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

1. $p$ is proper and $p^{-1}(U) \to U$ is an isomorphism,

2. $f'|_{p^{-1}(U)}$ 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 $p^{-1}(U)$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).