While solving the problems on Analytic continuation from Gamelin's book; I encountered this one- still unsolved:
Let $D= \{0 < |z| < \epsilon\}$ and suppose $f$ is holomorphic at $z_{0} \in D$ and $e^{w_{0}} = z_{0}$ . Show that $f$ has an analytic continuation along any path in $D$ starting at $z_{0}$ if and only if there is a holomorphic function $g(w)$ in the half plane $L\colon=\{\operatorname{Re}(w)<\operatorname{log}(\epsilon)\}$ such that $f (e^w ) = g(w)$ for $w$ near $w_{0}$.
In summary, the keywords for this exercise are monodromy theorem and covering map.
$"\implies"$-direction.
Assume that (the germ of) $f$ has an analytic continuation along any path in $D$ starting at $z_0$. Let $u\colon[0,1]\rightarrow L$ be a path in $L$ starting at $w_0$, i.e. $u(0)=w_0$. We will first construct an analytic continuation of the germ of $f\circ{\operatorname{exp}}$ at $w_0$ along the path $u$. To do so, consider the path $\hat{u}\colon={\operatorname{exp}}\circ u$ in $D$. By assumption, there exists a partition $0=t_0<t_1<\ldots<t_{n-1}<t_n=1$ of the interval $[0,1]$, domains $U_i\subset D$ with $\hat{u}([t_{i-1},t_i])\subseteq U_i$ for $i=1,\ldots,n$ and holomorphic functions $f_i\colon U_i\rightarrow\mathbb{C}$ for $i=1,\ldots,n$ such that
Now, for $i=1,\ldots n$, consider the preimages $W_i\colon=\operatorname{exp}^{-1}(U_i)$. Since the exponential function is a local homeomorphism, we can without loss of generality, possibly by shrinking the $U_i$, assume that the exponential function when restricted to $W_i$ is a homeomorphism for all $i$. Thus, we have $u([t_{i-1},t_i])\subseteq W_i$ for all $i$. Additionally, as a composition of holomorphic functions, the function $g_i\colon=(f_i\circ{\operatorname{exp}})\restriction_{W_i}$ is holomorphic for all $i$. Since the germs of $f$ and $f_1$ at the point $z_0$ agree, so do the germs of $f\circ {\operatorname{exp}}$ and $f_1\circ {\operatorname{exp}}$ at the point $w_0$. Hence, the germs of $f\circ {\operatorname{exp}}$ and $g_1$ agree at the point $w_0$. One finally checks that $g_i\restriction_{V_i}=g_{i+1}\restriction_{V_i}$ for all $i=1,\ldots n-1$, where $V_i$ denotes the connected component of $W_i\cap W_{i+1}$ containing the point $u(t_i)$. All in all, we have constructed an analytic continuation of the germ of $f\circ{\operatorname{exp}}$ at $z_0$ along the path $u$.
Now, since $L$ is simply connected, it follows from the monodromy theorem that there exists a globally defined holomorphic function $g$ on $L$ such that near $w_0$ the functions $f\circ{\operatorname{exp}}$ and $g$ are equal (i.e. a 'direct analytic continuation of the germ of $f\circ{\operatorname{exp}}$ to $L$'). A proof of this corollary of the monodromy theorem can be found after Corollary 7.4. in Forster's Lectures on Riemann Surfaces.
$"\impliedby"$-direction.
Assume that we are given a holomorphic function $g\colon L\rightarrow \mathbb{C}$ such that the germs of $g$ and $f\circ{\operatorname{exp}}$ at the point $w_0$ agree. Let $\gamma\colon [0,1]\rightarrow D$ be a path in $D$ such that $\gamma(0)=z_0$. Convince yourself that the exponential function $\operatorname{exp}\colon L\rightarrow D$ is a covering map. Every covering map has the path lifting property. Thus, there exists a path $\hat{\gamma}\colon [0,1]\rightarrow L$ with ${\operatorname{exp}}\circ \hat{\gamma}=\gamma$ and $\hat{\gamma}(0)=w_0$.
Since the exponential function is a local homeomorphism, we can find a family of open sets $\{W_i\}_{i\in I}$ in $L$ such that the following conditions hold:
We can choose the set $I$ to be finite, since the trace of the path $\hat{\gamma}$ is compact.
For $i\in I$ define $U_i\colon=\operatorname{exp}(W_i)$ and denote the holomorphic inverse of the exponential function on $W_i$ by $h_i$. Finally, for $i\in I$, set $f_i\colon=g\circ h_i \colon U_i\rightarrow\mathbb{C}$. One checks that the $(f_i,U_i)$ give an analytic continuation (of the germ) of $f$ along the path $\gamma$.