I have already asked this question on [stack exchange](https://math.stackexchange.com/questions/3918462/is-surjective-holomorphic-self-map-on-compact-complex-manifold-finite), but I didn’t get any answer.


Let $X$ be a compact connected complex manifold.



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> Let $f:X \to X$ be a **surjective** holomorphic map. Is it true that $f$ is a finite map (i.e., every point has finitely many preimages)?
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If $X$ is one dimensional then the answer to the above question is yes. If $X$ is a complex projective space then the answer is again yes.


If we don’t assume surjectivity then we can easily construct non-constant, non-finite self-maps (just consider the product manifold and $f$ to be projection on one of the coordinates).



Is every surjective holomorphic self-map on a compact complex manifold finite-to-one?
Let me give a sketch of proof for Gromov's claim in the case where $X$ is Kähler. More precisely, let me prove the following ${}$



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> **Proposition [G03, p.223].** Let $X$, $Y$ be two complex manifolds (not necessarily compact) of the same dimension and having the same even Betti numbers. If $X$ is Kähler, then every proper surjective holomorphic map $f \colon X \to Y$ is finite-to-one.
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*Proof.* By a result of Wells [**W74**, Theorem 3.1], the map $f \colon X \to Y$ induces an injection $$f^\* \colon H^{r}(Y, \, \mathbb{C}) \to H^{r}(X, \, \mathbb{C})$$ for all $r$. Using the duality $$H\_r(X, \, \mathbb{C})=\mathrm{Hom}(H^r(X, \, \mathbb{C}), \, \mathbb{C})$$ we infer that the induced map in homology $$f\_\* \colon H\_{r}(X, \, \mathbb{C}) \to H\_{r}(Y, \, \mathbb{C})$$ is surjective for all $r$.


If $f \colon X \to Y$ contracts a subvariety $Z$ of positive complex dimension $i$ to a point, then its fundamental class would give an element $[Z]$ in the kernel of $$f\_\* \colon H\_{2i}(X, \, \mathbb{C}) \to H\_{2i}(Y, \, \mathbb{C})$$


Now $Z$ is compact by the properness assumption and so, pairing with the Kähler  

form on $X$, we can deduce that $[Z]$ is non-zero. Hence $b\_{2i}(X) > b\_{2i}(Y)$, contradiction. $\square$


**Remark.** The first version of this answer did not assume that $X$ was Kähler (in fact, Gromov does not make this assumption). However, if $X$ is not Kähler it may happen that the fundamental class of a compact subvariety of $X$ is homologically trivial, see the example of Hopf's surface given in Michael Albanese answer to **[MSE]**: in this case, $X$ contains a torus $C$ whose fundamental class $[C]$ is zero in $H\_2(X, \, \mathbb{C})$, simply because the last group is trivial. So, if $X$ is not Kähler, the last part of the proof breaks down.


In fact, I do not know how to make the argument work without the Kähler assumption for $X$, so I asked a new [MO question](https://mathoverflow.net/questions/377399/about-a-claim-by-gromov-on-proper-holomorphic-maps) about this problem.


**References.**


**[MSE]** <https://math.stackexchange.com/q/1556561/456212>


**[G03]** Gromov, M: [On the entropy of holomorphic maps](https://www.ihes.fr/%7Egromov/wp-content/uploads/2018/08/1024.pdf), *Enseign. Math.* II. Sér. **49**, No. 3-4, 217-235 (2003). [ZBL1080.37051](https://zbmath.org/?q=an:1080.37051).


**[W74]** Wells, R. O. jun: [Comparison of de Rham and Dolbeault cohomology for proper surjective mappings](http://dx.doi.org/10.2140/pjm.1974.53.281), *Pac. J. Math.* **53**, 281-300 (1974). [ZBL0261.32005](https://zbmath.org/?q=an:0261.32005).


