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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