In the book "Lecture on Riemann surface" of Forster, in the page 23, there is a theorem as follows:
Suppose $X$ and $Y$ are Riemann surfaces, $p: Y\rightarrow X$ is an unbranched holomorphic map and $f:Z\rightarrow X$ is any holomorphic map. Then every lifting $g:Z\rightarrow Y$ of $f$ is holomorphic.
I understand the idea of the proof there, however in the proof, there is a claim :
Let $c$ be an arbitrary in $Z$. Let $b=g(c)$ and $a=p(b)=f(c)$. There exist open neighborhoods $V$ of $b$ and $U$ of $a$ such that $p|V\rightarrow U$ is biholomorphic.
I do not understand why such restriction should be biholomorphic. And I can't even find the definition of biholomorphic map on 2 open subsets of Riemann surfaces from the section 1 to there.
Could you please explain for me : Why $p|V\rightarrow U$ is biholomorphic ?