Lemma 66.17.8. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f, g : X \to Y$ be morphisms of algebraic spaces over $B$. Let $U \subset X$ be an open subspace such that $f|_ U = g|_ U$. If the scheme theoretic closure of $U$ in $X$ is $X$ and $Y \to B$ is separated, then $f = g$.

Proof. As $Y \to B$ is separated the fibre product $Y \times _{\Delta , Y \times _ B Y, (f, g)} X$ is a closed subspace $Z \subset X$. As $f|_ U = g|_ U$ we see that $U \subset Z$. Hence $Z = X$ as $U$ is assumed scheme theoretically dense in $X$. $\square$

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