Lemma 29.55.7. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Then the seminormalization of $X$ in $X^\nu$ is is the seminormalization of $X$. In a formula: $X^{sn} = X^{X^\nu /sn}$.

Proof. Let $f : Y \to X$ be as in (29.54.0.1) so that $X^\nu$ is the normalization of $X$ in $Y$. The seminormalization $X^{sn} \to X$ of $X$ is the initial object in the category of universal homeomorphisms $X' \to X$ inducing isomorphisms on residue fields. Since $Y$ is the disjoint union of the spectra of the residue fields at the generic points of irreducible components of $X$, we see that for any $X' \to X$ in this category we obtain a canonical lift $f' : Y \to X'$ of $f$. Then by Lemma 29.53.4 we obtain a canonical morphism $X^\nu \to X'$. Whence in turn a canonical morphism $X^{X^\nu /sn} \to X'$ by the universal property of $X^{X^\nu /sn}$. In this way we see that $X^{X^\nu /sn}$ satisfies the same universal property that $X^{sn}$ has and we conclude. $\square$

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