Why is the inverse of a bijective rational map rational?

Question

Let $f:X\to Y$ be a bijective rational map of an open dense subset $X$ of $\mathbb{C}\times\mathbb{C}$ onto an open dense subset $Y$ of $\mathbb{C}\times\mathbb{C}$. How to prove that the inverse map $f^{-1}:Y\to X$ is rational as well? Could you recommend any exact reference to a theorem which guarantees this?

edit Oct 30-31 I have looked carefully through the provided answers and references and still did not find any indication how to prove that the inverse map is rational. Theorem 12.83 in page 355 from [1] mentioned below does not contain the part of Proposition 2 below starting from 'hence an isomorphism onto its image'. Definition of an open immersion, see Definition 3.40 in page 83 from [1], does not say that an open immersion is an isomorphism onto its image. If there is a theorem that an open immersion in the sense of [1] is an isomorphism onto its image, then could you provide an exact reference to it (such a theorem seems to be more or less equivalent to the initial question)?

Related question: Isomorphism between varieties of char 0.

[1] U. Görz and T. Wedhorn, Algebraic Geometry I, Vieweg+Teubner Verlag-Springer Fachmedien Wiesbaden GmbH 2010.

Answer 1

Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that $X$ and $Y$ are algebraic varieties. Therefore we have a particular case of the following well-known statement in algebraic geometry.

Proposition 1. Let $k$ be an an algebraically closed field of characteristic zero and $f \colon X \to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism, that is $f^{-1} \colon Y \to X$ is a morphism as well.

This is in turn a consequence of the following version of Zariski's Main Theorem, see [Görz-Wedhorn, Algebraic Geometry I, Theorem 12.83 page 355].

Proposition 2. Let $f \colon X \to Y$ be a separated morphism of finite type such that $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$ is an isomorphism. Let $V$ be the open set of $X$ given by the points $x$ such that $\dim f^{-1}(f(x))=0$. Then the restriction $f_{|V} \colon V \to Y$ is an open immersion, hence an isomorphism onto its image. In particular, if $f$ is dominant and all fibres of $f$ are finite then $f$ is birational.

In fact, given a proper morphism $f \colon X \to Y$ with integral fibres, if $Y$ is normal then $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$ is an isomorphism, see again [Görz-Wedhorn, Exercise 12.29 page 365], hence Proposition 2 implies Proposition 1.

In your situation, $k= \mathbb{C}$ and $Y$ is smooth (hence normal), because it is an open dense subset of $\mathbb{C}^2$. So the previous results apply.

A related discussion can be found in this MathOverflow question: Isomorphism between varieties of char 0.

Edit October 31, 2014 (see comments). In Görz-Wedhorn's book an open immersion is defined as a morphism $j \colon V \to Y$ such that the underlying continuous map is a homeomorphism of $V$ onto the open set $U:=j(V)$ and the sheaf homomorphism $\mathscr{O}_Y \to j_* \mathscr{O}_V$ induces an isomorphism $\mathscr{O}_{Y|U} \cong j_* \mathscr{O}_V$ (of sheaves on $U$). This definition is equivalent to requiring that $j \colon Y \to X$ is an isomorphism onto the open subscheme $U:=j(V)$, see [EGA I, Chapitre I Proposition 4.2.2 a), page 122]. A related discussion is in this MSE question, see in particular Georges Elencwajg's answer.