Lemma 78.12.4. 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 78.12.0.1. 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 66.45.9. Since $f$ is separated so is the morphism $T \times _ Y X \to T$, see Morphisms of Spaces, Lemma 66.4.4. Thus the closedness of $Z$ follows from Morphisms of Spaces, Lemma 66.40.6. $\square$

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