Let $(X,p)$ be a Riemann domain over $\mathbb{C}^n$, that is, $X$ is a Hausdorff space with a local homeomorphism $p:X\rightarrow \mathbb{C}^n$. We call a set $A\subset X$ univalent if $p|_A :A\rightarrow p(A)$ is a homeomorphism. Also, a continuous map $s:p(A)\rightarrow X$ is called a section of $(X,p)$ over $p(A)$ if $p\circ s=id_{p(A)}$ holds.
In the book, Extension of Holomorphic Functions by Jarnicki and Pflug, it is claimed that a set $A\subset X$ is univalent if and only if there exists a section $s$ such that $s(p(A)) = A$.
I have shown that if $A$ is univalent then the given condition $s(p(A)) = A$ holds for some section $s$. But I find that the reverse implication is hard and I could not find any solution. Any help will be appreciated.
Thank you.
This has nothing to do with complex analysis and is an exercise in the elementary set theory:
Suppose that $A$ satisfies the stated condition (the existence of a section $s$). Then $p\circ s=id_{p(A)}$. Let's consider the composition $s\circ p$. For $a\in A$, $x=s(p(a))\in p^{-1}(p(a))$. I claim that $x=a$. Indeed, since $s(p(A))=A$, there has to be $y\in p(A)$ such that $s(y)=a$. But $ps(y)=y$, hence, $y=p(a)$. Thus, $x=s(p(a))=a$.
Thus, $p|_A: A\to p(A)$ is a continuous bijection with continuous inverse map $s$. Hence, $p|_A: A\to p(A)$ is a homeomorphism.