Lemma 31.36.2. Let S be a scheme. Let X, Y be schemes over S. Assume X is Noetherian and Y is proper over S. Given an S-rational map f : U \to Y from X to Y there exists a morphism p : X' \to X and an S-morphism f' : X' \to Y such that
p is proper and p^{-1}(U) \to U is an isomorphism,
f'|_{p^{-1}(U)} is equal to f \circ p|_{p^{-1}(U)}.
Proof. Denote j : U \to X the inclusion morphism. Let X' \subset Y \times _ S X be the scheme theoretic image of (f, j) : U \to Y \times _ S X (Morphisms, Definition 29.6.2). The projection g : Y \times _ S X \to X is proper (Morphisms, Lemma 29.41.5). The composition p : X' \to X of X' \to Y \times _ S X and g is proper (Morphisms, Lemmas 29.41.6 and 29.41.4). Since g is separated and U \subset X is retrocompact (as X is Noetherian) we conclude that p^{-1}(U) \to U is an isomorphism by Morphisms, Lemma 29.6.8. On the other hand, the composition f' : X' \to Y of X' \to Y \times _ S X and the projection Y \times _ S X \to Y agrees with f on p^{-1}(U). \square
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