Lemma 79.12.4. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. Let T be an algebraic space over S, and let (a, Z) be a pair as in 79.12.0.1. If f is separated, then Z is closed in T \times _ Y X.
Proof. A finite morphism of algebraic spaces is universally closed by Morphisms of Spaces, Lemma 67.45.9. Since f is separated so is the morphism T \times _ Y X \to T, see Morphisms of Spaces, Lemma 67.4.4. Thus the closedness of Z follows from Morphisms of Spaces, Lemma 67.40.6. \square
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