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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