Lemma 66.4.9. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X' \subset X$ be an open subspace. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Then $f$ factors through $X'$ if and only if $|f| : |Y| \to |X|$ factors through $|X'| \subset |X|$.

**Proof.**
By Spaces, Lemma 65.12.3 we see that $Y' = Y \times _ X X' \to Y$ is an open immersion. If $|f|(|Y|) \subset |X'|$, then clearly $|Y'| = |Y|$. Hence $Y' = Y$ by Lemma 66.4.8.
$\square$

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