Lemma 81.14.5. Let $S$ be a scheme. Let $X \to B$ and $Y \to B$ be morphisms of algebraic spaces over $S$. Let $U \subset X$ be an open subspace. Let $V \to X \times _ B Y$ be a quasi-compact morphism whose composition with the first projection maps into $U$. Let $Z \subset X \times _ B Y$ be the scheme theoretic image of $V \to X \times _ B Y$. Let $X' \to X$ be a $U$-admissible blowup. Then the scheme theoretic image of $V \to X' \times _ B Y$ is the strict transform of $Z$ with respect to the blowing up.
Proof. Denote $Z' \to Z$ the strict transform. The morphism $Z' \to X'$ induces a morphism $Z' \to X' \times _ B Y$ which is a closed immersion (as $Z'$ is a closed subspace of $X' \times _ X Z$ by definition). Thus to finish the proof it suffices to show that the scheme theoretic image $Z''$ of $V \to Z'$ is $Z'$. Observe that $Z'' \subset Z'$ is a closed subspace such that $V \to Z'$ factors through $Z''$. Since both $V \to X \times _ B Y$ and $V \to X' \times _ B Y$ are quasi-compact (for the latter this follows from Morphisms of Spaces, Lemma 67.8.9 and the fact that $X' \times _ B Y \to X \times _ B Y$ is separated as a base change of a proper morphism), by Morphisms of Spaces, Lemma 67.16.3 we see that $Z \cap (U \times _ B Y) = Z'' \cap (U \times _ B Y)$. Thus the inclusion morphism $Z'' \to Z'$ is an isomorphism away from the exceptional divisor $E$ of $Z' \to Z$. However, the structure sheaf of $Z'$ does not have any nonzero sections supported on $E$ (by definition of strict transforms) and we conclude that the surjection $\mathcal{O}_{Z'} \to \mathcal{O}_{Z''}$ must be an isomorphism. $\square$
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