Lemma 65.47.5. Let $S$ be a scheme. Let $X$ and $Y$ be algebraic spaces over $S$. Assume $X$ is reduced and $Y$ is separated over $S$. Let $\varphi $ be a rational map from $X$ to $Y$ with domain of definition $U \subset X$. Then there exists a unique morphism $f : U \to Y$ of algebraic spaces representing $\varphi $.

**Proof.**
Let $(V, g)$ and $(V', g')$ be representatives of $\varphi $. Then $g, g'$ agree on a dense open subspace $W \subset V \cap V'$. On the other hand, the equalizer $E$ of $g|_{V \cap V'}$ and $g'|_{V \cap V'}$ is a closed subspace of $V \cap V'$ because it is the base change of $\Delta : Y \to Y \times _ S Y$ by the morphism $V \cap V' \to Y \times _ S Y$ given by $g|_{V \cap V'}$ and $g'|_{V \cap V'}$. Now $W \subset E$ implies that $|E| = |V \cap V'|$. As $V \cap V'$ is reduced we conclude $E = V \cap V'$ scheme theoretically, i.e., $g|_{V \cap V'} = g'|_{V \cap V'}$, see Properties of Spaces, Lemma 64.12.4. It follows that we can glue the representatives $g : V \to Y$ of $\varphi $ to a morphism $f : U \to Y$ because $\coprod V \to U$ is a surjection of fppf sheaves and $\coprod _{V, V'} V \cap V' = (\coprod V) \times _ U (\coprod V)$.
$\square$

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