Here is my question :


Suppose you have a simple (analytic) closed curve $\gamma$ in an open simply connected domain $\Omega \neq \mathbb{C}$. Does there exist a conformal bijection $f : \Omega \rightarrow U \subset \mathbb{C}$, such that $\gamma$ is sent to the unit circle $S^1$ (the unit disc $D$ would then be contained in $U$) ?


The Riemann mapping theorem tells you that you can find a conformal map from the interior of $\gamma$ to $D$, and the conformal geometry of an annulus tells you that you can also find a conformal map from the annulus $\Omega - int(\gamma)$ to an annulus $A(r\_1,r\_2) = \{z \in \mathbb{C} | r\_1 < |z| < r\_2 \}$. But this does not answer the question...


Note that in the question I don't ask for the open set $U$ to be bounded by a circle (this would clearly be too restrictive). 


This question is somehow related to a previous question I asked, but I think this one is quite different though. The answer must be well-know, but I can not find it anywhere - neither from myself.



Conformal map and Jordan curve
If $G\neq\mathbb{C}$ is a s.c. domain, then a conformal map to a disc extends analytically beyond the boundary if and only if $G$ is bounded by an analytic Jordan curve. In particular, if you were to let your domain $\Omega$ depend on the curve $\gamma$, then the answer would be positive.


However, as stated, the answer is negative. Indeed, let $\Omega$ be a simply-connected domain; we shall construct an analytic curve $\gamma$ such that no conformal isomorphism of the interior of $\gamma$ to a disc extends to all of $\Omega$.


Let $G$ be any bounded simply-connected domain whose closure is contained in $\Omega$, such that $G$ has non-analytic boundary. Let $\phi$ be a conformal isomorphism to the unit disc $\mathbb{D}$ and let $\gamma$ be the preimage of a round circle under $\phi$; say
$$\gamma := \phi^{-1}( \partial B(0,1/2) )$$
(where $B(z,\delta)$ is the ball of radius $\delta$ centred at $z$).


Suppose $f:\Omega \to U\subset\mathbb{C}$ was a conformal isomorphism with $f(\gamma)=\partial\mathbb{D}$. Since a conformal isomorphism from the inside of $\gamma$ to a disc is unique up to post-composition by a Möbius transformation, it follows that there is Möbius $M$ such that $\phi=f\circ M$. But then $\phi$ extends analytically beyond the boundary of $G$, a contradiction.


(**Edited** as requested to provide further details and adjust notation slightly.)


