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I'm looking for some intuition regarding universal covers of topological spaces.

$\textbf{Setup:}$ For a topological space $X$ with sufficient adjectives we can construct a/the simply connected covering space of it by looking at equivalence classes of paths at a given base point. We then can put a topology in the standard way done by Hatcher - an open set around an equivalence class of paths, say $[\gamma]$ is the set of $[\gamma\cdot\eta]$ where $\eta$ is a path starting at $\gamma(1)$ contained in $U$ open in $X$.

Here are my questions:

Q: I find this topological space, as constructed above, non-intuitive. Certainly I dont know how I would manipulate it and make topological arguments in it. What is the 'right' way of thinking about the topology here? Or is this construction useful solely for proving the existence of simply connected covers?

Q: Often times it is tractable to construct a simply connected covering by ad-hoc methods (fancy guessing). The projective plane, torus, etc all spring to mind. By universality I know that the covering space obtained by any ad-hoc method is $\textit{the}$ universal covering space obtained by the above method, so there is an isomorphism of these two. Is there a standard way to see this isomorphism? Being really concrete, say in the cases of $\mathbb RP^2$, or $S^1\times S^1$.

In simple terms: how can I 'see' what the universal cover looks like from the general construction?

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There's no reason to expect the universal cover to "look like" anything at all-though when you accomplish an ad hoc construction, it looks an awful lot like the thing you constructed. "Look like" implies there would be something else you've already seen that is homeomorphic or at least homotopic to the abstract universal cover. In general, why should you have seen such a thing? –  Kevin Carlson Oct 18 '12 at 16:13

2 Answers 2

up vote 1 down vote accepted

I find it nicer to look at all the universal covers at all the base points. One way of doing this it to topologise the fundamental groupoid $\pi_1 X$ (assuming the usual local conditions) by adding neighbourhoods at the two end points of each path class. In this way $\pi_1 X$ becomes a topological groupoid. We have the source and target maps $s,t: \pi_1 X \to X$ and, depending on conventions, $s^{-1}(x)$, or $t^{-1}(x) $, is the universal cover of $X$ based at $x$.

Does this help one to "see" the universal cover?

One possible answer is that to "see" the universal cover one needs an algebraic model of a covering map. This is best supplied, IMHO, by a covering morphism of groupoids. An example of a covering morphism of groupoids is to take a group $G$ and to form the action groupoid for the action of $G$ on itself by multiplication. This gives a groupoid $\widetilde{G}$ with object set the set $G$, and depending on one's conventions, only one arrow $(h,g):hg \to g$ between any two objects, and composition law, $$(k,hg)(h,g)=(kh,g).$$ The covering morphism $\widetilde{G} \to G$ is $(h,g)\mapsto h$. This gives the "universal cover" of the group $G$. For more information, with different conventions, see this book, but that does not do topological groupoids.

Another way of thinking about such spaces is to say that a way of studying a space $X$ is to look at maps $Y \to X$ which are in some ways simpler.

But the origin of covering spaces was in the notion of Riemann surface, and so a way of making complex functions such as $\sqrt{z}$, or $ \log z$, single valued, by defining them on surfaces covering the complex plane minus one or more points. This "back to history" view might really help you to "see" what is wanted and what is going on!

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Suppose $p:\overline{C}\to C$ is a universal covering. By definition, around every point in $C$ is an open set that lifts up to $\overline{C}$. So, locally, $\overline{C}$ looks just like $C$. Suppose one wanted to cut up $C$ into small (contractible) patches and then stitch them together again to form $\overline{C}$ - the problem is that $\overline{C}$ is to be simply connected, so if (say) we started stitching patches around a nontrivial loop in $C$ when we wrap back around to the basepoint we can't stitch that last patch back to the original patch, instead we have to create a copy of the original patch and continue on from there.

Consider the space $C=\Bbb C\setminus\{0\}$. If one takes a counterclockwise loop from $-1$ around $0$ back to itself, then the last patch cannot be stitched to the first, so we should make a copy of the original patch to stitch it to. In the picture below, we've literally lifted the copy above the original:

$\hskip 2in$ quilt

If we continue this process indefinitely, then there will lots of copies of pieces of $C$ that are being stitched together. Given a point in $C$ in a patch, there will be many copies of that patch in our quilt, and so many lifts of that point - what allows us to distinguish between lifts of the same point is how we got to it from the original basepoint. Thus, we can interpret points in $\overline{C}$ as points in the original space $C$ but with a "memory" of how we got there from a basepoint.

This inspires us to formalize our construction by letting elements of $\overline{C}$ be paths in $C$, modulo endpoint-fixing homotopy. Points in $\overline{C}$ are specified by points in $C$ with a memory of how we got there from the basepoint, so if we got to $x\in C$ via a path $\gamma$ in $C$ and $U$ is any basic nbhd of $x\in C$, then the lift $\overline{U}$ of that nbhd is comprised of points $\overline{u}$, and to specify these $\overline{u}$ we must say which points of $C$ they are (done: they lie above $U$) and how we got to them. We got to these points in $\overline{U}$ by first travelling along $\gamma$ from the basepoint to $x$ and then wiggled around within $U$ itself.

As for your other question, try lifting the paths. Say that $D\to C$ is a covering, where $D$ is a familiar space you know well, and in particular you know $D$ is simply connected. Our construction of $\overline{C}$ is comprised of paths emanating from (say) $x\in C$. To see what the corresponding point of $D$ is, just lift the path from $C$ to $D$ and look at its endpoint! This is the isomorphism.

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