Lemma 74.11.18. The property \mathcal{P}(f) =“f is separated” is fpqc local on the base.
Proof. A base change of a separated morphism is separated, see Morphisms of Spaces, Lemma 67.4.4. Hence the direct implication in Definition 74.10.1.
Let \{ Y_ i \to Y\} _{i \in I} be an fpqc covering of algebraic spaces over S. Let f : X \to Y be a morphism of algebraic spaces over S. Assume each base change X_ i := Y_ i \times _ Y X \to Y_ i is separated. This means that each of the morphisms
is a closed immersion. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 73.9.3 hence \{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} is an fpqc covering of algebraic spaces. Moreover, each \Delta _ i is the base change of the morphism \Delta : X \to X \times _ Y X. Hence it follows from Lemma 74.11.17 that \Delta is a closed immersion, i.e., f is separated. \square
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