Lemma 73.12.1. The property $\mathcal{P}(f) =$“$f$ is an immersion” is fppf local on the base.

Proof. Let $f : X \to Y$ be a morphism of algebraic spaces. Let $\{ Y_ i \to Y\} _{i \in I}$ be an fppf covering of $Y$. Let $f_ i : X_ i \to Y_ i$ be the base change of $f$.

If $f$ is an immersion, then each $f_ i$ is an immersion by Spaces, Lemma 64.12.3. This proves the direct implication in Definition 73.10.1.

Conversely, assume each $f_ i$ is an immersion. By Morphisms of Spaces, Lemma 66.10.7 this implies each $f_ i$ is separated. By Morphisms of Spaces, Lemma 66.27.7 this implies each $f_ i$ is locally quasi-finite. Hence we see that $f$ is locally quasi-finite and separated, by applying Lemmas 73.11.18 and 73.11.24. By Morphisms of Spaces, Lemma 66.51.1 this implies that $f$ is representable!

By Morphisms of Spaces, Lemma 66.12.1 it suffices to show that for every scheme $Z$ and morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ is an immersion. By Topologies on Spaces, Lemma 72.7.4 we can find an fppf covering $\{ Z_ i \to Z\}$ by schemes which refines the pullback of the covering $\{ Y_ i \to Y\}$ to $Z$. Hence we see that $Z \times _ Y X \to Z$ (which is a morphism of schemes according to the result of the preceding paragraph) becomes an immersion after pulling back to the members of an fppf (by schemes) of $Z$. Hence $Z \times _ Y X \to Z$ is an immersion by the result for schemes, see Descent, Lemma 35.24.1. $\square$

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