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I'm working on problem IX.7.1 of Conway's Functions of a Complex Variable:

If $(X,\rho)$ is a covering space of $\Omega$ and $(\Omega, \pi)$ is a covering space of $Y$, prove that $(X,\pi\circ\rho)$ is a covering space of $Y$.

This is how I've gone about the problem: If $y \in Y$, there is some neighborhood $\Delta _y$ such that (calling $\{c_i\}_{I \in \mathcal{A}}$ the components of $\pi^{-1}(\Delta_y)$ in $\Omega$) each $c_i$ is open in $\Omega$ and $\pi:c_i \to \pi(c_i) = \Delta_y$ is a homeomorphism (calling this a fundamental neighborhood).

We can do the same thing for each preimage of $y$ under $\pi$ in $\Omega$, constructing a fundamental neighborhood for each w.r.t. the map $\rho: X \to \Omega$.

So the way that I've tried to construct a fundamental neighborhood about $y$ for the mapping $\pi \circ \rho$ is by intersecting the fundamental neighborhoods of each preimage of $y$ under $\pi$ with the the components $c_i$ (call these intersections $B_i$). Each of these intersections is a single intersection of open sets, so open. Then, take the fundamental neighborhood of $y$ w.r.t the map $\pi \circ \rho$ to be the intersection of each $\pi(B_i)$.

The problem that I'm having is that this is an arbitrary intersection of open sets, so this may not be open; further, I don't see how we can just take the interior, since the intersection may just be a singleton. Any ideas on how to ensure this results in an open set?

Thanks!

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