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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)

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(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. – John Salvatierrez Jul 5 '13 at 8:28
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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. – Vivek Shende Jul 5 '13 at 8:31
    
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. – Hans Jul 5 '13 at 8:55
    
@Hans: sure, I just thought the answers there could help with cooking up a counterexample. (I said it was a comment :) – John Salvatierrez Jul 5 '13 at 8:57
    
thank you, i see. unfortunately, i did not pose the question, as i wanted to... i will edit now! – Hans Jul 5 '13 at 9:06

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