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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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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. – Jason Starr Jun 14 '12 at 10:47
Also see Sándor's answer to this question.… – Karl Schwede Jun 14 '12 at 12:02
@ Jason Starr, Thank you very much for the answer. I think that this answers my question. – tarosano Jun 14 '12 at 12:54
@ Karl Schwede, thank you very much for teaching me the related question. – tarosano Jun 14 '12 at 12:55
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. – roy smith Jun 15 '12 at 16:25

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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@ 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. – tarosano Jun 14 '12 at 12:58
Hi, I edited the answer. I think you can make it work in $\mathbb{Q}$-factorial case as well. – J.C. Ottem Jun 14 '12 at 15:58
@Ottem -- What if the varieties / the morphism are not projective? – Jason Starr Jun 14 '12 at 16:35

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