Let $f:X\to Y$ be a morphism of algebraic varieties over $\mathbb C$. Assume that


a) $f$ is bijective on $\mathbb C$-points


b) $X$ is connected


c) $Y$ is normal.


Does it imply that $f$ is an isomorphism? If not, are there stronger reasonable conditions under which it is true?



Bijection implies isomorphism for algebraic varieties
I am just expanding the comment above. The statement is local on $Y$, so assume that $Y$ is connected. Since $Y$ is normal, this is equivalent to assuming that $Y$ is irreducible. Grothendieck's formulation of Zariski's Main Theorem is EGA $\textrm{III}\_2$, Théorème 4.4.3, p. 136. According to that theorem, there exists a factorization of $f$, $f=\overline{f}\circ i$ where $i:X\hookrightarrow\overline{X}$ is a dense open immersion, and where $\overline{f}:\overline{X}\to Y$ is a finite morphism. Identify $X$ with the image of $i$.


Since $f$ is bijective on points, there exists a unique irreducible component $\overline{X}\_0$ of $\overline{X}$ that dominates $Y$. The restriction of $\overline{f}$ to $\overline{X}\_0$ is a finite, surjective morphism that is generically one-to-one. Since the characteristic is $0$, this morphism is birational (in positive characteristic, it might be purely inseparable but not birational). Thus, by the classical form of Zariski's Main Theorem, $\overline{f}$ restricts to an isomorphism from $\overline{X}\_0$ to $Y$, cf. Mumford's *Red Book of Varieties and Schemes*, p. 288. 


By way of contradiction, assume that $\overline{X}\_0$ is a proper subset of $\overline{X}$. Denote by $Z \subset \overline{X}$ the union of all irreducible components different from $\overline{X}\_0$. Since $X$ is a dense open in $\overline{X}$, also $X\cap Z$ is a dense open in $Z$. Thus, also $(X\cap Z) \setminus (\overline{X}\_0\cap Z)$ is a dense open subset of $Z$. Denote by $W$ the image in $Y$ of this open subset. Since $f$ is injective, the constructible subset $W$ is disjoint from the dense open $V=f(X\cap \overline{X}\_0)$ of $Y$. Thus, the closure of $W$ is disjoint from $V$. Since $\overline{f}$ is finite, this closure equals the image of $Z$. Thus, $X\cap \overline{X}\_0$ is disjoint from $\overline{X}\_0\cap Z$, i.e., $X\cap \overline{X}\_0\cap Z$ is empty. Thus, the open subsets $\overline{X}\_0\setminus (\overline{X}\_0\cap Z)$ and $Z\setminus (Z\cap \overline{X}\_0)$ of $\overline{X}$ pullback to disjoint open subsets of $X$ that cover $X$. Since $X$ is connected, this is a contradiction. 


Therefore, $\overline{X}\_0$ equals all of $\overline{X}$. Since $\overline{X}\_0\to Y$ is an isomorphism, it follows that $f$ is an open immersion. Since $f$ is also surjective, $f$ is an isomorphism.


