Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic? Are the fibers of a surjective holomorphic submersion $\mathbb{C}^n\to\mathbb{C}$ all homeomorphic?
For $n=1$ this means that a surjective entire function $\mathbb{C}\to\mathbb{C}$ without critical points assumes each value infinitely often. Is this obvious?
 A: The answer is no for $n=1$.
Lemma: Suppose $f$ is an entire function with $f^{-1}(z_0)$ finite non-empty for some $z_0 \in \mathbb{C}$. Then $f$ is surjective.
Proof: By Picard, $f$ misses at most one value. Up to translating $f$ by a scalar (which obviously preserves the hypothesis), we may assume $f$ misses $0$. Then $f = e^g$ for some entire $g$. By assumption, $z_0 \neq 0$, so $\exp^{-1}(z_0)$ is infinite, so by Picard, $g^{-1}(\exp^{-1}(z_0)) = f^{-1}(z_0)$ is infinite, which is a contradiction.
Now take $q(z) := \sum_{n \geq 1} \frac{z^n}{n\cdot n!}$ and $f(z) := ze^{q(z)}$. I claim this $f$ yields a contradiction.
Observe that $f(z) = 0$ if and only if $z = 0$. Therefore, the lemma applies, and we see $f$ is surjective. Moreover, we see that not all fibers of $f$ are in bijection: $f^{-1}(0)$ is a singleton, but by (great) Picard (applied to $f(\frac{1}{z})$), $f^{-1}(z)$ is infinite for any other value of $z$.
Finally, let's check that $f$ is a submersion. We have:
$$df = (1+zq')e^q dz.$$
We have $1+zq' = e^z$ by choice of $q$, so this is $e^{z+q}dz$, which is clearly a nowhere vanishing differential.
