# Proper holomorphic map from the unit disc to the bidisc

Let $F\colon \mathbb{D}\rightarrow\mathbb{D}^2$ be a proper, holomorphic map. It is not very difficult to see that every proper holomorphic map from $\mathbb{D}$ to $\mathbb{D}$ is a finite "Blaschke product". Now denote $F=(F_1,F_2)$. It is easy to see that if you put at least one of the $F_i$'s, $i=1,2$, to be a finite Blaschke product, then $F$ becomes proper. My question here is:

Does there exist a proper map $F\colon \mathbb{D}\rightarrow\mathbb{D}^2$ which is not of the above type?

PS: In the above everywhere assume that the functions are not constant.

-

Yes: just take any pair of non-proper holomorphic functions from $\mathbb D$ to $\mathbb D$ that are continuous on the closed unit disk and satisfy $\max(|F_1|,|F_2|)\equiv 1$ on the boundary.
For example, begin with $h(z)=g^{-1}(\sqrt{g(z)})$ where $g$ is a Möbius map of $\mathbb D$ onto upper half-plane. The function $h$ satisfies $|h|\equiv 1$ on a subarc $\Gamma\subset \partial\mathbb D$ that is mapped by $g$ onto $[0,\infty]$. Elsewhere $|h|<1$. Let $\phi:\mathbb D\to\mathbb D$ be a Möbius map that exchanges $\Gamma$ and its complement. Then $F=(h,h\circ \phi)$ meets the requirements.
To make the previous paragraph more concrete, one can take $g(z)=i\frac{1+z}{1-z}$, $\Gamma=\partial \mathbb D\cap \{\operatorname{Im}z\le 0\}$, and $\phi(z)=-z$.
@Abelvikram This happens to be the case here, but it's not important for the argument. The important point is that the range of each $F_k$ is a region contained in the unit disk that shares a boundary arc with the disk. This ensures $|F_k|=1$ on a subarc. We can always stretch this subarc with a Möbius transformation of the disk, so that two of them cover the circle. –  ˈjuː.zɚ79365 Jun 17 '13 at 12:16