# Necessary and sufficient conditions for $X$ to have a universal covering space

We know that being path connected, locally path connected, and semilocally simply connected are sufficient conditions for a space $$X$$ to have a universal cover. But are those conditions necessary? I would assume not.

Is there a succinct way to describe some possible set of both necessary and sufficient conditions for $$X$$ to have a universal cover?

Related—if my hypothesis is correct, does there exist a simple example of a space that is not path connected, locally path connected, and semilocally simply connected that does have a universal cover?

To clarify (based on comments): I mean "universal cover" to mean a simply connected covering space where "covering space" is as defined in Hatcher in the beginning of section 1.3 (pg. 56).

• You'll have to clarify what you mean by 'universal cover'. For well-behaved spaces, one could postulate several different definitions, which are all equivalent thanks to the theory of covering spaces. Outside the realm of such spaces, this theory could entirely break down. Commented Dec 10, 2021 at 7:40
• Note that every simply connected space is a universal cover of itself. Regardless of whether it is locally path connected or semilocally simply connected (although it is always path connected). Commented Dec 10, 2021 at 7:42
• @JeroenvanderMeer I apologize, I was not aware there was more than one definition. I meant "universal cover" to mean a simply connected covering space where "covering space" is as defined in Hatcher the beginning of 1.3 (pg. 56). Commented Dec 10, 2021 at 7:42
• @freakish Duh, thanks for that. Exactly what I was looking for in regards to the second part of my question. Commented Dec 10, 2021 at 7:44
• @isaiahtx7 also, there's a theorem in Spanier's Algebraic Topology: Corollary 14, pg. 83: A connected, locally path-connected space $X$ has a simply connected covering space if and only if it is semi-locally simply connected. Commented Dec 10, 2021 at 7:48

There is not much hope for necessary and sufficient conditions for general spaces.

Here is a first necessary condition:

1. If $$X$$ has a simply connected covering space, then $$X$$ must be path connected (because simply connected implies path connected and $$X$$ is the continuous image of a path connected spaces).

So we might as well focus on path-connected spaces. Now if $$X$$ is locally path-connected, then "semilocally simply connected" is a necessary and sufficient condition.

Theorem 1: Let $$X$$ be path-connected and locally path connected. Then $$X$$ admits a simply connected covering space if and only if $$X$$ is semilocally simply connected.

If $$X$$ is NOT locally path connected, then you quickly run into trouble. Here is a second necessary condition that holds in general:

1. If $$X$$ has a simply connected covering space, then $$X$$ must be semilocally simply connected in the unbased sense that for every $$x\in X$$ there is an open set $$U$$ of $$x$$ such that for all $$y\in U$$, the homomorphism $$\pi_1(U,y)\to \pi_1(X,y)$$ induced by inclusion is trivial. In other words, every loop in every path component of $$U$$ contracts in $$X$$.

But having this property will still not guarantee having a universal covering space. For instance, the space $$W$$ illustrated below is semilocally simply connected in the unbased sense but it does not have a simply connected covering space.

So from here, you have to decide what you're interested in. If you're only interested in the fundamental group, then there is a hack. The locally path-connected coreflection of a space $$X$$ is the space $$lpc(X)$$ with the same underlying set but with topology generated by all path components of all open sets in $$X$$. The result $$lpc(X)$$ is locally path connected and the continuous identity function $$lpc(X)\to X$$ is a homeomorphism if and only if $$X$$ is locally path connected. Moreover, $$lpc(X)\to X$$ induces an isomorphism on $$\pi_1$$ (and on all homotopy, homology, and cohomology groups). For example, $$lpc(W)$$ is the same set but where the circles no longer approach the disk.

Since $$lpc(X)$$ is a refined version of $$X$$ with the same fundamental group, if you want to know about the subgroup lattice of $$\pi_1(X,x)$$, you might as well just use $$\pi_1(lpc(X),x)$$. Now $$X$$ won't have a universal cover just because $$lpc(X)$$ does. And this is the point. You replace $$X$$ with something that suits your purposes better. It's not too hard to check that if $$X$$ is semilocally simply connected in the unbased sense, then $$lpc(X)$$ is semilocally simply connected in the usual sense.

Combining the Theorem and the fact that $$lpc(X)$$ is locally path connected you get:

Theorem 2: For any path-connected space $$X$$, the following are equivalent:

1. $$X$$ is semilocally simply connected in the unbased sense,
2. $$lpc(X)$$ is semilocally simply connected,
3. $$lpc(X)$$ has a simply connected covering space

When $$lpc(X)$$ does have a universal covering $$p:E\to lpc(X)$$, identifying $$\pi_1(X,x)=\pi_1(lpc(X),x)$$ means that $$\pi_1(X,x)$$ acts freely and transitively on the fibers of $$p$$. Hence $$E$$ still provides a space to understand $$\pi_1(X,x)$$.

There is a lot of fascinating mathematics out there surrounding these ideas. It is true that you can classify which subgroups of $$\pi_1(X,x)$$ correspond to coverings using topologies on $$\pi_1$$. For instance, in this post. Lastly, I'll share that although many spaces like the earring space, Menger Cube, or the space pictured below don't have simply connected covering spaces, they do have simply connected generalized covering spaces. Generalized universal covering maps have all the same lifting properties of covering maps but they may not be local homeomorphisms. If you think of covering maps as "unwinding" a space according to homotopy classes of loops, then generalized covering maps do the same thing but allow for arbitrarily small structures to be unwound.

For well-behaved spaces, one could postulate a few different definitions of 'universal covering', which are equivalent by virtue of the theory of covering spaces. Notably, you can conceive of a universal covering as a simply connected covering space, but 'universal' could also mean the 'greatest' covering space, in the sense that it itself is not covered by anything bigger.

Outside the realm of well-behaved spaces, the theory of covering spaces breaks down, and the two notions become distinct. Consider for instance the Warsaw circle: this space is simply connected, and therefore the identity map forms a universal covering in the first sense, but it is not universal in the sense of being the largest covering, as there are nontrivial coverings as well: just connect up multiple Warsaw circles into a loop.

Another example would be the Hawaiian earring, which fails to be semilocally simply connected and because of that cannot possess a simply connected covering. However, there are nontrivial coverings of this space. I'm not sure if there is a 'universal' one in the sense of being the biggest --- maybe string together infinitely many Hawaiian earrings in some way or other? (Edit: Exercise 1.6 of Hatcher's book also looks at coverings of Hawaiian earrings.)

• A concrete example of a space that is both path connected and locally path connected, but does not admit neither simply connected covering nor universal covering is an infinite product of copies of $S^1$ (see here: math.stackexchange.com/questions/2888873/… ). Commented Dec 10, 2021 at 7:51
• @freakish Thanks, that's a clean example! At the opposite end of the spectrum, it may enjoy you to know that there rae are also path connected and locally path connected spaces with crazy $\pi_1$'s which do not admit any covering at all. A cool example is outlined here: mathoverflow.net/a/111388. Commented Dec 10, 2021 at 8:01
• That post also contains some interesting remarks about how $\pi_1$ can still be used to detect coverings provided that one topologises the group. (It depends on what exactly counts as a covering though. The Warsaw circle could be said to have coverings despite $\pi_1$ being $0$. These coverings, however, are not path connected.) Commented Dec 10, 2021 at 8:09