Sections of holomorphic functions

Question

This question is motivated by Show the function $f: \mathbb{C}_{\pi}\to\mathbb{C}$ such that $(f(z))^2=z$ and $\text{re($f(z)$)$>0$}$ is continuous on $\mathbb{C}_{\pi}$ and Branch of the n-th root of a holomorphic function is holomorphic and generalizes them.

Let $U \subset \mathbb C$ be open and $f : U \to \mathbb C$ be holomorphic. A function $g : V \to U$ defined on an open $V \subset \mathbb C$ is called a section function for $f$ if $f(g(z)) = z$ for all $z \in V$. Note that we do not assume that $g$ is continuous. Also note that section functions must be injective.

On an open $V \subset \mathbb C$ section functions exist if and only if $V \subset f(U)$. For the if-part observe that the product $P(V,f) = \prod_{z \in V} f^{-1}(z)$ is non-empty and that the elements of $P(V,f)$ can be identified with section functions for $f$. The only-if-part is trivial.

Note that $P(V,f)$ is in general uncountable (this happens if infinitely many fibers $f^{-1}(z)$ with $z \in V$ have more than one element). Most of the corresponding section functions are not continuous and thus not holomorphic, i.e. they are fairly uninteresting.

Such section functions are considered e.g. for the entire functions $f(z) = z^n$ and $f(z) = e^z$. It is well-known that there exists holomorphic section functions (holomorphic branches of the $n$-th root resp. logarithm). So the following question arises:

Is a continuous section function automatically holomorphic?

Answer 1

The answer is yes provided $f'(g(z)) \ne 0$ for all $z \in V$. We can prove the following variant of the inverse function theorem:

Let $g : V \to U$ be a section function for $f : U \to \mathbb C$. Then $g$ is complex differentiable in $z_0$ if and only if $g$ is continuous in $z_0 \in V$ and $f'(g(z_0)) \ne 0$. In that case we have $g'(z_0) = \dfrac{1}{f'(g(z_0))}$.

Proof.

1. If $g$ is complex differentiable in $z_0$, then it is clearly continuous in $z_0$. Moreover, the chain rule gives us $f'(g(z_0) g'(z_0) = 1$ which shows that $f'(g(z_0)) \ne 0$ and $g'(z_0) = \dfrac{1}{f'(g(z_0))}$.

2. Let $g$ be continuous in $z_0 \in V$ and $f'(g(z_0)) \ne 0$. For $z \ne z_0$ we have $$\frac{g(z)-g(z_0)}{z - z_0} = \frac{g(z)-g(z_0)}{f(g(z)) - f(g(z_0))} = \left( \frac{f(g(z)) - f(g(z_0))}{g(z)-g(z_0)}\right)^{-1} \tag{1}$$ Note that $g(z) \ne g(z_0)$ because $g$ is injective. The continuity of $g$ in $z_0$ means that $g(z) \to g(z_0)$ as $z \to z_0$, thus $$\lim_{z \to z_0} \frac{f(g(z)) - f(g(z_0))}{g(z)-g(z_0)} = \lim_{w \to g(z_0)} \frac{f(w) - f(g(z_0))}{w-g(z_0)} = f'(g(z_0)).$$ Since $f'(g(z_0)) \ne 0$, we conclude from $(1)$ $$\lim_{z \to z_0} \frac{g(z)-g(z_0)}{z - z_0} = \frac{1}{f'(g(z_0))}.$$