Construct a simply connected covering space of the space $X \subset \mathbb{R}^3$ that is the union of a sphere and diameter.
Okay, let's pretend for a moment that I've shown, using van Kampen's theorem or some other such method, that X has the fundamental group $\mathbb{Z}$, and I have in mind a covering space that consists of a row of spheres each connected to the next by a line segment, along with a map from that row of spheres to X that takes the line segments to the diameter, and the spheres to the sphere. It's not hard to sow that this choice is simply connected.
My question is not how to do all of that. My question is, at this point, how do I show for certain that it is a covering space for X? The only definition I can find for a covering space, hence the only way to be certain that I have one in hand, is this:
A covering space of X is a space X' with a map $p: X' \rightarrow X$ so that there exists an open cover $U_\alpha$ of X so that for each $\alpha$, $p^{-1}(U_\alpha)$ is a disjoint union of open sets in X', each of which is mapped by p homemorphically onto $U_\alpha$.
First problem: What is the best/easiest way, handed some space X in general and with a covering space X', p in mind, to find an open cover that might fit this bill? I'm sure it isn't that hard for this particular case, but it's easy to imagine (and I think I even have on another problem in the same set) a more difficult case.
Second problem: A generic open cover might consist of uncountably infinitely many sets. How can I be certain that every single one of the $U_\alpha$'s has a preimage that's a disjoint union of sets, each of which is homeomorphic by p to $U_\alpha$? I've had some trouble with correctly showing two spaces to be homeomorphic by a map in the past, and I don't want to make a mistake of that kind again.
I may be worrying over nothing here, but whenever geometric intuition comes into algebraic topology I'm never completely sure just how much explanation is considered enough for that intuition, or how to translate it into a more rigorous proof of the idea. So far I've made the mistake of not explaining enough, and I'd prefer to err on the other side from now on.
(I know my "X'" should be X-tilde, but don't know how to code for that)
\tilde X
. Also try\widetilde X
$\widetilde X$ $\endgroup$