# Any non-constant surjective holomorphic map between connected compact complex manifolds of equal dimension is a ramified finite-sheet covering

Any non-constant surjective holomorphic map between connected compact complex manifolds of equal dimension is a ramified finite-sheet covering.

Since there are some disagreements,can you give some mild condition to draw the conclusion? Thanks in advance!

• I do not think this is true without a flatness assumption. The map $\pi \colon X \to \mathbb{P}^2$ obtained by blowing-up a point is a surjective, holomorphic map between two smooth complex surfaces, but it is not a finite-sheet covering since it contracts the exceptional divisor, which is a curve in $X$. – Francesco Polizzi Nov 6 '13 at 13:17
• The qualification that the covering is "ramified" is important here. Checking Wikpedia, the standard terminology is "branched covering" rather than "ramified", that is, a covering over a dense subspace of the base. With this qualification, your example is not yet a counterexample for your blowup is bijective outside the blowup point in the base projective plane. – user2529 Jul 20 '14 at 6:38

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 $p$.