Conformal map from the punctured unit disc onto the unit disc? I remember seeing this statement, I don't remember where (maybe in Lang's Complex). Is this true or do I have a faulty memory. It was always somewhere in the back of my mind but I never believed it. Is it true that there is a conformal map from the punctured unit disc onto the unit disc?
 A: I know this is an old question, which I stumbled upon, but it seems to me that the answers currently address two questions. Often in complex analysis "conformal map" is used equivalently to "biholomorphic map", so let me avoid any confusion by using the terms "conformal isomorphism" and "locally conformal".


*

*There is no conformal isomorphism between the unit disc and the punctured disc, since one is simply-connected and the other is not. (Soarer's answer).

*There is a locally conformal map from a simply-connected domain (and hence from the disc) onto the punctured disc, namely the exponential map. Of course, this is true for any domain omitting at least two points, due to the uniformisation theorem.


However, neither answer addresses the original question as I understand it, namely: is there a locally conformal map from the punctured disc onto the unit disc.
We may instead ask: is there a locally conformal map $\phi$ from the unit disc $D$ onto itself that is not a conformal isomorphism? If such a map exists, let $z$ and $w$ be points with $\phi(z)=\phi(w)$, we may assume without loss of generality that $z=0$, and $\phi\colon D\setminus\{0\}\to D$ has the desired property. 
Of course, by the Riemann mapping theorem, we can just look for a locally conformal map from a simply-connected domain $U$ onto another simply-connected domain $V$. To see that such a map exists is a nice exercise. You should imagine drawing a simply-connected domain on a three-sheeted cover of the plane (with two different branch points) whose projection to the plane is simply-connected. 
I believe that this cover can be realised by a cubic polynomial $p$ with two critical points. That is, there is a simply-connected domain $V$ containing neither critical point such that $p\colon V\to p(V)$ is not injective, and $p(V)$ is simply-connected.
A: No, because punctured disc is not simply connected.
A: There is a surjective conformal map, though (in opposition to bijective, which is what Soarer has in mind in his answer). Just compose an holomorphic bijection from the disc to the upper half-plane with $z\mapsto e^{i z}$.
(I made a video showing the images under the map $z\mapsto\exp\frac{z-1}{z+1}$, which is a surjection from the unit disk to the punctured unit disk, of the circles centered at $0$. You can see in it how it manages to avoid the origin.) (The file will not last forever in that location... if someone can upload it to some site or another, it would be great!)
Later In any case, this is the Mathematica code I used:
Animate[
 ParametricPlot[
   With[{z = r Exp[I theta]},
   Through[{Re, Im}[Exp[(z - 1)/(z + 1)]]]
   ],
  {theta, 0, 2 \[Pi]},
  ImageSize -> Medium, PlotRange -> {{-1, 1}, {-1, 1}},
  PlotPoints -> 1000, Ticks -> False
  ],
 {r, 0, 0.999, 0.001}
 ]


