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Let $K \subset \mathbb{C}$ be compact and let $A(K)$ be the space of functions that are continuous on $K$ and holomorphic on $K^\circ$ (the case $K = \emptyset$ is allowed). If $f$ is injective and if we put $L := f(K)$, is it true that the inverse function $f^{-1}: L \rightarrow K$ belongs to $A(L)$?

From the compactness of $K$ it follows that $f^{-1}$ is continuous. To show that $f^{-1}$ is holomorphic at $z \in L^\circ$ I need to show that $f$ is holomorphic in a neighborhood of $f^{-1}(z)$ or equivalently that $f^{-1}(z)$ is an interior point of $K$ but I don't see why this is true.

Does anyone have an idea?

Bruno Krams
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    by the open mapping theorem for holomorphic functions, $f$ sends a small ball around $z$ interior to $K$ into an open (relative to the plane of course) set in $L$ containing $f(z)$; because $f$ is injective, $f'$ has no zeroes on that open ball, so the inverse is holomorphic on the open set containing $f(z)$ above; note though that $L$ depends on $f$, so in general there won't be many such functions for given $K,L$ when they have nonempty interiors (eg when they are closed discs only special Mobius transforms work etc) – Conrad Dec 28 '20 at 13:11
  • Thank you. This settles the question for the case where the preimage of $x \in L$ is an interior point of $K$. But is it possible that the preimage of an interior point of $L$ is a boundary point of $K$? – Bruno Krams Dec 29 '20 at 12:57
  • no because of the fact that (nonconstant) analytic functions are open so if $z$ is interior to $K, f(z)$ must be interior to $L$ – Conrad Dec 29 '20 at 14:47
  • But I'm interested in the other way: If $x$ is interior to $L$ is it true that $f^{-1}(x)$ is interior to $K$? I can not use that $f^{-1}$ is analytic because that is what I want to prove. – Bruno Krams Dec 29 '20 at 17:58
  • That is true for any topological homeomorphism in the plane and $f$ is such since inverse is continuous because of compacity for example – Conrad Dec 29 '20 at 21:37
  • Ah. Thank you. Do you have a reference for this result? If found this related question but the answers given there assume that $f$ is a homeomorphism of $\mathbb{C}$. – Bruno Krams Dec 31 '20 at 08:46
  • the result follows for example from the Jordan curve theorem here - let $f^{-1}=g$; as noted $g$ is continuous and injective on $L$; if $w \in L$ interior, there is a small closed disc $D$ centered at $w$ and contained in the interior of $L$; if $C$ is its boundary, $g(C)=J$ is a Jordan curve in the plane hence (Jordan curve theorem here) $g$ sends the interior of $D$ into the interior of $J$ (this here meaning the bounded component of the complement of $J$ in the plane) so in particular in an open plane set; this being in $K$ it must, of course, be in the interior of $K$ and we are done – Conrad Dec 31 '20 at 17:22
  • Again thank you very much. This is exactly what I was hoping for. If you posted your comments as an answer I'd be glad to accept & upvote it – Bruno Krams Jan 01 '21 at 13:32
  • no problem - added it and happy to be of help – Conrad Jan 01 '21 at 15:19

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Let $K_o, L_o$ the interiors of $K, L$ respectively; then we show that $K_o$ is non-empty if and only if $L_o$ is so, and in this case, $f$ is an analytic isomorphism between them, so $g=f^{-1}$ is analytic on $L_o$.

By hypothesis since $K$ compact and $f:K \to L$ continuous bijective, it follows that $g:L \to K$ is continuous and bijective.

If $z \in K_o$, there is a small disc $D_z \subset K_o$ centered at $z$ and by the open mapping for holomorphic functions, it follows that $f(D_z)$ is open in the plane and it is in $L$ by our choices, so it is in $L_o$.

Conversely, if $w \in L_o$ there is a small closed disc $\overline D_w \subset L_o$ centered at $w$ and if $C_w$ is its circle boundary, $g(C_w)$ is a Jordan curve in the plane $J$ with $U$, its interior in the plane, a bounded (simply connected) open set, hence by the Jordan Curve Theorem, $g(D_w)=U$ homeomorphically. But $U$ is in $K$ hence it is in $K_o$.

This shows that $f$ restricted to $K_o$ is a homeomorphism onto $L_o$ with inverse $g$ and now by the inverse function theorem, $g$ is analytic if $f$ is so, so we are done!

Conrad
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