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 $\Bbb C$.



When is the morphism to a variety from its normalization a homeomorphism?
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](http://stacks.math.columbia.edu/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](http://stacks.math.columbia.edu/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.


