3
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ $\tilde X$ can be done via \tilde X. Also try \widetilde X $\widetilde X$ $\endgroup$ – Stefan Hamcke Sep 22 '14 at 17:03
4
$\begingroup$

Without information about $X$ you can't really write an "explicit" universal cover. What you can do is write a generic $\widetilde X$ using some general construction (for example the one found in Hatcher, $\widetilde X = \{ (x, \gamma) \mid \gamma : I \to X, \gamma(0) = x \} / \sim$). But that's not a very maniable construction.

If you're able to compute explicitly the fundamental group and find for example a "small" generating set and simple relations between the generators, it's easier to find the universal cover -- as you did in this example. Basically for every generator you cut your space in two in the middle of the loop, and then you glue an infinite number of spaces in a string (like you did). Then if there are relations you identify things. But this technique starts to become very complicated when there are many generators, or when it's not obvious how to "cut out" the space (think about $\mathbb{RP}^2$, what do you cut?)...

The upshot of all that is that describing the universal cover explicitly isn't easy in general. Working with the properties of the universal cover is easier, namely: it's simply connected and it's a covering space of the original space... (Which implies all the nice lifting properties you know). Knowing that it exists is a powerful theorem, and one isn't usually too concerned with how the universal cover looks (well, I guess it depends on applications).

As for checking whether a map is a covering map: it's enough to check the covering condition for a basis of the topology. So in your example, you can check it for very small open sets that are homeomorphic to either a segment, a disk, or a disk with a segment glued in the middle (a picture will make all that easier). For these open sets, it's more or less easy to check the covering condition, if you've defined your covering space correctly.

$\endgroup$
4
$\begingroup$

For a space $X$ which does admit a universal cover there is a quite explicit construction but you need to take account of all base points of $X$, as Najib states, and which can be given in the following form.

First one forms the fundamental groupoid $\pi_1(X)$. There are two functions $s,t:\pi_1(X) \to X$, namely the source and target. As a set, the universal cover $\tilde{X}_x$ of $X$ at a point $x \in X$ is just $s^{-1}(x)$, the Star of $\pi_1(X)$ at $x$, or, equivalently, the Costar, $t^{-1}(x)$. Under the given local conditions on $X$ the topology on $X$ can be lifted to a topology on $\pi_1(X)$, making it in fact a topological groupoid. This method of using lifted topologies via covering morphisms of groupoids, in particular covering morphisms of $\pi_1(X)$, is dealt with in Chapter 10 of Topology and Groupoids. In particular, 10.5.8 shows how to topologise $\pi_1(X)$ (actually it deals with $\pi_1(X)/N$ where $N$ is a normal subgroupoid of $\pi_1(X)$).

Now you may be able to see how to construct the universal cover of you example, at a specific point of it.

Alternatively, the homotopy type of the example you give is that of $S^2 \vee S^1$, and the universal cover of that at the wedge point is easily seen to be a copy of $\mathbb R$ with an $S^2$ glued to each integer point. Can you see from this how to get the universal cover of your example at a particular point, e.g. one end of the given diameter?

Update September 24, 2014: Here is a description of the universal cover. Consider the real line $\mathbb R$ with a blob at each integer point. Now expand each blob to a $2$-sphere. That is the universal cover at some point of your example $X$.

To see how this fits with the construction, star at one point $a$ of your $2$-sphere $S$ with $b$ as its opposite point. Now $S$ is simply connected so there is one path class in $\pi_1S(a,x)$ for each $x \in S$. However from $\pi_1X(a,b)$ we get more path classes going back along the diameter joining $b$ to $a$. These give the line joining the first sphere to the next one.

I also think this is a good example to compare with the more usual one of the universal cover of $S^2 \vee S^1$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.