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If the 'right' one is bijective then this holds, however assume that $g,f$ and $w$ are holomorphic and $w$ is not bijective. Perhaps it is only onto.

edit: Assume further that they are non-constant. Ideally we are talking about them on compact Riemann surfaces (so the fact they are onto is given by virtue of being non-constant).

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Suppose for a moment that $g = 0$, and $z(0) = 0$ (just to give you a silly example so that you can adjust your hypotheses). – t.b. Aug 1 '11 at 2:11
What if $g$ and $z$ are both constants? Then there's not much you can conclude about, say $f$, even knowing a lot about $w$. – Jason DeVito Aug 1 '11 at 2:11
@GottfriedLeibniz: Here's another simple counterexample. Let $X$ be the Riemann sphere—so a compact Riemann surface, like you ask. Let $f(x) = x^{1/2}$, $g(x) = x^2$ and $z = w = \mathrm{id}$. Then $f$ is not even continuous, so how could it be holomorphic? However, I think you will find no counterexamples when $f$ is a surjective continuous map and the rest are holomorphic. (This intuition comes from thinking of Riemann surfaces as locally ringed spaces, but I may be mistaken!) – Zhen Lin Aug 1 '11 at 2:50
up vote 5 down vote accepted

On $\mathbb{C}$, holomorphic functions coincide with analytic functions. Wikipedia therefore intimates that (a) two holomorphic functions composed are holomorphic, and (b) the functional inverse of a holomorphic function is also holomorphic provided the function's derivative vanishes nowhere.

Without loss of generality, suppose $g$ and $f$ are two of the three holomorphic functions. If $z$ is holomorphic and $z'$ vanishes nowhere, it then follows that $w=z^{-1}\circ g\circ f$ is holomorphic. If $w$ is holomorphic and $w'$ vanishes nowhere, then $z=g\circ f\circ w^{-1}$ is holomorphic. However, if one of the functions on the other side, $z$ or $w$, is the third holomorphic function and has a derivative that vanishes somewhere, then it stands to reason the fourth function might not be holomorphic.

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