A continuous local section of a holomorphic function is holomorphic In a comment to A simple proof that a real differentiable local section of a holomorphic function is holomorphic it was remarked that there is the following more general result:

Let $U, V \subset \mathbb C$ be open and $s : U \to \mathbb C, f : V \to \mathbb C$ be functions such that $s(U) \subset V$ and $(f \circ s)(z) = z$ for all $z \in U$ (this means that $s$ is a local section of $f$). If $s$ is continuous and $f$ is holomorphic, then $s$ is holomorphic.

This theorem allows for example to define the concept of branches of the logarithm and the $n$-th root as follows:

*

*A branch of the logarithm on a domain $U$ is a continuous function $\ell : U \to \mathbb  C$ such that $e^{\ell(z)} = z$ for all $z \in U$.


*A branch of the $n$-the root on a domain $U$ is a continuous function $\rho : U \to \mathbb  C$ such that $\rho(z)^n = z$ for all $z \in U$.
By the above theorem such branches are automatically holomorphic.
The purpose of this question is to give a proof of this theorem to obtain a standard reference in this forum.
 A: First note that $s$ is injective on $U$.
Fix $z \in U$ and let $s(z)=w \in V$. Assume now that $w$ is not a critical point of $f$ and note that $f(w)=z$ hence there is a small neighborhood $V_w$ of $w$ st $f(V_w)=U_z$ is a small neighborhood of $z$ (which we can take included in $U$) st $f$ is injective on $V_w$ and hence there is $g:U_z \to V_w$ analytic inverse of $f$ there.
Now $s$ is continuous so $s^{-1}(V_w)$ contains a neighborhood $U'_z$ of $z$ which again we can take included in $U_z$; then for all $y \in U'_z$ we have $s(y) \in V_w$ so $(g\circ f) (s(y))=s(y)$ by the definition of $g$; but also $f(s(y))=y$ by hypothesis, so we get that $g(y)=s(y), y \in U'_z$, hence $s$ is analytic on $U'_z$
This shows that $s$ is analytic on $U-A$ where $A$ is the set of points $z \in U$ for which $s(z) \in V$ is a critical point of $f$. Now the set $B$ of (all) critical points of $f$ in $V$ is discrete since $f$ is clearly nonconstant by hypothesis and $s$ is injective and continuous, hence $A=s^{-1}(B \cap s(U))$ is discrete too, so each point in $A$ is an isolated singularity for continuous $s$, hence is removable, so $s$ is analytic on $U$
