Bijection implies isomorphism for algebraic varieties 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?
 A: 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.
