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Let $f:G_1 \rightarrow G_2$ be a local biholomorphism and $g:G_2 \rightarrow G_1 $ continious

such that $(f\circ g)(z)=z$

Proof that $g$ is a biholomorphism

I know that $g$ is biholomorphic if $g' \gt 0$ for all $z\in G_2$ and $g^{-1}$ is continious.

How do I know that $g$ is already a holomorphic function ?



Can someone give me a hint/ or explain why $g$ is holomorphic if $g$ is continous and satisfies the condition above ?

XPenguen
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  • This looks like a special case of http://math.stackexchange.com/questions/485384/fz-and-g-circ-fz-holomorphic-implies-gz-holomorphic-as-well. – Martin R May 31 '16 at 17:44
  • Duplicate of http://math.stackexchange.com/questions/1283713/let-fu-to-v-be-a-bijective-holomorphic-function-show-that-inverse-of-f-is? – Martin R May 31 '16 at 17:48
  • @MartinR In my case it´s a local biholomorphism. – XPenguen May 31 '16 at 18:12

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