Conformal mapping of disk, surjective, not injective

Question

Is there an example of a conformal mapping of the disk onto itself which is not injective? If not, how may we prove there does not exist such a map?

This came up in the answer to this question.

Answer 1

I do not know a simple example, but here is a construction that works.

Let $D^2$ be the open unit disk, and let $U$ be a $C^\infty$ manifold diffeomorphic to the open unit disk. Let $f\colon U \to D^2$ be a $C^\infty$ map with the following properties:

1. $f$ is surjective but not injective.

2. $f$ is a local diffeomorphism.

It should be clear geometrically that such a map exists. (See this question.) Then the complex structure on $D^2$ lifts via $f$ to a complex structure on $U$, under which $f$ becomes a holomorphic map.

Since $U$ is simply connected, the uniformization theorem states that $U$ must be conformally equivalent to either the Riemann sphere, the complex plane, or the open unit disk. Since $f$ is a bounded, non-constant holomorphic function on $U$, it follows from Liouville's Theorem that $U$ must be conformally equivalent to the open unit disk, so there exists a biholomorphism $g\colon D^2 \to U$. Then the composition $f\circ g$ is the desired map.

Of course, the uniformization theorem is only an existence statement, so this proof isn't constructive. In practice, though, many of the usual techniques for approximating Riemann maps should work fairly well for approximating $g$.