How can I actually find the universal covering space of $S^1$ using the path classes construction?

Question

According to Introduction to Topological Manifolds (John M. Lee, p. 298), one can easily build the universal cover of a given topological space $X$ by picking a point $x_0$ at random, and then considering all the possible paths from this base point to any other point in the space. If we call this set $A(X; x_0)$, then our universal covering space would be $A(X; x_0)/\sim$, where the symbol $\sim$ stands for the path equivalence relation.

But my question is: how do you actually use this construction in practice? The proof of this theorem relies on giving the covering space a topology in a pretty unintuitive way, and I'm afraid that's the trickiest part when trying to make use of it.

Suppose I want to find the universal covering space of $S^1$ (forgetting for a moment that we already know it's $\mathbb{R}$): if I choose a point $x_0\in S^1$, then I can say that any two paths that start from $x_0$ and end at different points on the circle cannot be equivalent, and the same is true for any two paths that end at the same point but after having gone around the circle a different number of times. This gives me a rough idea that the covering space must have "the same number of points" as $\mathbb{R}$... but how do I show it's $\mathbb{R}$ using solely this construction?

Answer 1

This is unfortunately a non-answer, and so should really be a comment, but it's too long.

I don't think your main question has a useful answer. The definition of the universal covering space of $X$ using paths is tremendously useful for many things:

• The proof of existence of a universal covering space, assuming the correct hypotheses on $X$ (path connected, locally path connected, semilocally simply connected);
• The classification of pointed covering spaces in terms of subgroups of $\pi_1$; other classifications, ie. unpointed covering spaces in terms of conjugacy classes of subgroups; and regular covering spaces in terms of normal subgroups;
• The definition of the deck transformation action of $\pi_1 X$ on its universal covering space uses the path construction directly.

and so on.

But, I don't think that definition is particularly useful for "finding" what familiar space the universal covering space is homeomorphic to. If you can guess what the universal covering space is homeomorphic to, and prove your guess is right, all with a brief formula like $x \mapsto e^{2\pi i x}$, what better way than that?

One can certainly prove that $\mathbb R$ is homeomorphic to the path space construction: the map $\mathbb R \mapsto S^1$ defined by $t \mapsto e^{its}$ is a covering space, and $\mathbb R$ is simply connected; the map from the path space construction to $S^1$ is a covering space, and it is simply connected; now apply the uniqueness of a simply connected covering space to deduce that $\mathbb R$ and the path space construction are homeomorphic.

Try your line of thought out on other examples: How would you use the path construction to deduce that the universal covering space of the projective plane is $S^2$? How would you use it to deduce that the universal covering space of the torus, or the Klein bottle, is $\mathbb R^2$? How would you use it to deduce that the universal covering space of the wedge of two circles is the infinite tree with constant valence 4? In each case, the useful, practical thing to do is to prove that the guessed universal covering space is simply connected and to directly write down the covering map. Why hamstring yourself by insisting on using a tool which is designed for theoretical use, i.e. for abstract existence and classification results, when the mathematical reality of $\mathbb R$ is staring at you in the face?