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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.

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.

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.

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.

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.

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.