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Let $f:U\rightarrow U$ be a homeomorphism of the disc. Then it is not true that f must extend to a continuous map $f:\overline U\rightarrow\overline U$ (there is an example on this site). ($U$ is unit disc.)

My question is, is the statement true if we insist $f$ be holomorphic?

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2 Answers 2

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Yes, this is true; in fact, any such $f$ must be a Möbius transformation. First, let us suppose $f(0)=0$. Then the Schwarz lemma implies $|f(z)|\leq |z|$ for all $z\in U$, with equality for some $z\neq 0$ iff $f(z)=cz$ for some $c$ with $|c|=1$. But we can also apply the Schwarz lemma to $f^{-1}$ to learn that $|f^{-1}(z)|\leq|z|$, and letting $z=f(w)$ gives $|w|\leq |f(w)|$. Thus in fact $|f(z)|=|z|$ for all $z$ and $f$ is of the form $f(z)=cz$.

In the general case where $f(0)\neq 0$, we may choose a bijective Möbius transformation $g:U\to U$ such that $g(f(0))=0$. The previous paragraph now shows that $g\circ f$ is a Möbius transformation, and hence so is $f$.

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    $\begingroup$ Thanks – to apply your argument, it is important to know that a bijective holomorphic map is automatically a biholomorphism!! $\endgroup$
    – delgato
    Apr 27, 2016 at 14:26
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This is, in fact, true in a much more general setting. A theorem by Carathéodory from 1913, states that: if $\Omega$ is a simply connected domain whose boundary is a Jordan curve, then the Riemann map $f : \Omega \to U$ extends to a homemorphism $\bar\Omega \to \bar U$.

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