Lemma 81.14.1. Let $S$ be a scheme. Let $X \to Y$ be a morphism of algebraic spaces over $S$. If $(U \subset X, f : V \to X)$ is an elementary distinguished square such that $U \to Y$ and $V \to Y$ are separated and $U \times _ X V \to U \times _ Y V$ is closed, then $X \to Y$ is separated.
Proof. We have to check that $\Delta : X \to X \times _ Y X$ is a closed immersion. There is an étale covering of $X \times _ Y X$ given by the four parts $U \times _ Y U$, $U \times _ Y V$, $V \times _ Y U$, and $V \times _ Y V$. Observe that $(U \times _ Y U) \times _{(X \times _ Y X), \Delta } X = U$, $(U \times _ Y V) \times _{(X \times _ Y X), \Delta } X = U \times _ X V$, $(V \times _ Y U) \times _{(X \times _ Y X), \Delta } X = V \times _ X U$, and $(V \times _ Y V) \times _{(X \times _ Y X), \Delta } X = V$. Thus the assumptions of the lemma exactly tell us that $\Delta $ is a closed immersion. $\square$
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