proof.**
Choose a scheme $V$ and a smooth surjective morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a smooth surjective morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. By assumption the morphism of schemes $U \to V$ is flat and locally of finite presentation. Hence $U \to V$ is open by Morphisms, Lemma 29.25.10. By construction of the topology on $|\mathcal{Y}|$ the map $|V| \to |\mathcal{Y}|$ is open. The map $|U| \to |\mathcal{X}|$ is surjective. The result follows from these facts by elementary topology.
$\square$

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