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Let $f$ be a morphism between two irreducible varieties, and one-to-one. Is $f$ actually a homeomorphism onto its image?

Here the varieties are equipped with Zariski topology.

I know if the varieties are projective then it is true. (Because projective varieties are complete, so $f$ maps closed sets to closed sets)

Akatsuki
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    With respect to what topology (Zariski or complex)? Since you tagged both topology and algebraic geometry, it's not clear to me which world you are working in. – Remy Dec 30 '15 at 23:51
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    Note that irreducibility is essential, otherwise there are fairly simple counterexamples, like the punctured line plus a point mapping to the line—or slightly more complicated connected examples in dimension 2. – Andrew Dudzik Dec 31 '15 at 01:57
  • @Remy Zariski topology – Akatsuki Dec 31 '15 at 03:57
  • @Slade Yes. I have added this condition. I used the term in Hartshorne's book where varieties are always irreducible. – Akatsuki Dec 31 '15 at 04:59
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    "Variety" is a singularly terrible word, because depending on the text it may or may not mean: irreducible, reduced, over an algebraically closed field (and I have even seen it used to describe schemes over $\mathbb{Z}$ in some cases). Reduced-ness isn't really important here, having an algebraically closed base might be. Anyway, this is a really interesting and challenging problem, and I hope it gets more traction. – Andrew Dudzik Dec 31 '15 at 06:00

2 Answers2

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Let $Y\subset \mathbb P^2(\mathbb C)$ be the singular cubic $y^2z=x^2z+x^3$ and consider the morphism $$f:X=\mathbb P^1(\mathbb C)\setminus \{(-1:1)\}\to Y:(u:v)\mapsto (u^2v-v^3:u^3-uv^2:v^3) $$ That morphism $f$ is a bijective continuous map with source $X$ homeomorphic to $\mathbb C$ but is not a homeomorphism in the classical topology because $Y$ is compact and $X\cong \mathbb C$ is not.

Note however that in the Zariski topology $X$ and $Y$ are homeomorphic, because over $\mathbb C$ every bijection between irreducible curves is (quite counterintuitively!) a homeomorphism in the Zariski topology!

Georges Elencwajg
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This example comes from a book written by a professor of mine. The map $f$ is actually injective, but it is not an embedding.

enter image description here

here condition (i) is injectivity, while condition (ii) is that the differential of $f$ has always rank equal to $1 = \dim \Bbb{R}$.

Crostul
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