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.
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.