I work over field of charateristic zero. I have a morphism $f\colon X\to Y$ of smooth varieties of the same dimension which is proper and dominant. Is it true that $f=g\circ h$ where h is birational and $g$ is finite? I try to use Stein factorisation, but I am not sure why connectedness imply that my morphism is birational.
Intutively it is clear: over some open my morphism is finite and so the composition of finite and trivial one.
By Stein factorization, we may assume $f$ is a dominant rational map of varieties of the same dimension, which has connected fibers. Hartshorne's exercise 3.22(e) (Chevalley's theorem on fiber dimension) implies that there is a dense open subset of $Y$ over which $f$ has $0$-dimensional fibers. But then, $0$-dimensional connected schemes are points, so $f$ is generically one-to-one.
It follows from Stein Factorization that quasi-finite + proper = finite, and thus over this open set $f$ is finite of degree $1$. Hence $f$ is an isomorphism over this set, so it is birational.