The Stacks project

Proof. Let $T' \subset V$ be the complement of the maximal open over which $f|_ U$ is an isomorphism. Then $T', T$ are closed in $V$ and $T \cap T' = \emptyset $. Since $V$ is a spectral topological space, we can find constructible closed subsets $T_ c, T'_ c$ with $T \subset T_ c$, $T' \subset T'_ c$ such that $T_ c \cap T'_ c = \emptyset $ (choose a quasi-compact open $W$ of $V$ containing $T'$ not meeting $T$ and set $T_ c = V \setminus W$, then choose a quasi-compact open $W'$ of $V$ containing $T_ c$ not meeting $T'$ and set $T'_ c = V \setminus W'$). By Lemma 38.33.2 we may, after replacing $Y$ by a $V$-admissible blowing up, assume that $T_ c$ and $T'_ c$ have disjoint closures in $Y$. Set $Y_0 = Y \setminus \overline{T}'_ c$, $V_0 = V \setminus T'_ c$, $U_0 = U \times _ V V_0$, and $X_0 = X \times _ Y Y_0$. Since $U_0 \to V_0$ is an isomorphism, we can find a $V_0$-admissible blowing up $Y'_0 \to Y_0$ such that the strict transform $X'_0$ of $X_0$ maps isomorphically to $Y'_0$, see Lemma 38.31.3. By Divisors, Lemma 31.34.3 there exists a $V$-admissible blow up $Y' \to Y$ whose restriction to $Y_0$ is $Y'_0 \to Y_0$. If $f' : X' \to Y'$ denotes the strict transform of $f$, then we see what we want is true because $f'$ restricts to an isomorphism over $Y'_0$. $\square$