----
> [!theorem] Theorem. ([[constructing universal covers]])
> Let $(X, x_{0})$ be a [[pointed set|based]] [[topological space]]. Does $X$ admit a [[universal cover]]?
>
*A necessary condition.* If $X$ admits a [[universal cover]] $p: \widetilde{X} \to X$, then it is [[semilocally simply connected]]. Also, there is a [[bijection]] $
\widetilde{X} \leftrightarrow \left\{\begin{align}&\text{path homotopy classes of paths in }X \\ &\text{starting at } x_{0} \text{ and ending anywhere} \end{align} \right\}.$
>
*A (more stringent) sufficient condition.* If $X$ is [[path-connected]] and [[locally connected, locally path-connected|locally path-connected]] in addition to being [[semilocally simply connected]], then it admits a [[universal cover]] $\widetilde{X}$ *defined* as the 'path space' $\widetilde{X}:= \left\{\begin{align}&\text{path homotopy classes of paths in }X \\ &\text{starting at } x_{0} \text{ and ending anywhere} \end{align} \right\}$
when this set is endowed with a suitable [[topological space|topology]], with the [[covering space|covering map]] $p: \widetilde{X} \to X$ given by $p([\gamma]):=\gamma(1)$.
^theorem
> [!proof]+ Proof. ([[constructing universal covers]])
> *A necessary condition.* Let $U \ni x_{0}$ be a [[neighborhood]] of $x_{0}$ [[evenly covered]] by $p$, so $p ^{-1}(U)= \coprod_{\beta}V_{\beta}$. Fix some $\beta$. Then for any [[parameterized curve|loop]] $\gamma:I \to \textcolor{Skyblue}{U}$ based at $x_{0}$ there is a [[the homotopy lifting lemma|unique lift to a loop]] $\tilde{\gamma}$ *in $V_{\beta}$* starting at $\tilde{x}_{0}=V_{\beta} \cap p ^{-1}(x_{0})$ [^1]. Thus the [[homomorphism of fundamental groups induced by a continuous map|pushforward]] [[group homomorphism|homomorphism]] of [[fundamental group|fundamental groups]]$ p_{*}: \pi_{1}(\widetilde{X}, x_{0}) \to \pi_{1}(X,x_{0})$
maps $[\tilde{\gamma}]$ to $[p \circ \tilde{\gamma}]=[\gamma]$. But since $\widetilde{X}$ is [[simply connected]], $\tilde{\gamma}$ is [[homotopy|nulhomotopic]]. So $\gamma$ is too— hence $X$ is [[semilocally simply connected]].
>
Next for deriving the [[bijection]]: since $\widetilde{X}$ is [[simply connected]], for any point $y \in \widetilde{X}$ there is [[in a simply connected space, any two paths having the same initial and final points are path homotopic|exactly one homotopy class]] $[\alpha]$ of [[parameterized curve]] from $\tilde{x}_{0}$ to $y$, and $[p \circ \alpha]$ gives a preferred [[homotopy|homotopy class]] of [[parameterized curve]] from $x_{0}$ to $p(y)$. Thus we can recover $y$ as the endpoint of the [[lifting|lift]] of $[p \circ \alpha]$ starting at $\tilde{x}_{0}$. This defines a [[bijection]] $
\widetilde{X} \leftrightarrow \left\{\begin{align}&\text{homotopy classes of paths in }X \\ &\text{starting at } x_{0} \text{ and ending anywhere} \end{align} \right\}.$
![[CleanShot 2024-06-10 at
[email protected]|500]]
>
>
*A sufficient condition.* Next we wish to *define* a [[topological space]] $\widetilde{X}$ as $\widetilde{X}:= \left\{\begin{align}&\text{homotopy classes of paths in }X \\ &\text{starting at } x_{0} \text{ and ending anywhere} \end{align} \right\}$
and [[covering space|covering map]] $p: \widetilde{X} \to X$ given by $p([\gamma]):=\gamma(1)$. We must (a) construct a [[topological space|topology]] on $\widetilde{X}$, (b) show $p$ is [[continuous]] in this [[topological space|topology]], (c) show that $p$ is a [[covering space|covering map]] in this [[topological space|topology]], and finally (d) show that $\widetilde{X}$ is [[simply connected]].
>
**(a).** Consider the collection of subsets [^3] $\mathscr{U}:= \left\{ U \subset X : \begin{align} &
U\text{ is open in } X, \text{ is path-connected, and} \\ &\pi_{1}(U,x) \xrightarrow{\iota_{*}} \pi_{1}(X,x) \text{ is trivial for all } x \in U
\end{align} \right\}.$
Here, $\pi_{1}(U,x)$ and $\pi_{1}(X,x)$ are the [[fundamental group|fundamental groups]] of $U$ and $X$ based at $x$ respectively, and $\iota_{*}$ denotes the [[inclusion map|inclusion map's]] [[homomorphism of fundamental groups induced by a continuous map|pushforward homomorphism]].
