This is a re-post of a question I asked a month ago on MSE, but unfortunately didn't receive any answers. I'm hoping someone could help me with it. Here it goes:

Recently I've been self-studying the theory of covering spaces from "Introduction to Topological Manifolds", by John M. Lee. At the end of Chapter 11, there is an explicit construction of the universal covering space for a connected, locally path connected and semilocally simply connected topological space $X$.

Very briefly, the idea is as follows: take $x_{0}\in X$, define $P(X,x_{0})$ to be the set of all paths in $X$ which start at $x_{0}$, and let $\tilde{X}$ be the quotient set of $P(X,x_{0})$ by the relation: $\alpha \sim \beta$ if and only if $\alpha(1)=\beta(1)$ and $\alpha\simeq \beta$ are path-homotopic. We define $q\colon \tilde{X}\to X$ to be the map $q([\alpha]):=\alpha(1)$. We later define a topology on $\tilde{X}$ by means of a basis, which turns $\tilde{X}$ into a simply connected, locally path connected space and $q$ into a covering map.

However, some authors suggest that an alternative way to construct $\tilde{X}$ is to give $P(X,x_{0})$ the compact-open topology (that is, the subspace topology it inherits as a subset of $\mathcal{C}([0,1],X)$), and $\tilde{X}$ the quotient topology induced by the canonical map $\pi \colon P(X,x_{0})\to \tilde{X}$. Nevertheless, I haven't been able to find a clear proof that $q\colon \tilde{X}\to X$ is the universal cover of $X$.

What I'm trying to do is imitate the proof in Lee's book, but using this alternative topology: most of the details of the proof are rather set-theoretic, so I've managed to reduce the problem of rewriting the proof to just having to prove these 4 facts:

- $\tilde{X}$ is a path connected topological space.
- $q\colon \tilde{X}\to X$ is continuous.
- $q\colon \tilde{X}\to X$ is open.
- For every $\alpha \in P(X,x_{0})$ and every open set $U\subseteq X$ such that $\alpha(1)\in U$, the subset

$$ [\alpha \cdot U]=\{ [\alpha \cdot \beta]\colon \beta \text{ is a path in } U \text{ such that } \beta(0)=\alpha(1) \} $$

is open in $\tilde{X}$.

Here are my ideas so far:

$P(X,x_{0})$ is path connected (and hence $\tilde{X}$ is too): let $\tilde{x}_{0}=[c_{x_{0}}]$ be the class of the constant path at $x_{0}$ and $\alpha\in P(X,x_{0})$ be arbitrary. Define $F\colon [0,1]\times [0,1]\to X$ by $F(t,s)=\alpha(ts)$. This is a continuous map, so by properties of the CO-topology, the map $f\colon [0,1]\to \mathcal{C}([0,1],X)$ given by $f(t)(s)=F(t,s)$ is continuous. Since $f(0)(s)=x_{0}$, $f(1)(s)=\alpha(s)$ and $f(t)(0)=x_{0}$, $f$ is a path in $P(X,x_{0})$ from $c_{x_{0}}$ to $\alpha$. Therefore, $P(X,x_{0})$ is path-connected.

Let $\operatorname{ev}\colon P(X,x_{0})\to X$ be the map defined by $\operatorname{ev}(\alpha)=\alpha(1)$. This map is continuous (again, this is a general property of the CO-topology) and passes continuously to the quotient, inducing precisely the map $q$, so $q$ is continuous.

As of openness of $q$, I'm not sure how to proceed: what I'm sure of is that, if $\operatorname{ev}$ is an open map (which I think it is), then $q$ is too. I believe that it should be possible to prove that basic open subsets of $P(X,x_{0})$ are mapped into open sets of $X$.

