Inverse functions on $A(K)$ 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?
 A: 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!
