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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