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Let $X$ be a (irreducible) variety, $\tilde{X}$ its normalization, $\pi\colon \tilde{X}\to X$ the natural map. Is there a "nice" characterization of when $\pi$ is a homeomorphism? E.g. is it enough to know that $\pi$ is injective?

If it makes things simpler, I'm most interested in the case that $X$ is an affine toric variety over $\Bbb C$.

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It's easy to check that injectivity is enough. Indeed, normalisation is finite, hence closed. Since it's also dominant, it's surjective. A surjective closed injection is a homeomorphism.

Remark. We can actually say much more: since $\pi$ is a finite morphism (hence affine), we can reduce to the case of ring maps $A \to B$. If your ground field $k$ is algebraically closed, then normalisation does not induce residue field extensions (in general, it could).

Hence, by Tag 0BRD, the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is a universal homeomorphism (from what we already wrote, this is just the observation that the assumptions are stable under base change). Then Tag 0CND gives the following criterion:

Proposition. Let $A \subseteq B$ be a ring extension. Then the following are equivalent.

  1. $\operatorname{Spec} B \to \operatorname{Spec} A$ is a universal homeomorphism inducing isomorphisms on residue fields,
  2. Every finite subset $E \subseteq B$ is contained in an extension $A[b_1,\ldots,b_n] \subseteq B$ such that $b_i^2$ and $b_i^3$ are in $A[b_1,\ldots,b_{i-1}]$ for all $i \in \{1,\ldots,n\}$.

Since in our case the extension $A \to B$ is finitely generated, we actually get $B = A[b_1,\ldots,b_n]$ of the form above.

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  • $\begingroup$ If I understand correctly, your proposition implies that if B is the integral closure of an affine $\Bbb C$-algebra A, then $Spec B \to Spec A$ is a universal homeomorphism iff A is seminormal. Is this right? $\endgroup$
    – Avi Steiner
    Commented May 10, 2017 at 4:04
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    $\begingroup$ @AviSteiner: no, it's somehow opposite to that. Seminormal means if $a \in \operatorname{Frac}(A)$ satisfies $a^2, a^3 \in A$, then $a \in A$. In our case, we get $b_1^2, b_1^3 \in A$, but $b_1 \not\in A$. I think the condition is probably that the seminormalisation of $A$ is normal, but I don't know the terminology well enough to say this with full conviction. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 10, 2017 at 4:28
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    $\begingroup$ In EGA varieties whose normalization morphism is bijective are called (geometrically) unibranch. There is also a wikipedia entry on that with references. $\endgroup$
    – Friedrich Knop
    Commented May 10, 2017 at 7:47
  • $\begingroup$ @FriedrichKnop: ah, I somehow did not make that connection. It's never explicitly stated that unibranch boils down to this, but it seems to follow immediately from the definition (plus the fact that localisation commutes with normalisation). $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 10, 2017 at 10:01

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