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Let $V, \tilde{V}$ be smooth algebraic varieties over $\mathbb{C}$ and $f \colon \tilde{V} \rightarrow V$ a projective (or proper) birational morphism. Assume that the exceptional locus $E \subset \tilde{V}$ has codimension $\ge 2$.

Question Is $f$ an isomorphism?

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    $\begingroup$ Yes, such an $f$ is an isomorphism. Consider the pullback map on relative differentials, $f^*:f^*\Omega^1_V \to \Omega^1_{\tilde{V}}$. This is a map of locally free sheaves of the same rank. It is an isomorphism if and only if the associated determinant is an isomorphism, i.e., it is everywhere nonzero considered as a section of the associated Hom sheaf. This Hom sheaf is invertible, so this section is zero on a Cartier divisor. Your hypotheses imply this Cartier diviser is empty. Hence $f^*$ is everywhere an isomorphism. $\endgroup$
    – Jason Starr
    Commented Jun 14, 2012 at 10:47
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    $\begingroup$ Also see Sándor's answer to this question. mathoverflow.net/questions/31696/… $\endgroup$
    – Karl Schwede
    Commented Jun 14, 2012 at 12:02
  • $\begingroup$ @ Jason Starr, Thank you very much for the answer. I think that this answers my question. $\endgroup$
    – tarosano
    Commented Jun 14, 2012 at 12:54
  • $\begingroup$ @ Karl Schwede, thank you very much for teaching me the related question. $\endgroup$
    – tarosano
    Commented Jun 14, 2012 at 12:55
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    $\begingroup$ I think this is Zariski's "main theorem", factorial or smooth case, as in Shafarevich BAG vol. 1, p.120, or Mumford's red book, SLN 1358, 2nd ed. p.210. $\endgroup$
    – roy smith
    Commented Jun 15, 2012 at 16:25

1 Answer 1

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Yes. Suppose $f$ contracts a curve $C$. Then for any ample divisor $D$, we have $D\cdot C>0$. But $D=f^*f_*D$ by your hypotheses on the exceptional locus, and so $D\cdot C=f_*D\cdot f_*C=0$, a contradiction.

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  • $\begingroup$ @ John, thank you for the comment. Actually, I'm not assuming that $f$ is extremal. I thought contractions like flopping contractions might cause a problem, but it seems there does not exist such ones between smooth ones. $\endgroup$
    – tarosano
    Commented Jun 14, 2012 at 12:58
  • $\begingroup$ Hi, I edited the answer. I think you can make it work in $\mathbb{Q}$-factorial case as well. $\endgroup$
    – J.C. Ottem
    Commented Jun 14, 2012 at 15:58
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    $\begingroup$ @Ottem -- What if the varieties / the morphism are not projective? $\endgroup$
    – Jason Starr
    Commented Jun 14, 2012 at 16:35
  • $\begingroup$ And then you need to invoke something like Zariski's Main Theorem to conclude that a proper birational quasi-finite map between smooth (or normal) varieties is an isomorphism. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 23, 2020 at 6:48

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