Lemma 29.49.9. Let $X$ and $Y$ be schemes. Assume $X$ reduced and $Y$ separated. Let $\varphi $ be a rational map from $X$ to $Y$ with domain of definition $U \subset X$. Then there exists a unique morphism $f : U \to Y$ representing $\varphi $. If $X$ and $Y$ are schemes over a separated scheme $S$ and if $\varphi $ is an $S$-rational map, then $f$ is a morphism over $S$.
Proof. Let $(V, g)$ and $(V', g')$ be representatives of $\varphi $. Then $g, g'$ agree on a dense open subscheme $W \subset V \cap V'$. On the other hand, the equalizer $E$ of $g|_{V \cap V'}$ and $g'|_{V \cap V'}$ is a closed subscheme of $V \cap V'$ (Schemes, Lemma 26.21.5). Now $W \subset E$ implies that $E = V \cap V'$ set theoretically. As $V \cap V'$ is reduced we conclude $E = V \cap V'$ scheme theoretically, i.e., $g|_{V \cap V'} = g'|_{V \cap V'}$. It follows that we can glue the representatives $g : V \to Y$ of $\varphi $ to a morphism $f : U \to Y$, see Schemes, Lemma 26.14.1. We omit the proof of the final statement. $\square$
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