This question is partly motivated by my answer to [this question](https://math.stackexchange.com/questions/95570/does-constant-modulus-on-boundary-of-annulus-imply-constant-function/95581#95581) on math.stackexchange.


Let $\Omega$ be a bounded $n$-connected domain in the plane, bounded by $n$ pairwise disjoint Jordan curves. 


It was proved by Ahlfors that by solving an extremal problem (related to so-called *analytic capacity*), one can obtain a function $f$ holomorphic in $\Omega$ with the following properties :


1. $f$ is an $n$-to-$1$ branched covering of $\Omega$ onto the unit disk $\mathbb{D}$,
2. $f$ extends continuously to the boundary of $\Omega$, and maps each boundary curve homeomorphically onto the unit circle.


See e.g. Krantz's *Geometric function theory: explorations in complex analysis*, theorem 4.5.9.


My question is :


**Is there another (perhaps more intuitive) way to see that such a function necessarily exists?**



On holomorphic branched coverings of a domain in the plane to the unit disk
I now know a lot more about this question than I did when I asked it, so for the sake of completness let me add :


By applying the Riemann mapping theorem $n$ times, we can assume that $\Omega$ is bounded by *analytic* Jordan curves. In this case, we have the following theorem due to Bieberbach, which says that there are many proper holomorphic maps onto the unit disk.


**Theorem**
Let $\Omega$ be a domain in $\mathbb{C}\_{\infty}$ bounded by disjoint analytic Jordan curves $E\_1, E\_2, \dots, E\_n$, and let $p \in X$. For each $j$, fix a point $\alpha\_j \in E\_j$. Then there is a unique proper holomorphic map $f:\Omega \to \mathbb{D}$ such that $f(\alpha\_j)=1$ for all $j$ and $f(p)$=0.


For more information about proper holomorphic mappings, see the preprint [there](http://arxiv.org/abs/1307.5042).


