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Let $k$ be an algebraically closed field of characteristic $0$ and let $f:X \rightarrow Y$ be a map between smooth varieties over $k$ that is bijective on closed points. Musr $f$ be an isomorphism?

The restriction to fields of characteristic $0$ is necessary to rule out the Frobenius map, and the restriction to smooth varieties is necessary to rule out resolutions of singularities of curves. My guess is that there will be counterexamples with $X$ and $Y$ algebraic surfaces, but I do not know a rich enough store of examples of surfaces to find one.

Edit: I assert nothing about the maps on tangent spaces (unlike the question that someone has claimed is a duplicate of this one), and would be happy for a counterexample where that is the issue (but as I said, I insist that my varieties be smooth, which rules out the counterexamples in the purported duplicate question).

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    $\begingroup$ @IgorRivin: none of the answers to that question address the question I asked (and the questions are definitely different). The only example that is given is singular. $\endgroup$
    – Julia
    Commented Jul 8, 2018 at 3:39
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    $\begingroup$ Under your conditions, $f$ must be an isomorphism. Apply Zariski's main theorem to factor $f$ into an open immersion and a finite mapping. This finite mapping is birational, hence an isomorphism because $Y$ is normal (this is all you need for conditions on $Y$), so $f$ is an open immersion. The only bijective open immersions are isomorphisms, giving you the result. $\endgroup$
    – dhy
    Commented Jul 8, 2018 at 4:32

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