Analytic continuation gives a covering space (and not just a local homeomorphism)

Question

Let $\mathcal{G}$ be the space of germs of holomorphic functions defined on open subsets of $\mathbb{C}$, topologized in the usual way. There is a natural map $p\colon \mathcal{G} \rightarrow \mathbb{C}$ taking a germ of a holomorphic function defined on a neighborhood of a point $z_0 \in \mathbb{C}$ to $z_0$. Let $M$ be a path component of $\mathcal{G}$. One can view $M$ as being the "maximal analytic continuation" of any of its constituent germs. It is basically trivial that the restriction of $p$ to $M$ is a local homeomorphism onto some open subset $U \subset \mathbb{C}$. However, I have seen it asserted in many places that the map $p\colon M \rightarrow U$ is not just a local homeomorphism, but is actually a covering space. Since in general the fibers of $p$ might be infinite, this does not come for free.

Is this true, and if so does anyone know a reference that proves it cleanly? For instance, I can't quite find it in Forester's book, though he spends a lot of time constructing $\mathcal{G}$.

Answer 1

Consider the map $f\colon z \mapsto z^3-z$. Then there is a path component $M$ of $\mathcal G$ consisting of germs of the inverse map to $f$. More precisely $M$ is isomorphic to $\mathbb C \setminus \{ \pm \sqrt{ \frac{1}{3}}\}$, the domain of $f$ minus the points where its derivative vanishes, $M$ maps to $U = \mathbb C$ by $f$, and the germ at any point is the germ of a local inverse of $f$ at that point.

But $M \to \mathbb C$ is not a covering space since it has degree $3$ over most points but only degree $1$ over $ \pm - \frac{2}{3} \sqrt{\frac{1}{3}}$.