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?



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.