Is every surjective holomorphic self-map on a compact complex manifold finite-to-one? I have already asked this question on stack exchange, but I didn’t get any answer.
Let $X$ be a compact connected complex manifold.

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)?

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).
 A: 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 ${}$

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.

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 about this problem.
References.
[MSE] https://math.stackexchange.com/q/1556561/456212
[G03] Gromov, M: On the entropy of holomorphic maps, Enseign. Math. II. Sér. 49, No. 3-4, 217-235 (2003). ZBL1080.37051.
[W74] Wells, R. O. jun: Comparison of de Rham and Dolbeault cohomology for proper surjective mappings, Pac. J. Math. 53, 281-300 (1974). ZBL0261.32005.
