Lemma 35.25.4. The property $\mathcal{P}(f)=$“$f$ is universally open” is fppf local on the source.

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{ X_ i \to X\} _{i \in I}$ be an fppf covering. Denote $f_ i : X_ i \to X$ the compositions. We have to show that $f$ is universally open if and only if each $f_ i$ is universally open. If $f$ is universally open, then also each $f_ i$ is universally open since the maps $X_ i \to X$ are universally open and compositions of universally open morphisms are universally open (Morphisms, Lemmas 29.25.10 and 29.23.3). Conversely, assume each $f_ i$ is universally open. Let $Y' \to Y$ be a morphism of schemes. Denote $X' = Y' \times _ Y X$ and $X'_ i = Y' \times _ Y X_ i$. Note that $\{ X_ i' \to X'\} _{i \in I}$ is an fppf covering also. The morphisms $f'_ i : X_ i' \to Y'$ are open by assumption. Hence by the Lemma 35.25.3 above we conclude that $f' : X' \to Y'$ is open as desired. $\square$

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