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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.

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    $\begingroup$ EGA I, Chapitre I Proposition 4.2.2 a), page 122. $\endgroup$ – Francesco Polizzi Oct 30 '14 at 9:10
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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.

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    $\begingroup$ Thanks you very much! My question is what is a REFERENCE to this proposition or to the initial question being a particular case. In the discussion of the question Isomorphism between varieties of char 0 I could not find any indication how it is proved that the inverse map is rational. The version of the Zarisky main theorem is not sufficient because of the assumption that the initial map is birational. This answer does not prove anything on the rationality of $f^{-1}$. $\endgroup$ – Mikhail Skopenkov Sep 30 '14 at 11:35
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    $\begingroup$ I have looked carefully through the provided references and did not find any indication to the proof that the inverse map is rational. Theorem 12.83 in page 355 from [1] does not contain the part of Proposition 2 starting from 'hence an isomorphism onto its image'. Definition 3.40 from [1] does not tell that an open immersion is an isomorphism onto its image. Finally, I cannot see immediately which assumptions of Proposition 2 are not satisfied by the map $C\to C$, $z\mapsto z^2$ with no rational inverse. Could you explain what am I missing? [1] Görz-Wedhorn, Algebraic Geometry I. $\endgroup$ – Mikhail Skopenkov Oct 29 '14 at 19:15
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    $\begingroup$ (2) The map $z \to z^2$ does not satisfy the assumption that $\mathcal{O}_Y \to f_* \mathcal{O}_X$ is an isomorphism. The point is that its fibres are not integral, since it is a (ramified) double cover. In fact, $f_* \mathcal{O}_X$ is a rank $2$ vector bundle in this case, containing $\mathcal{O}_X$ as a direct summand. $\endgroup$ – Francesco Polizzi Oct 29 '14 at 19:43
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    $\begingroup$ thank you very much! Point (2) is clear, thanks. Point (1) turns us back to the initial question: what is the exact REFERENCE to the proof that the inverse map is rational? Theorem 12.83 in page 355 from [1] does NOT tell this because Definition 3.40 from the same book [1] does not say that an open immersion is an isomorphism onto its image. The authors of [1] use the term immersion in their own sense (they GIVE a formal definition DIFFERENT from the one from Glossary_of_scheme_theory). If these defs are equivalent then a reference to the proof of equivalence is very much wanted. $\endgroup$ – Mikhail Skopenkov Oct 29 '14 at 19:55
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    $\begingroup$ Yes, the two definitions of open immersion are equivalent. A reference is Grothendieck EGA I, Chapitre I Proposition 4.2.2 a), page 122. See also this MSE question and the beautiful answer of Georges Elencwajg: math.stackexchange.com/questions/70293/… $\endgroup$ – Francesco Polizzi Oct 30 '14 at 9:09

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