I am thinking of the importance of having a "simply connected domain" in Riemann mapping theorem. So, is there some simple examples for non simply connected domains like annulus that violates the theorem? This is not homework so please don't downvote for not having any attempts. I'm just curious about it and wanted to know some examples in order to understand. Any hint/help is appreciated.

  • $\begingroup$ When Riemann mapping theorem violates? I mean what is the expression which you think is contradiction of Riemann mapping theorem (with adding a non simply connected domains as well)? $\endgroup$ – Nosrati Sep 12 '17 at 8:43
  • $\begingroup$ MyGlasses: There is a non simply connected domain that cannot be conformally mapped onto the unit disk. $\endgroup$ – Extremal Sep 12 '17 at 12:40

Let $D_1$ and $D_2$ be domains in $ \mathbb C$ and $f:D_1 \to \mathbb C$ a holomorphic and injective function with $f(D_1)=D_2$.

Then it is easy to see:

$D_1$ is simply connected $ \iff $ $D_2$ is simply connected .

Try a proof !


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