About a claim by Gromov on proper holomorphic maps

Question

At p. 223 of his paper [G03], Mikhail Gromov makes the following claim:

Let $X$, $Y$ be two complex manifolds (not necessarily compact or Kähler) of the same dimension and having the same even Betti numbers. Then every proper surjective holomorphic map $f \colon X \to Y$ is finite-to-one.

In my answer to MO question 377353, I proposed a proof for this claim. However, checking the details I realized that such a proof requires that $X$ is Kähler.

In fact, the last step of my argument uses in an essential way the fact that the fundamental class of a compact subvariety of $X$ is non-trivial in homology, and this is in general false when $X$ is not Kähler. For instance, in his answer to [MSE], Michael Albanese notices that the standard Hopf's surface $X$ contains a compact torus $C$ whose fundamental class $[C]$ is zero in $H_2(X, \, \mathbb{C})$, simply because the last group is trivial.

So, let me ask the

Question. Is Gromov's claim true without the Kähler assumption for $X$? If so, what is a proof?

References.

[G03] Gromov, M: On the entropy of holomorphic maps, Enseign. Math. II. Sér. 49, No. 3-4, 217-235 (2003). ZBL1080.37051.

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

Answer 1

I think I've got it. This is only a sketch, and I am not an expert on singular complex analytic things, but it seems about right.

Let $f:X\to Y$ be a proper holomorphic map between complex manifolds of the same dimension $n$, and assume that some fiber has positive dimension. I want to find a rationally nontrivial element of the kernel of $f_\ast :H_{2k}(X)\to H_{2k}(Y)$ for some $k$.

Let $S\subset X$ be the union of all fibers having positive dimension. I presume that $S$ is a closed analytic subset of $X$. Let $Z$ be an irreducible component of $S$ of maximal dimension. Let $m$ be the dimension of $f(S)$. Let $k<n-m$ be the codimension of $Z$ in $X$. Choose a holomorphic map $j:D\to X$, where $D$ is the closed unit disk in $\mathbb C^k$, such that $j(0)\in Z$ and $j(D\backslash 0)\cap S=\emptyset$.

$f\circ j:D\to Y$ gives a relative $2k$-cycle for the pair $(Y,Y\backslash f(S))$, which I'll call $\Delta$. The $(2k-1)$-cycle $\partial \Delta$, given by the restriction of $f\circ j$ to the sphere $\partial D$, represents zero in $H_{2k-1}(Y\backslash f(S))$, because $H_{2k}(Y,Y\backslash f(S))=0$, because $2n-2m$, the (real) codimension of $f(S)$ in $Y$, is greater than $2k$. So there is a $2k$-chain $c$ in $Y\backslash f(S)$ with $\partial c=\partial\Delta $. In fact, the whole cycle $\Delta-c$ can be taken to be in a little ball, so that it is trivial in $H_{2k}(Y)$.

Because $f$ is finite to one over the complement of $f(S)$, we can arrange for $c$ to have a sort of branched cover, a chain $\tilde c$ in $X\backslash S$ whose boundary is a branched cover of $\partial\Delta$.

Now I want a chain $\tilde \Delta$ in $X$, an analytic chain that is a branched cover of $\Delta$ and has the same boundary as $\tilde c$. For this, make the fiber product of $f:X\to Y$ and $f\circ j:D\to Y$. In this fiber product let $\tilde D$ be the closure of the preimage of $D\backslash 0$. The resulting map $\tilde D\to X$ gives the desired $\tilde \Delta$.

The cycle $\tilde \Delta-\tilde c$ represents an element of $H_{2k}(X)$ which must have infinite order because its intersection number with the fundamental class of $Z$ is positive, because the only part of $\tilde \Delta-\tilde c$ that intersects $Z$ is the analytic part $\tilde \Delta$ (and it does intersect at least once, at $j(0)\in Z$).