The germ projection of a complete analytic function is a covering map

Let $p:\mathcal{R}({\xi }_0,\mathrm{\Omega })\to \mathrm{\Omega }$ be the germ projection of a complete analytic function and fix $z\in \mathrm{\Omega }$. Choose a disc $D$ centred at $z$ with $\bar{D}\subseteq \mathrm{\Omega }$. Every germ $\xi \in p^{-1}(z)$ continues along every path in $D$: concatenate such a path with one from the original base point to $z$ that produces $\xi$. Since $D$ is simply connected, On a simply connected domain, pathwise continuation glues to one holomorphic function gives a holomorphic representative $f_{\xi }$ on all of $D$.

The sets $N(f_{\xi },D)$ are pairwise disjoint. If two met, their representatives would agree as germs at one point of $D$; continuation back to $z$ inside $D$ would make their centre germs equal. They also cover $p^{-1}(D)$, because any germ over a point of $D$ can be continued inside $D$ back to $z$ and hence lies on the sheet determined by that centre germ. On each sheet, $p$ is the chart homeomorphism of The germ projection is a local biholomorphism. Thus $D$ is evenly covered in the sense of Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings, and the germ projection is a covering map.