Conformal map from unit disc into itself

Question

Is it true that any conformal (derivative is not zero) map $f:D(0,1) \rightarrow D(0,1) $ (not necessarily surjective) is in fact a homography (Möbius transformation)?

Answer 1

The Riemann mapping theorem tells you that there is a bijective conformal map between $D(0,1)$ and any non-empty, simply connected open set in $\mathbb C$ other than $\mathbb C$ itself.

If we pick a non-empty simply connected open $U \subset D(0,1)$ whose boundary is not a circle, then a bijective conformal map $D (0,1) \to U $ cannot be a Mobius transformation, because Mobius transformations map circles to circles or straight lines. Composing this map with the inclusion of $U$ into $D(0,1)$, you get the counterexample you want.