algebraic-topologygeneral-topology
Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $X$:
I don't understand why it follows that $\mathcal{U}$ is a basis for $X$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.
Lemma: Given the commutative diagram
$$\begin{array}{ccccccccc} \widetilde{X} & \\ \downarrow{\small{p}} & {\searrow}^{q} \\ X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$
where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.
Proof:
$q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ to $p_1(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.
$q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in ${p_{1}}^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$
If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.
There is not much hope for necessary and sufficient conditions for general spaces.
Here is a first necessary condition:
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:
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:
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
A basis for a topology $\mathcal{T}$ on a space $X$ is a subset $\mathcal{B} \subset \mathcal{T}$ such that for each $U \in \mathcal{T}$ and each $x \in U$ there exists $B \in \mathcal{B}$ such that $x \in B \subset U$.
A space $X$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $\mathcal{P}$ of path connected open sets forms a basis for $X$).
Hatcher shows that in a locally path connected semilocally simply-connected space $X$ the subset $\mathcal{U} \subset \mathcal{P}$ of all $U \in \mathcal{P}$ such that $\pi_1(U) \to \pi_1(X)$ is trivial also forms a basis for $X$. Note that this part does not use that $X$ is path connected.