1
$\begingroup$

Let $f: X \to Y$ be a morphism between two quasiprojective, irreducible varieties over the complex numbers, such that the image of $f$ is Zariski dense in $Y$ and there is a Zariski dense subset $U$ (not necessarily open) of $X$ such that we have $\{x\}=f^{-1}(f(x))=\{y \in X: f(y)=f(x)\}$ for all $x \in U$.

Is $f$ then always a birational morphism? If not, is there a simple counterexample?

(Sorry for the first version of the question)

Improve this question
$\endgroup$
9
  • $\begingroup$ (I don't have enough rep to comment) You might find this question to be of interest mathoverflow.net/questions/73321/…. If you allow Y to be non-normal then you can finde bijective morphisms which are not isomorphisms. $\endgroup$
    – John Salvatierrez
    Commented Jul 5, 2013 at 8:28
  • 3
    $\begingroup$ Consider $X = Y = E$ an elliptic curve, $f$ the map $\times 2$, and $U$ any infinite collection of points so that no two differ by 2-torsion. $\endgroup$
    – Vivek Shende
    Commented Jul 5, 2013 at 8:31
  • $\begingroup$ Hi Carmelo, I am aware of this. When I say "birational morphism" i do not mean "isomorphism". A birational morphism is a morphism, which is invertible as a rational map. $\endgroup$
    – Hans
    Commented Jul 5, 2013 at 8:55
  • $\begingroup$ @Hans: sure, I just thought the answers there could help with cooking up a counterexample. (I said it was a comment :) $\endgroup$
    – John Salvatierrez
    Commented Jul 5, 2013 at 8:57
  • $\begingroup$ thank you, i see. unfortunately, i did not pose the question, as i wanted to... i will edit now! $\endgroup$
    – Hans
    Commented Jul 5, 2013 at 9:06

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .