In topology, a quotient map is a surjective map $\pi:X\to Y$ such that $V\subseteq Y$ is open in $Y$ if and only if $\pi^{-1}(V)$ is open in X. This definition has the following nice property: If $\rho:X\to Z$ is a continuous map and $f:Y\to Z$ is any map of sets such that $\rho=f\circ \pi$, then $f$ is continuous.

What is the corresponding concept of quotient map in the category of quasi-projective varieties? Is it only $\textit{surjective regular map}$?

A more concrete question: let $\pi:X\to Y$ be a surjective regular map of quasi-projective varieties. Is it true that for any regular map $\rho:X\to Z$, a map of sets $f:Y\to Z$ such that $\rho=f\circ\pi$ is necessarily a regular map?

  • $\begingroup$ There are different notions of quotient maps in AG. Keyword: "geometric invariant theory". $\endgroup$ – paf Dec 22 '16 at 18:38
  • $\begingroup$ @paf well, say I only care about quotients of $X$ by a finite group. Is surjective regular map the correct notion of quotient map? (in the sense of my original question). $\endgroup$ – Marco Flores Dec 22 '16 at 18:49
  • 1
    $\begingroup$ In the affine case, the inclusion of the ring of invariants $R^G \subset R$ into a ring $R$ corresponds to the quotient of $\operatorname{Spec}(R)$ by $G$. A surjective regular map is definitely NOT the right idea; consider e.g. the blowup of the plane at a point. $\endgroup$ – Tabes Bridges Dec 22 '16 at 19:35

The answer to the concrete question is no. For example, take $X=\mathbb{P}^1=Z$ with $\rho$ the identity. Let $Y$ be a projective singular rational curve with a cusp and let $\pi:X\to Y$ be the normalization map. Then, $\pi$ is a bijection and thus we get a map of sets $f:Y\to Z$ with $\rho=f\circ\pi$, but $f$ is not regular.

  • $\begingroup$ Thank you. Do you think the answer would change if I require $X,Y$ and $Z$ to be smooth? $\endgroup$ – Marco Flores Dec 23 '16 at 5:44
  • $\begingroup$ At least over complex numbers, if $Y$ is normal, $f$ is indeed regular. $\endgroup$ – Mohan Dec 27 '16 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.