The Stacks project

Proof. Being locally of finite presentation is Zariski local on the source and the target, see Morphisms, Lemma 29.21.2. It is a property which is preserved under composition, see Morphisms, Lemma 29.21.3. This proves (1), (2) and (3) of Lemma 35.26.4. The final condition (4) is Lemma 35.14.1. Hence we win. $\square$

Proof. Being locally of finite type is Zariski local on the source and the target, see Morphisms, Lemma 29.15.2. It is a property which is preserved under composition, see Morphisms, Lemma 29.15.3, and a flat morphism locally of finite presentation is locally of finite type, see Morphisms, Lemma 29.21.8. This proves (1), (2) and (3) of Lemma 35.26.4. The final condition (4) is Lemma 35.14.2. Hence we win. $\square$

Proof. Being an open morphism is clearly Zariski local on the source and the target. It is a property which is preserved under composition, see Morphisms, Lemma 29.23.3, and a flat morphism of finite presentation is open, see Morphisms, Lemma 29.25.10 This proves (1), (2) and (3) of Lemma 35.26.4. The final condition (4) follows from Morphisms, Lemma 29.25.12. Hence we win. $\square$

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 35.28.3 above we conclude that $f' : X' \to Y'$ is open as desired. $\square$