This is related to Forster, Riemann surfaces Exercise 7.1

Let $X$ be a riemann surface and $a\in X$ be a point with $\phi\in O_{X,a}$ where $O_{X,a}$ is the set of holomorphic germs at $a$. Let $Y$ be another riemann surface. Suppose $(Y,p,f,b)$ is a maximal analytic continuation of $a$ of $X$ in the following sense. $p:Y\to X$ unbranched holomorphic map of $X$ and $p(b)=a$( or $b$ is a fiber of $a$ under $p$) s.t. pushforward on stalkwise $p_\star(f_b)=\phi\in O_{X,a}$ as $p$ is local homeo. Maximal is in the sense that for another analytic continuation $(Z,q,g,c)$ of $\phi$, $\exists F:Z\to Y$ compatible with projection map $p,q$(i.e fiber preserving holomorphic map).

Ex 7.1 $X,Y$ are riemann surfaces. $p:Y\to X$ holomorphic unbranched covering map, $f:Y\to C$ holomorphic,$\phi\in O_{X,a}$. Then $(Y,p,f,b)$ is maximal in the above sense iff for any $b_1\neq b_2,b_i\in p^{-1}(a),p_\star(f_{b_1})\neq p_\star(f_{b_2})$.

$\textbf{Q1:}$ I think maximal here is a very bad terminology. Maximal should mean irredundant/minimal against the necessary information to maximally extend the germ, if I understand Ex 7.1 correctly. If I have the image in the germ agrees, I can locally remove a closed ball $K$ containing $b_1$ and consider $\tilde{Y}=Y-K$ and $\tilde{Y}\to Y$ is canonical embedding fiber preserving holomorphic map obviously. Then I consider the pair $(\tilde{Y},p|_{\tilde{Y}},f|_{\tilde{Y}},b)$ instead $(Y,p,f,b)$. The converse direction is just reversing argument for contrapositive statement.

$\textbf{Q2:}$ I do not think I need covering here. All I need is unbranched holomorphic map. Is this correct?


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