>
*Claim:* $\mathscr{U}$ is a [[basis for a topology|basis]] [[topology generated by a basis|generating]] the [[topological space|topology]] on $X$. To show this, we'll invoke: [[condition for obtaining a basis from a topology]]. Let $V$ be an open set in $X$ containing $x \in X$.
>
>1. As $X$ is [[semilocally simply connected]], for any $x \in X$ there is a [[neighborhood]] [^2] $U' \ni x$ for which $\pi_{1}(U', x) \xrightarrow{\iota'_{*}}\pi_{1}(X,x)$ is [[group homomorphism|trivial]].
>2. Because $X$ is [[locally connected, locally path-connected|locally path-connected]], there is a neighborhood $x \in U \subset V \cap U'$ which is [[path-connected]].
>3. Moreover, $\pi_{1}(U,x) \xrightarrow{\iota_{*}} \pi_{1}(X,x)$ is [[group homomorphism|trivial]] because it factors through the trivial map $\pi_{1}(U', x) \xrightarrow{\iota'_{*}}\pi_{1}(X,x)$.
>4. It remains to be shown that the inclusion pushforward is trivial *for all* elements of $U$, and not just $x$. Take $y \in U$ and a [[parameterized curve]] $u:I \to U$ from $x$ to $y$. Then we have a commutative square
> ```tikz
> \usepackage{tikz-cd}
> \usepackage{amsmath}
> \begin{document}
> % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB120sB9ARgAUAVVIBPAJQgAvqXSZc+Qij7kqtRizadu-AQA0xkmXOx4CRFXzX1mrRBy69BBgB5HZIDKcVEyV6jaa9tpOwqRu0mowUADm8ESgAGYAThAAtkhkIDgQSADMARp2IEw8nADE0h4p6Ugq2bmIAEyFtmylFVVJqRmIWTl1xiA1vQUNSC3qbcHsODAuOMA4yVg0WIxSkVJAA
> \begin{tikzcd}
> {\pi_1(U,y)} \arrow[d, "u_\#"] \arrow[r] & {\pi_1(X,y)} \arrow[d, "u_\#"] \\
> {\pi_1(U,x)} \arrow[r, "\text{trivial}"] & {\pi_1(X,x)}
> \end{tikzcd}
> \end{document}
> ```
>
where the vertical maps are the [[fundamental group]] [[group isomorphism|isomorphisms]] outlined in [[change of basepoint isomorphism|this result]], and the bottom horizontal map (inclusion pushforward for $(U,x)$ to $(X,x)$) is [[group homomorphism|trivial]], hence the top horizontal map (inclusion pushforward for $(U,y)$ to $(X,y)$ must be trivial also. This completes the proof of *claim*.
>
$\mathscr{U}$ is useful because, if $x,y \in U \subset \mathscr{U}$ then there exists [[parameterized curve]] from $x$ to $y$ in $U$, and all such [[parameterized curve|paths]] are [[path homotopy|path homotopic]] *in $X$*. [^4] Now, for a given $[\alpha] \in \widetilde{X}$ and $U \in \mathscr{U}$ such that $\alpha(1) \in U$, define $(\alpha, U):= \left\{ [\beta] \in \widetilde{X} : \begin{align}
& [\beta]=[\alpha * \alpha'] \text{ for some path } \\
& \alpha' \text{ in } U \text{ starting at } \alpha(1)
\end{align} \right\}.$
![[CleanShot 2024-06-10 at
[email protected] |500]]
>
We claim that such sets form a [[basis for a topology|basis]] for a [[topological space|topology]] on $\widetilde{X}$.
>1. (Covering) This obviously holds because we define one set per element of $\widetilde{X}$
>2. (Nestling) We must show $[\beta] \in (\alpha_{0}, U_{0}) \cap (\alpha_{1}, U_{1})$ has a [[neighborhood]] of the required form. Let $W \subset U_{0} \cap U_{1}$ be a [[neighborhood]] of $\beta(1) \in U_{0} \cap U_{1}$ in the collection $\mathscr{U}$, which exists because $\mathscr{U}$ is a [[basis for a topology|basis for the topology]] on $X$. ![[CleanShot 2024-06-10 at
[email protected]]]
It will be enough to show that the [[neighborhood]] $(\beta, W)$ of $\beta$ lives in the intersection $(\alpha_{0}, U_{0}) \cap (\alpha_{1}, U_{1})$, i.e., that any [[equivalence class|class]] $[\gamma]$ of [[parameterized curve|paths]] from $x_{0}$ to a point in $W$ that is obtained by 'lengthening $\beta