If $X$ is the Hawaiian earring, assume universal cover $\widetilde{X}$ exists. Let $p$ be the "interesting point" of $X$ (namely, the origin). $q$ be the lift of $p$ in $\widetilde{X}$. If $U$ is an evenly covered neighborhood of $p$, then it lifts to a neighborhood $V$ of $q$ such that the projection $U \to V$ by restricting the covering map is a homeomorphism.
$i : U \hookrightarrow X$ be the inclusion map. We can look at the composition $\pi_1(V) \stackrel{\cong}{\to} \pi_1(U) \stackrel{\pi_1(i)}{\to} \pi_1(X)$ where first map is the induced map of projection. This map is the same as $\pi_1(V) \stackrel{\pi_1(j)}{\to} \pi_1(\widetilde{X}) \to \pi_1(X)$ where $j$ is the inclusion $V \hookrightarrow \widetilde{X}$.
But as $\widetilde{X}$ is simply connected, the composition is the zero map, hence so is $\pi_1(i)$. Thus, $U$ is a neighborhood of $p$ which makes $X$ semilocally simply connected at $p$. Contradiction, as every neighborhood of $p$ contains a small circle, and a loop going around that small circle can never be nullhomotoped in $X$.
There is not much hope for necessary and sufficient conditions for general spaces.
Here is a first necessary condition:
- 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:
- 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:
- $X$ is semilocally simply connected in the unbased sense,
- $lpc(X)$ is semilocally simply connected,
- $lpc(X)$ 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.
Best Answer
Hint:
(If you want the directions to match up, you'll need to cross each set of crosslinks in the middle, but the directions of each circle is not topologically distinguishable anyway, and it would make the figure more complex ...).