Any non-constant surjective holomorphic map between connected compact complex manifolds of equal dimension is a ramified finite-sheet covering I'm reading this site:holomorphy of inverse map
There is a statement made by Colin Tan at the last answer made by himself.
Any non-constant surjective holomorphic map between connected compact complex manifolds of equal dimension is a ramified finite-sheet covering.
I am amazing about this result.I cannot prove it.Is this satement right?Can you prove it or give me some reference about this topic?
Since there are some disagreements,can you give some mild condition to draw the conclusion?
Thanks in advance!
 A: The correct statement is the following:

Proposition. Let $X$, $Y$ be irreducible complex spaces. Then every holomorphic, finite surjection $\pi \colon X \longrightarrow Y$ is an analytic (in general ramified) covering map. 

For a proof of this standard result, see [Grauert-Remmert, Coherent Analytic Sheaves, page 179].
From the Proposition one can deduce the following

Corollary. Let $X$, $Y$ be irreducible, compact complex manifolds of the same dimension. Then any  holomorphic, flat surjection $\pi \colon X \longrightarrow Y$ is an analytic covering map.

Indeed, since $\dim X = \dim Y$ one has that the dimension of the general fibre of $\pi$ is $0$, and since $\pi$ is flat all the fibres must have dimension $0$. Finally,  $X$ and $Y$ are compact so we deduce that all the fibres are finite, hence $\pi$  is a finite map and one can apply the Proposition.   
Without the flatness assumption the statement of the Corollary is clearly false, as shown by the following example. Let $X$ be the blow-up of $\mathbb{P}^2$ at a point $p$ and $\pi \colon X \longrightarrow \mathbb{P}^2$ the blow-up map. Then $\pi$ is birational, hence the general fibre consists of a single point. However, $\pi$ is not an analytic covering, since there is a fibre of dimension $1$: in fact, $\pi$ contracts the exceptional divisor $E \subset X$ to the point $p$.
