8
$\begingroup$

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?

  • 1
    Is what true? You haven't made a statement, nor asked a comprehensible question.2011-07-12
  • 0
    Ok let me edit that.2011-07-12

2 Answers 2

12

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}
 ]

punctured disk map

  • 0
    Thank you, I think this is what I was looking for.2011-07-12
  • 0
    amazing video. thank you!2011-07-12
  • 5
    Technically this doesn't answer the actual question of whether there exists a surjection in the opposite direction...2011-07-12
  • 0
    @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.2011-07-12
  • 0
    @Theo: I could not convince Mathematica to generate a reasonably sized animated GIF :(2011-07-12
  • 0
    :( But tanks for trying! Jim Belk managed to do it [here](http://math.stackexchange.com/questions/42877/if-the-minkowski-sum-of-two-convex-closed-sets-is-a-euclidean-ball-then-can-the/42897#42897) (with source), but his thing is much smaller in size and probably much less complex (I think2011-07-12
  • 0
    [Here](http://i.stack.imgur.com/WXJLQ.gif) is my attempt (though I used an older version of *Mathematica*); would this be suitable?2011-07-16
  • 0
    @J.M.: great! Can you upload it and add a link to my answer?2011-07-16
2

No, because punctured disc is not simply connected.

  • 0
    Could you say more (why do holomorphic functions preserve the property of being simply-connected?)2011-07-12
  • 0
    I think what Soarer had in mind was to apply the Riemann mapping Theorem, but it does not apply in my case.2011-07-12
  • 0
    I had in mind a bijective conformal map, which is just a biholomorphism. Sorry for the confusion.2011-07-13