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Riemann mapping theorem

Let $\Omega \subseteq \mathbb{C}$ be a region and $\mathbb{D}$ be the unit disk, $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$. We define the equivalence relation of the set of the regions in $\mathbb{C}$ as follows: Two regions $\Omega_1,\Omega_2 \subseteq \mathbb{C}$ are conformally equivalent ($\Omega_1 \sim \Omega_2$) when there is $f\in H(\Omega_1)$ ($f$ analytic on $\Omega_1$) with $f:\Omega_1 \rightarrow \Omega_2$ is bijective (one to one).

Then we have: $\Omega \sim \mathbb{D} \Leftrightarrow \Omega$ is simply connected and $\Omega \neq \mathbb{C}$.

My questions revolves around the proof. Can someone describe me in descriptive steps (an outline in words) how the proof works?

Thank you very much for your time and patience, I appreciate it very much!

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The statement is not correct: $\Omega$ is simply connected. –  timur Jun 7 '12 at 14:05
    
Sorry, yes, the "not" should not be there! –  Chris Jun 7 '12 at 14:10
    
Please see Section 3 of the Wikipedia page. en.wikipedia.org/wiki/Riemann_mapping_theorem You might also be interested in the approach via normal families - see, for example, Section 23 of Taylor's on-line notes: math.unc.edu/Faculty/met/complex.pdf –  Sam Nead Jun 7 '12 at 14:15
    
@Chris In your question above the letter $\,\mathbb{D}\,$ has two definitions... –  DonAntonio Jun 7 '12 at 23:11
    
Where? I defined it as the set of open unit disks. –  Chris Jun 7 '12 at 23:24
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up vote 2 down vote accepted

I've found following resource:

http://www.dancalloway.com/lets_have_a_word/riemann-mapping-theorem-explained

It's really very good and fitted my needs well.

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