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Let the riemann mapping theorem be

Let $D\subset \mathbb{C}$ be a simply-connected domain and $z_0 \in D$. Then there exists a unique conformal map $f$ from $D$ onto the open unitdisk $\mathbb{D} = \lbrace z \in \mathbb{C} \ | \ |z| < 1 \rbrace$ with $f(z_0) = 0$.

A text Im reading now uses this for mapping $\hat{\mathbb{C}} \setminus \overline{B}$ conformal onto $\hat{\mathbb{C}} \setminus \overline{\mathbb{D}}$ with $f(\infty) = \infty$ (where $\hat{\mathbb{C}} = \mathbb{C} \cup \lbrace \infty \rbrace$ and $B$ is a simply-connected domain that contains the point $0$). How can one show that this is always allowed by the riemann mapping theorem?

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  • $\begingroup$ What is $B$???? $\endgroup$ – David C. Ullrich Nov 8 at 16:50
  • $\begingroup$ Sorry. I edited the question. $\endgroup$ – Arjihad Nov 8 at 17:34
  • $\begingroup$ Ok. Not sure what aspect of this you're troubled about. First, $\hat\Bbb C\setmminus \overline B$ is simply connected, just because $\overline B$ is connected... $\endgroup$ – David C. Ullrich Nov 8 at 20:45
  • $\begingroup$ $\hat{\Bbb C}\setminus\overline B$. $\endgroup$ – David C. Ullrich Nov 9 at 2:40
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    $\begingroup$ Let $g(z) = 1/z$. Show that $g(\hat {\Bbb C} \setminus \overline B)$ is simply connected. Apply the theorem to it with $z_0 = 0$, then consider $g\circ f\circ g$ $\endgroup$ – Paul Sinclair Nov 9 at 3:31

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