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Let us denote the open unit disk by $\Delta$ and write $\cong$ for conformal equivalence. Then the problem follows:

Suppose that $G\subset\hat{\mathbb C}=\mathbb C\cup\{\infty\}$ is a region such that $\hat{\mathbb C}\setminus G$ has $n\,(n\geqslant 2)$ components $K_1,\cdots,K_n$. If each $K_i\,(i=1,\cdots,n)$ is not a singleton, then $$G\cong\Delta\setminus(\overline E_1\cup\cdots\cup\overline E_{n-1}),$$ where $E_1,\cdots,E_{n-1}$ are simply connected regions in $\Delta$ such that $\overline E_i\subset\Delta\,(i=1,\cdots,n)$ and their boundaries are smooth closed Jordan curves.

My professor left this assignment to me when he was giving a lecture about classification of finitely connected regions in $\hat{\mathbb C}$ while I tried but yet had not found a proof.

I first considered the case that $n=2$, say, that $\hat{\mathbb C}\setminus G$ has two components $A$ and $B$. By Riemann mapping theorem it follows that $G\cup A\cong\Delta$ and if we write $f$ for the conformal equivalence that carries $G\cup A$ onto $\Delta$, then $G\cong\Delta\setminus f(A)$ since $f$ is bijective. So the problem is reduced to whether there is a simply connected region $E\subset\Delta$ such that $\partial E$ is a smooth closed Jordan curve and $\Delta\setminus f(A)\cong\Delta\setminus\overline E$, but the only thing we know about $f(A)$ is that it is compact and connected. Now I would like to ask how to construct such an $E$ with a closed smooth Jordan curve as its boundary and also, the same question for a general $n$... Thanks in advance...

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For the case $n=2$, there are two components $K_1$, $K_2$ of $\hat{\mathbb C}\setminus G$. Denote $G_1:=\hat{\mathbb C}\setminus K_1$, then by Riemann mapping theorem, there is a biholomorphic map $f_1\colon G_1\to\Delta$ that sends the set of prime ends of $G_1$ (i.e., of $K_1$) homeomorphically onto $\partial\Delta=S^1$. Set $G_2:=\hat{\mathbb C}\setminus f_1(K_2)$ and again by Riemann mapping theorem there is a biholomorphism $f_2\colon G_2\to\Delta$ that sends the set of prime ends of $G_2$ (i.e., of $f_1(K_2)$) homeomorphically onto $\partial\Delta=S^1$. Define $f:=f_2|_{\Delta\setminus f_1(K_2)}\circ f_1|_{G}$. Apparently this map $$ f\colon G\to\Delta\setminus E_1 $$ is biholomorphic, where $E_1=f_2(\hat{\mathbb C}\setminus\Delta)\subset\Delta$ is a compact subset, and its boundary $\partial E_1=f_2(\partial\Delta)$ is a closed analytic Jordan curve. Thus the interior $E_1^\circ$ is simply connected.

Inductively, the argument for a general $n$ is similar.

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