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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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