Let $X,Y$ be smooth irreducible algebraic varieties over $\mathbb{C}$, let $\pi: X \to Y$ be a morphism which is injective and surjective on closed points of $X,Y$. Then how to show $\pi$ is an isomorphism of $X,Y$ as varieties? 


This result is not true if $X,Y$ are not smooth. One can consider the example $\mathbb{P}^1 \to C$, where $C$ is a plane curve defined by $y^2-x^3=0$, because the map $t \mapsto (t^2,t^3)$ satisfies the condition, but $\mathbb{P}^1$ cannot isomporhic to $C$ which is singular at origin. 



Set isomorphism of smooth varieties implies isomorphism as varieties
One way to prove this is to show that for each closed point $x\in X$ the differential $T\_\pi(x):T\_xX\rightarrow T\_{\pi(x)}Y$ is surjective, that is $\pi$ is a smooth morphism.


Let $\pi:X\rightarrow Y$ be a morphism or relative dimension $r$ of smooth varieties over an algebraically closed field. Assume that the relative cotangent sheaf $\Omega\_{X/Y}$ is locally free of rank $r$ on $X$. We have an exact sequence
$$\pi^{\*}\Omega\_{Y}\rightarrow\Omega\_{X}\rightarrow\Omega\_{X/Y}\mapsto 0.$$
Let $k(x)$ be the residue field at a closed point $x$. Tensorizing we get  

$$\pi^{\*}\Omega\_{Y}\otimes k(x)\rightarrow\Omega\_{X}\otimes k(x)\rightarrow\Omega\_{X/Y}\otimes k(x)\mapsto 0.$$
Since $X$ and $Y$ are smooth and $\Omega\_{X/Y}$ is locally free of rank $r$ these three vector spaces are of dimension $dim(Y),dim(X),r$ respectively. So the first map is injective and we have 
$$0\mapsto\pi^{\*}\Omega\_{Y}\otimes k(x)\rightarrow\Omega\_{X}\otimes k(x)\rightarrow\Omega\_{X/Y}\otimes k(x)\mapsto 0.$$
For any closed point $x\in X$ we have $k(x)\cong k$. Therefore we can identify the injective map $\pi^{\*}\Omega\_{Y}\otimes k\rightarrow\Omega\_{X}\otimes k$ with the map between the cotangent spaces $\mathfrak{m}\_y/\mathfrak{m}^2\_y\rightarrow\mathfrak{m}\_x/\mathfrak{m}^2\_x$ where $y = \pi(x)$. Dualizing we have that the differential $T\_\pi(x):T\_xX\rightarrow T\_yY$ is surjective. Therefore $\pi$ is a smooth morphism.


In particular, if any fiber of $\pi$ is just one point, then $\Omega\_{X/Y}$ is locally free of rank $r = 0$ and $\pi$ is a smooth morphism of relative dimension zero. Finally, since $\pi$ is surjective it is an isomorphism.


