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Actually, this is an exercise on Rudin's Real and Complex Analysis:

Suppose $\Omega_1, \Omega_2$ are plane regions, $f$ and $g$ are nonconstant complex functions in $\Omega_1$, $\Omega_2$ resp. and $f(\Omega_1) \subset \Omega_2$, so that $h=g \circ f$ can be defined. If we know that $f$ and $h$ are holomorphic, is $g$ holomorphic as well? What if we know that $g$ and $h$ are holomorphic?

The easiest example will be a constant function, but the problem does not allow this. I found it difficult to find counter examples. Can anyone help me?

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    Lieber @Ben, um Missverständnisse vorzubeugen: ich selber bin ein großer Anhänger des Wirtinger Kalküls. Er scheint leider außerhalb des Kreises der Spezialisten der komplexen Analysis in mehreren Veränderlichen nicht so breit bekannt zu sein, wie er es verdient.Ich freue mich deshalb um so mehr, dass Sie diesen Kalkül erwähnen und beherrschen!2012-10-03

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I suspect Rudin's problem was intended to be a little broader in scope.

Let me deal with the second part first by giving an counterexample where $g,h$ are homomorphic, but $f$ is not.

Define the function $\lambda: \mathbb{C}\setminus \{0\} \to \mathbb{C}$ by $\lambda(z) = \log |z| + i \arg z$, where I am taking $\arg: \mathbb{C}\setminus \{0\} \to [-\pi, \pi)$. This is the principal branch of $\log$ extended to include the ray $\{(x,0)| x < 0\}$. $\lambda$ is not continuous on this ray, hence not analytic on $\mathbb{C}\setminus \{0\}$. The point being that $e^{\lambda(z)}= z$ (on the appropriate region).

Then take $\Omega_1 = \mathbb{C}\setminus \{0\}$, $\Omega_2 = \mathbb{C}$. Let $g:\Omega_2 \to \Omega_2$ be given by $g(z) = e^z$, and let $h:\Omega_1 \to \Omega_2$ be given by $h(z) = z$. Again, it is easy to see that $f = \lambda$ satisfies the equation $h(z) = g(f(z))$, for $z \in \Omega_1$. Thus $f$ need not be holomorphic.

Now for the first part. It turns out that if $f$ and $h$ are holomorphic, then $g$ is too. We must show that $g$ is differentiable on $f(\Omega_1)$. To simplify life, take $\Omega_2 = f(\Omega_1)$. Since $f$ is not constant, $\Omega_2$ is a region (Rudin, The Open Mapping Theorem).

First we show that $g$ is continuous. Let $\hat{w} \in \Omega_2$, then $\hat{w} = f(\hat{z})$ for some $\hat{z} \in \Omega_1$. By Rudin, Theorem 10.32, we can write $f(z) = \hat{w}+(\phi(z))^m$ in some neighborhood $V$ of $\hat{z}$, for some integer $m\geq 1$, and an invertible, analytic map $\phi$. Note that $\phi(\hat{z}) = 0$, and $|\phi(z)| = |f(z)-\hat{w}|^\frac{1}{m}$. Now let $w_k \to \hat{w}$, with $w_k \in f(V)$, and choose $z_k \in V$ such that $f(z_k) = w_k$. Then we have $|\phi(z_k)| \to 0$, and since $\phi$ is invertible and analytic, we have $z_k \to \hat{z}$. Then we have $g(w_k) = h(z_k)$, and since $h(\hat{z}) = g(\hat{w})$, we have that $g$ is continuous at $\hat{w}$.

Now suppose $f'(\hat{z}) \neq 0$. Then by Rudin, Theorem 10.30, $f$ has a local analytic inverse $\psi$ such that $\psi(f(z)) = z$ in a neighborhood $U$ of $\hat{z}$. This in turn gives $f(\psi(f(z))) = f(z)$, and since $f$ is non constant, $f(U)$ is a region. Then for $w \in f(U)$, we have $f(\psi(w)) = w$. Thus we have $g(f(\psi(w)) = g(w) = h(\psi(w))$ in $f(U)$. Hence $g$ is analytic at $f(\hat{z})$.

By Rudin, Theorem 10.18, we see that the zeros of a non-constant analytic function are isolated. In particular, the zeros of $f'$ are isolated. So, suppose $f'(\hat{z}) = 0$. Then $f'(z) \neq 0 $ for $z \neq \hat{z}$ in a neighborhood of $\hat{z}$. Hence $g$ is analytic for $z \neq \hat{z}$ in this neighborhood. Then Morera's theorem, along with Rudin, Theorem 10.13 and continuity of $g$ shows that $g$ is also analytic at $f(\hat{z})$.

Hence $g$ is analytic on $\Omega_2$.

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    I think I have fi$x$ed it, but the reasoning is more delicate. It turns out that I was wrong about the first part.2012-10-07
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if $\Omega_2$ is strictly larger than $f(\Omega_1)$, then $g \circ f$ doesn't depend of what $g$ does outside $f(\Omega_1)$, so you can pick $g$ non holomorphic there and still have $f$ and $h$ holomorphic.

As for the other question, if $g$ is many-to-one, such as $g(z) = z^2$, you can switch the arguments given to $g$ as you want as long as they get sent to the same image : here if $f$ is any function such that for any $z$, $f(z) \in \{-z;z\}$, then $g = g \circ f$, where $g$ and $g$ are holomorphic, but $f$ may not be.

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    definitely. It's better to assume f(omega1) = omega2. Thank you very much.2012-10-04