Lemma 29.54.11. Let $X$ be an irreducible, geometrically unibranch scheme. The normalization morphism $\nu : X^\nu \to X$ is a universal homeomorphism.

**Proof.**
We have to show that $\nu $ is integral, universally injective, and surjective, see Lemma 29.45.5. By Lemma 29.54.5 the morphism $\nu $ is integral. Let $x \in X$ and set $A = \mathcal{O}_{X, x}$. Since $X$ is irreducible we see that $A$ has a single minimal prime $\mathfrak p$ and $A_{red} = A/\mathfrak p$. By Lemma 29.54.4 the stalk $A' = (\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A$ in the fraction field of $A_{red}$. By More on Algebra, Definition 15.106.1 we see that $A'$ has a single prime $\mathfrak m'$ lying over $\mathfrak m_ x \subset A$ and $\kappa (\mathfrak m')/\kappa (x)$ is purely inseparable. Hence $\nu $ is bijective (hence surjective) and universally injective by Lemma 29.10.2.
$\square$

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