Lemma 76.39.3. Let $S$ be a scheme. Let $f : X \to B$ be a morphism of algebraic spaces over $S$. Let $U \subset B$ be an open subspace. Assume

$B$ is quasi-compact and quasi-separated,

$U$ is quasi-compact,

$f : X \to B$ is proper, and

$f^{-1}(U) \to U$ is an isomorphism.

Then there exists a $U$-admissible blowup $B' \to B$ such that the strict transform $X'$ of $X$ maps isomorphically to $B'$.

**Proof.**
By Lemma 76.39.1 we may assume that $X \to B$ is flat and of finite presentation. After replacing $B$ by a $U$-admissible blowup if necessary, we may assume that $U \subset B$ is scheme theoretically dense. Then $f$ is finite by Lemma 76.37.4 and an open immersion by Lemma 76.37.5. Hence $f$ is an open immersion whose image is closed and contains the dense open $U$, whence $f$ is an isomorphism.
$\square$

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