I'm struggling with the proof of the following theorem by Ahlfors, on page 289:
Theorem l. Two analytic continuations $\overline{\gamma_1}$ and $\overline{\gamma_2}$ of a global analytic function $\mathbf{f}$ along the same arc $\gamma$ are either identical, or $\overline{\gamma_1}(t) \neq \overline{\gamma_2}(t)$ for all t.
Here $\gamma:[a,b] \to \mathbb C$ is an arc, and $\overline{\gamma_i}:[a,b] \to \mathfrak{S}_0 (\mathbf{f})$ are analytic continuations of $\mathbf{f}$ along $\gamma$, where $\mathfrak{S}_0(\mathbf{f})$ is the Riemann surface associated with a global analytic function $\mathbf{f}$. This means that $\pi \circ \overline{\gamma_i}=\gamma$, as well that the functions $\{ \overline{\gamma_i} \}$ are continuous on $[a,b]$ in the topology of $\mathfrak{S}_0 (\mathbf{f})$.
Right after stating the theorem Ahlfors claims:
The proof is a triviality. Because $\pi$ is a local homeomorphism the image of $\overline{\gamma_1}-\overline{\gamma_2}$ cannot contain a point of the zero section without being contained in it.
I tried following his guidance in order to prove the theorem:
Suppose there exists some point $t_0 \in [a,b]$ such that $\overline{\gamma_1}(t_0)-\overline{\gamma_2}(t_0)= \mathbf{0}_{\gamma(t_0)}$. Since $\pi$ is a local homeomorphism we may find for each of the germs $\{\overline{\gamma_1}(t)-\overline{\gamma_2}(t) : t \in [a,b] \}$ a neighbourhood $\Delta_t \subset \mathfrak{S}_0 (\mathbf{f})$ which is homeomorphic to some open set $\Omega_t \subset \mathbb C$, which includes the point $\gamma(t)$. The section $\left( \pi \big|_{\Delta_{t_0}} \right)^{-1}\in \Gamma( \Omega_{t_0}, \mathfrak{S})$ vanishes at $\gamma(t_0)$, hence by a previous statement from the book, the only section in $\Delta_{t_0}$ is the zero section.
Since $(\overline{\gamma_1}-\overline{\gamma_2})([a,b])$ is compact, we can restrict the cover $\{ \Delta_t \}_{t \in [a,b]} $ to a finite cover, to which we join $\Delta_{t_0}$ (this clearly induces a finite subcover of $\gamma([a,b])$ as well). Since $\gamma([a,b])$ is connected we can form (finite) chains from $\Omega_{t_0}$ to any other element of the subcover with consecutive sets having nontrivial intersection. Repeating the argument from before we obtain the desired result.
Is the proof above correct? If not, please help me fix it. Thanks!
The idea is sound, but the way you built the argument feels like Weierstrass-style proof translated into sheaves. In the earlier question we essentially proved the following
Indeed: open because the zero germ is determined by $(0,\Omega)$, which determines only other zero germs. The complement is also open because a nonzero germ at $\zeta$ is determined by $(f,\Omega)$ where $f$ does not vanish in $\Omega\setminus\{\zeta\}$, hence it defines only other nonzero germs in $\Omega$.
The rest is general topology: if $A$ is a clopen subset of topological space $Y$, then every connected subset of $Y$ is either contained in $A$ or is disjoint from it. This applied to the image of $\gamma$ under any continuous map into $\mathfrak{S}$, in particular to the analytic continuation.