Conformal mappings of $\mathbb{D}$ onto $\mathbb{D}$ which is not one-to-one? A holomorphic (analytic) function $f:\Omega\to\mathbb{C}$ is called a conformal mapping if $f'$ is nonvanishing. Every one-to-one holomorphic function is conformal, but the converse is not true. (e.g. $\exp$.)
My question is:


*

*Is there any conformal mapping of $\mathbb{D}$ onto $\mathbb{D}$ which is not one-to-one?


(or


*

*If $f:\mathbb{D}\to\mathbb{D}$ is conformal and onto, then does it follow that $f$ is one-to-one?)

*Find all conformal mapping of $\mathbb{D}$ onto $\mathbb{D}$ such that $f(0)=0$.
 A: Start with the map $f(z)=z^3$ in the upper half-disk $D=\{ z:|z|<1,\Im z>0\}$.
It is "conformal" in the sense that $f'(z)\neq 0$ and maps $D$ surjectively
on the punctured unit disk $U^*=\{ z:0<|z|<1\}$. Let $G_0$ be the Riemann surface
$f(D)$. Its boundary consists of two intervals $[0,1]$ and $[-1,0]$ and the "arc of circle" $\{ e^{i\theta}:0<\theta<3\pi\}$. Notice that when tracing $[0,1]$ from
$0$ to $1$, $G_0$ stays on the left side. (This is because $D$ is on the left of $[0,1]$ the map on this interval is increasing.)
Now consider a simply connected domain $G_1$ in the unit disk which a) contains $0$, and b) whose oriented boundary contains the oriented interval from $1$ to $1/2$ (so that when you trace the interval from $1$ to $1/2$,  $G_1$ is on your left side. Such domain is easy to construct: for example take the
union of the lower half-disk with the disk $|z|<1/4$.
Then you can paste $G_0$ and $G_1$ along the interval $(1/2,1)$.
The result is a simply connected Riemann surface $G$ spread over the unit disk.
By the Uniformization theorem, one can map the unit disk conformally onto $G$.
This conformal map satisfies all your conditions.
Remark. In fact one can avoid using the difficult Uniformization theorem here, because our surface is in fact a polygon: if you choose $G_1$ as I described,
then this is a hexagon bounded by four straight segments and two arcs of circles. The existence of a conformal map onto such a hexagon can be proved
without appeal to the Uniformization theorem.
