Fibres of a Holomorphic Function 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.

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 $q$...
