Riemann Theorem states that for any simply-connected domain in $\mathbb{C}$ (which is not whole $\mathbb{C}$) there exists biholomorphic map onto the open unit disk.

I find it hard to show that we cannot replace "simply-connected" with "connected". Any ideas?

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    $\begingroup$ Consider the annulus $A = \{z \in \mathbb C \mid \frac{1}{4} < |z| < \frac{3}{4}\}$. What happens if you assume that there is a biholomorphic map from $A$ onto the open unit disk? $\endgroup$ – qaphla Jul 16 '15 at 18:49
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    $\begingroup$ Can you show that anything homeomorphic to a simply connected set must be simply connected? $\endgroup$ – David C. Ullrich Jul 16 '15 at 18:49
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    $\begingroup$ Oh, that's you! We got disconnected - DF and I are still wondering what proof of (2) you had in mind in that other thread... $\endgroup$ – David C. Ullrich Jul 16 '15 at 18:55
  • $\begingroup$ Hi! I posted comment there but deleted the topic too early i think so you didnt have chance to see it. I was terribly wrong, forgot about one piece of proof. I mean in showing uniqueness of antiderivative we used CG theorem. It is not needed in showing existence though. Sorry about that. Im complex analysis noob. $\endgroup$ – luka5z Jul 16 '15 at 18:57
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    $\begingroup$ @luka5z No problem. Glad we sorted that out - if there was a proof as you said I wanted to know about it... $\endgroup$ – David C. Ullrich Jul 16 '15 at 19:02

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