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Suppose that a holomorphic function $f$ from a regon $G$ to another region $H$ has a right inverse and this $f$ is locally bijective (that is the derivative of $f$ does not vanish).

Then is it true that this right inverse of $f$ is actually the 2-sided inverse (simply the inverse function) of $f$? I think it is true...but somewhat confused.

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    $\begingroup$ Might there not be regions in the complex plane where $e^z$ is not injective, but still a continuous branch of $log z$ may be defined on the image region? Is that the sort of thing you ask about? $\endgroup$ – GEdgar Dec 15 '17 at 10:53
  • $\begingroup$ That is right. Is it possible? $\endgroup$ – Keith Dec 15 '17 at 11:44
  • $\begingroup$ Does "region" mean "connected open set" (more commonly called a domain), or just an arbitrary open set? (Pretty sure it's the former because for the latter it's easy to find counterexamples.) $\endgroup$ – Daniel Fischer Dec 15 '17 at 13:46
  • $\begingroup$ I interpret your question as "Suppose $G,H$ are connected open sets and $f:G\to H,$ $g:H\to G$ are both holomorphic. If $f\circ g(z) = z$ on $H,$ is it true that $g\circ f(z) = z$ on $G?$" Is this what you mean? $\endgroup$ – zhw. Dec 16 '17 at 23:03
  • $\begingroup$ @zhw yes and there must be another condition that f is localy bijective $\endgroup$ – Keith Dec 17 '17 at 7:11
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Suppose $G,H$ are connected open sets and $f:G\to H, g:H\to G$ are holomorphic. If $f\circ g(z) = z$ on $H,$ then $g\circ f(z) = z$ on $H.$

Proof: For any $z\in H,$ $(g\circ f)(g(z)) = g(f\circ g(z)) = g(z)$ for all $z\in H.$ Another way to say that is $g\circ f (w) = w$ for all $w\in g(H).$ But $g$ is clearly non constant, so $g(H)$ is a nonempty open subset of $G$ by the open mapping theorem. By the identity principle, $g\circ f (w) = w$ everywhere in $G,$ and we're done.

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