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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?

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Is what true? You haven't made a statement, nor asked a comprehensible question. – Gerry Myerson Jul 12 '11 at 5:50
Ok let me edit that. – user786 Jul 12 '11 at 5:50
up vote 12 down vote accepted

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:

   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}

punctured disk map

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Thank you, I think this is what I was looking for. – user786 Jul 12 '11 at 6:29
amazing video. thank you! – user786 Jul 12 '11 at 7:39
Technically this doesn't answer the actual question of whether there exists a surjection in the opposite direction... – Vladimir Sotirov Jul 12 '11 at 7:44
@Vladimir: come to think about it, this is true. Mariano's example was so nice that I forgot about my original question :) I would even add that it hold even when the map is locally injective. I just can't represent it. – user786 Jul 12 '11 at 8:23
@Theo: I could not convince Mathematica to generate a reasonably sized animated GIF :( – Mariano Suárez-Alvarez Jul 12 '11 at 19:38

No, because punctured disc is not simply connected.

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Could you say more (why do holomorphic functions preserve the property of being simply-connected?) – Vladimir Sotirov Jul 12 '11 at 7:42
I think what Soarer had in mind was to apply the Riemann mapping Theorem, but it does not apply in my case. – user786 Jul 12 '11 at 8:24
I had in mind a bijective conformal map, which is just a biholomorphism. Sorry for the confusion. – Soarer Jul 13 '11 at 3:41

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