Lemma 76.25.2 (Normalization commutes with smooth base change). Let $S$ be a scheme. Let
be a fibre square of algebraic spaces over $S$. Assume $f$ is quasi-compact and quasi-separated and $\varphi $ is smooth. Let $Y_ i \to X_ i' \to X_ i$ be the normalization of $X_ i$ in $Y_ i$. Then $X_2' \cong X_2 \times _{X_1} X_1'$.
Proof. The base change of the factorization $Y_1 \to X_1' \to X_1$ to $X_2$ is a factorization $Y_2 \to X_2 \times _{X_1} X_1' \to X_1$ and $X_2 \times _{X_1} X_1' \to X_1$ is integral (Morphisms of Spaces, Lemma 67.45.5). Hence we get a morphism $h : X_2' \to X_2 \times _{X_1} X_1'$ by the universal property of Morphisms of Spaces, Lemma 67.48.5. Observe that $X_2'$ is the relative spectrum of the integral closure of $\mathcal{O}_{X_2}$ in $f_{2, *}\mathcal{O}_{Y_2}$. If $\mathcal{A}' \subset f_{1, *}\mathcal{O}_{Y_1}$ denotes the integral closure of $\mathcal{O}_{X_2}$, then $X_2 \times _{X_1} X_1'$ is the relative spectrum of $\varphi ^*\mathcal{A}'$ as the construction of the relative spectrum commutes with arbitrary base change. By Cohomology of Spaces, Lemma 69.11.2 we know that $f_{2, *}\mathcal{O}_{Y_2} = \varphi ^*f_{1, *}\mathcal{O}_{Y_1}$. Hence the result follows from Lemma 76.25.1. $\square$
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