Conformal map from $\mathbb C\setminus \{0\}$ to the open unit disk.

Question

Is there a conformal/analytic map from $\mathbb C\setminus \{0\}$ to the open unit disk?

I think the answer is no but I'm not sure how to prove it. I know that the former is not simply connected and the latter is simply-connected, but I'm not sure why this is useful or relevant since analytic maps are not necessarily homeomorphisms.

Suppose one existed from the disk to the punctured plane. Draw a circle around zero in the punctured plane. The inverse image of this circle should be some closed curve in the disc. Now the argument principle applied to the conformal map along this curve should show that conformal map has a zero, which should be impossible since zero is not in it’s range. — Blake, Jun 15 '18 at 06:28

score 9 · Answer 1 · answered Jun 15 '18 at 06:28

9

There's the constant map ...

Otherwise a holomorphic $f:\Bbb C-\{0\}\to D$ is bounded, so has a removable singularity at $0$, so is a restriction of a holomorphic $g:\Bbb C\to D$. Now apply Liouville.

answered Jun 15 '18 at 06:28

Angina Seng

158,341

I like this better than what my comment was suggesting – Blake Jun 15 '18 at 06:30
I see. (of course I meant "onto" the unit disk) – Jun 15 '18 at 06:49
@user1488 Another argument. Take $f$ as above, and apply Liouville to $h(z)=f(\exp(z))$. – Angina Seng Jun 15 '18 at 06:54
@LordSharktheUnknown Why does $f$ being bounded imply it has a removable singularity at $0$? – user5826 Nov 05 '19 at 01:22
@AlJebr That's a standard theorem in complex analysis. – Angina Seng Nov 05 '19 at 05:15

Conformal map from $\mathbb C\setminus \{0\}$ to the open unit disk.

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