Let $U\subset \mathbb{C}$ and $U$ open. Let $f:U\rightarrow \mathbb{C}$ and $f$ holomorphic. Let $p\in U$. Let $\gamma$ be path starting from $p$.
definition $f$ can be analytically continued along path $\gamma$ iff there exists an indexed finite set of tuples $(D_i,f_i)_{i\in I}$ where $D_i$ is a an open disc and $f_i$ is a holomorphic function on it such that for all $i,j \in I$ $f_i|_{D_i\cap D_j}=f_j|_{D_i\cap D_j}$ and for all $i\in I$ $f_i|_{D_i\cap U}=f$.
$V(f):=\{q\in \mathbb{C}: \ \exists $ a path $ \gamma $ starting at $p$ and ending at $q $ such that $f$ can be continued along it$\}$
The question is: can we perform analytic continuation along any path in $V$ starting from a point $p$?
So far, I have only encountered functions for which this is possible. For example: meromorphic functions, sections of functions as $z\rightarrow z^n$ and solutions of linear ode's on $\mathbb{C}$.
(edit: I want to stress that the resulting extension of $f$ could be multivalued, so that is not the issue.)
Best Answer
Okay, I found a counterexample myself. Define $z\mapsto \log(z)$ as a local inverse around 1 of $z\mapsto e^z$ such that $\log (1)=0$. Define $$ g(z)=\frac{1}{z-2 \pi i}. $$ The set of points for which there exists a path starting at 1 along which we can analytically continue the function $g(\log(z))$ is $\mathbb{C}\backslash \{0\}$. However, we cannot analytically continue this along the path $t\mapsto e^{2t\pi i}$, since there $\log(z)$ reaches $2\pi i$ and $g$ explodes on that value.