Question on a theorem on Riemann surfaces

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 ?

Thanks.

• The statement of the theorem in the first box has several typos. Please look back at the book and make fixes (an $X$ is missing and some $X$'s should be $Y$'s and vice versa, probably). – KCd Jun 22 '13 at 15:53
• As far as the definition of "biholomorphic" goes, connected open subsets of Riemann surfaces are themselves Riemann surfaces (Ex. 1.5 (b)), and you may use Definition 1.9 on the subsets. – fuglede Jun 22 '13 at 17:19

I follow Foster's book, as well. Use Corollary 2.5 at pag.11: as $p:Y\rightarrow X$ is a (non constant) unbranched holomorphic map, then it is locally a homeomorphism. Let us denote this homeomorphism by $p|_{V}:V\rightarrow U$. This implies, in particular, that $p|_{V}$ is injective. Now Corollary 2.5 implies that $p|_{V}:V\rightarrow U$ is biholomorphic. The proof of this result uses thm. 2.1. at pag.10.