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The Riemann mapping theorem states that for every non-empty, simply connected, open set $\Omega \subset \mathbb{C}$, which is not all of $\mathbb{C}$ there exists a function $f$ that maps $\Omega$ onto the open unit disc $D := \{ z \in \mathbb{C} : | z | < 1 \}$ and is biholomorphic.

I am currently reading a proof (about the construction of the Faber polynomials) that seems to assume that in the case that the complement of $\Omega$ is compact, the map $f$ fulfills the following property. For every open set $A \subseteq \mathbb{C}$ that contains the boundary of $\Omega$ there exists $\varepsilon > 0$ such that the set $S_\varepsilon := \{ z \in \mathbb{C} : | z | = 1 - \varepsilon \}$ fulfills $f^{-1}(S_\varepsilon) \subseteq A$.

In other words, $f^{-1}$ maps curves that are close to the boundary of $D$ to curves that are close to the boundary of $\Omega$.

If you prove the theorem by constructing a harmonic function that maps the boundary of $\Omega$ onto the boundary of $D$ (which requires that the boundary of $\Omega$ can be given by a smooth curve), the statement above is clear. Is the statement, however, also true in general? Is there a simple proof that shows that a biholomorphic function that maps $\Omega$ onto $D$ has the above property?

Edit Added the requirement that $\Omega^\textrm{c}$ should be a compact set.

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    $\begingroup$ There seems to be no connection between $A$ and $\Omega$. I think you want the closure of $A$ to be compact and contained in $\Omega$. But even then it cannot be true, as you can always postcompose an automorphism of D and the latter act transitively on D. $\endgroup$ – Zero Sep 25 '19 at 13:01
  • $\begingroup$ @Zero: Thank you for the comment. There was a mistake in my formulation. $A$ is supposed to contain the boundary of $\Omega$. I have corrected the question and hope that it now makes more sense. $\endgroup$ – H. Rittich Sep 25 '19 at 13:27
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This is not true of $\Omega$ (and thus $\partial\Omega$ is unbounded. Take, for example $\partial\Omega$ to be the positive ray, and for $A$ a halfplane containing this ray. To make the statement true, you have either to consider bounded domains, or to require $A$ to be open as a subset of the Riemann sphere.

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  • $\begingroup$ Thank you. The paper actually assumes that the complement of $\Omega$ is compact and thus $\partial \Omega$ should be bounded. I suppose. What would be the argument for the statement to be true, if we assume $\partial \Omega$ is bounded? $\endgroup$ – H. Rittich Sep 25 '19 at 14:17
  • $\begingroup$ It's pure topology. This statement is true for any homeomorphism. $\endgroup$ – Alexandre Eremenko Sep 25 '19 at 19:42
  • $\begingroup$ How does this topological argument look like? Are there any keywords I can search for? $\endgroup$ – H. Rittich Sep 26 '19 at 17:05

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