Fibres of a Holomorphic Function

Question

Let $f$ : $U\rightarrow V$ be a proper holomorphic map where $U$ and $V$ are open subsets of $\mathbb{C}$ with $V$ connected. Show that the cardinality of the fibres of $f$, i.e. $f^{-1}(\{z\})$ counted with the multiplicities are the same for each $z \in$ $V$. This looks like the property of covering maps and so I was trying to prove if $f$ is a local homeomorphism or a covering map, but to no avail. Thanks for any help.

Answer 1

This is very simple using some complex analysis.

Since $f$ is proper, given $p\in V$ and small enough $\epsilon>0$ there exists a cycle $\Gamma\subset U$ such that if $|p-q|<\epsilon$ then all the zeroes of $f-q$ lie "inside" $\Gamma$, and in fact such that if $z$ is a zero of $f-q$ then the index of $\Gamma$ about $z$ is $1$ (also the index of $\Gamma$ about any point of $\Bbb C\setminus U$ is $0$.).

Details added on request: If $\epsilon>0$ is small enough then $\overline{D(p,\epsilon)}\subset V$; since $f$ is proper this shows that $K=f^{-1}(\overline{D(p,\epsilon)})$ is a compact subset of $U$. Hence by a nameless result that appears in most books on complex analysis because it's needed a lot, there exists a cycle $\Gamma\subset U\setminus K$ with index $1$ about every point of $K$ and index $0$ about every point of $\Bbb C\setminus K$. (With apologies for knowing one particular book better than the others, this is Lemma 10.5.5 in Complex Made Simple.)

Hence if $|p-q|<\epsilon$ the number of zeroes of $f-q$ is $$\frac1{2\pi i}\int_\Gamma\frac{f'(z)}{f(z)-q}\,dz.$$That integral depends continuously on $q$...

Answer 2

Here's a useful lemma:

Lemma. Let $X,Y$ be locally compact Hausdorff topological spaces, $f : X \to Y$ continuous, open, proper, surjective and with discrete fibers. Let $K$ be a neighborhood of $y \in Y$, $y_1, \ldots, y_n$ its preimages (a finite number, by properness and discreteness) and $K_i$ a neighborhood of each $y_i$. Then there exist disjoint open neighborhoods $V_i \subseteq K_i$ of the $y_i$ and an open neighborhood $V \subseteq K$ of $y$ such that $f^{-1}(V)$ is the disjoint union of the $V_i$.

It is nothing deep (you can prove it). It is similar to the proof that a proper surjective local homeomorphism is a covering map. The only difference is that we don't have local injectivity here.

Note that proper implies closed, so proper+open+ $Y$ connected implies surjective.

Now let $X=U$, $Y=V$ as in the question. By the lemma, outside of the set $B$ of branch points (images of points where the derivative $f'$ vanishes) $f$ is a covering map (by the lemma + we now have local injectivity at the preimages). In particular, the size of fibers is constant on connected components of $Y - B$, say equal to $n$.

If $f$ proper, the set of branch points is closed (by closedness) and discrete (by properness) in $V$. In particular, $Y-B$ is connected.

It remains to check what happens in a neighborhood of a branch point $y$. Take $V$ and $V_i$ as in the lemma. It suffices to check that the number of preimages is constant for each of the restrictions $f : V_i \to V$. On such $V_i$, $f$ has the form $y + a_{m_i}(z-y_i)^{m_i} + a_{m_i+1} (z-y_i)^{m_i+1} + \cdots$ by Taylor-expansion, for some $m_i$ ($a_{m_i} \neq 0$). We want $\sum m_i=n$. W.l.o.g. we may assume $y_i=y=0$. We have that $$a_{m_i} z^{m_i} + a_{m_i+1} z^{m_i+1} + \cdots = g(z)^{m_i}$$ for some holomorphic $g$ which is a homeomorphism between open neighborhoods of $0$. Replacing $f(z)$ by $(f \circ g^{-1}) (z) = z^{m_i}$ does not change the number of preimages. So the fibers have cardinality $\sum m_i$ in a neighborhood of $y$, including at $y$. Comparing this to any point different from $y$ in that neighborhood, we conclude that $\sum m_i=n$.