Let $[\alpha \cdot U]$ defined as above, we need to see that $\pi^{-1}([\alpha \cdot U])$ is open in $P(X,x_{0})$. Take any $\gamma \in \pi^{-1}([\alpha \cdot U])$, so that $\gamma \simeq \alpha \cdot \beta$, where $\beta$ is a path in $U$ starting at $\alpha(1)$. Let $V$ be the path component of $U$ which contains $\gamma(1)$ (and therefore, it contains $\alpha(1)$). Then $\gamma\in [\{1\},V]$, where $[\{1\},V]$ is the basic open subset of $P(X,x_{0})$ given by all paths which end at some point of $V$. From here I don't know how to continue: I think this open set is contained in $\pi^{-1}([\alpha \cdot U])$, but I haven't got an idea to prove it (certainly, since we know "a posteriori" that $\tilde{X}$ is simply connected, every element of $[\{1\},V]$ is homotopic to $\alpha\cdot \eta$, where $\eta$ is a path in $V$ from $\alpha(1)$ to the endpoint of such element).

So far, are my ideas on the right track? Can someone help me especially with the last two statements?

Thank you in advance!

**SOME INSIGHTS I MADE AFTER ASKING THE QUESTION** (these appear as edits on the MSE post)

First, what I thought about the third statement:

- I think this could be a valid proof for openness of $\operatorname{ev}$: let $A=[K_{1},U_{1}]\cap \dots \cap[K_{n},U_{n}]$ be a basic open set of $\tilde{X}$, we need to prove that $\operatorname{ev}(A)$ is an open subset of $X$. Take $x\in \operatorname{ev}(A)$, so that there exists some $\alpha \in A$ such that $x=\alpha(1)$. Let $B\subseteq X$ be the intersection of all $U_{i}$'s such that $1\in K_{i}$ (if none of the $K_{i}$ contains $1$, then $B:=X$), so $x\in B$, and let $V$ be the path component of $B$ that contains $x$ (so $V$ is open in $X$). I claim that $V\subseteq \operatorname{ev}(A)$.

To see this, let $y\in V$. For every $j$ such that $1\notin K_{j}$, let $a_{j}=\max \{t\colon t\in K_{j}\}<1$, and let $a=\max a_{j}$. By continuity, one can choose some $a<b<1$ such that $\alpha([b,1])\subseteq V$. Take any path $\gamma$ in $V$ from $\alpha(b)$ to $y$ and define

$$ \beta(t)= \begin{cases} \alpha(t), & 0\leq t \leq b \\ \gamma\left( \dfrac{t-b}{1-b} \right), & b\leq t \leq 1. \end{cases} $$

Then $\beta$ is continuous, and since $\gamma([0,1])\subseteq B$ and $\beta = \alpha$ up to $t=b$, we have that $\gamma \in A$ and $y=\operatorname{ev}(\beta)\in \operatorname{ev}(A)$, which proves that $V\subseteq \operatorname{ev}(A)$. Hence, $\operatorname{ev}$ is open, and so is $q$.

Secondly, what I thought about the last statement:

The main idea is to prove that "translations by a path" reduce the study of such sets to just seeing that $[c_{x_{0}}\cdot U]$ is open in $\tilde{X}$.

For any $x\in X$, let $P(X,x)$ be the space of all paths in $X$ starting at $x$ (equipped with the compact-open topology), and let $Q(X,x)$ be the quotient of $P(X,x)$ by path homotopy, and denote by $\phi_{x}\colon P(X,x)\to Q(X,x)$ the natural map. Notice that $\tilde{X}=Q(X,x_{0})$.

Given $x,y\in X$, and any path $\gamma$ from $x$ to $y$, one can define a mapping $\Lambda\colon P(X,y)\to P(X,x)$ given by $\Lambda_{\gamma}(\alpha)=\gamma\cdot \alpha$. We prove that $\Lambda_{\gamma}$ is a continuous map.

Let $\alpha \in P(X,y)$, and suppose $[K,U]$ is a subbasic set of $P(X,x)$ such that $\Lambda_{\gamma}(\alpha)=\gamma \cdot \alpha \in [K,U]$. Divide $K$ into $K_{1}=K\cap [0,1/2]$ and $K_{2}=K\cap [1/2,1]$. Given any $t\in K_{2}$, we have $\gamma \cdot \alpha(t)=\alpha(2 t -1)\in U$, so $K_{2}'=\{ 2s-1 \colon s\in K_{2} \}$ is a compact subset of $[0,1]$ such that $\alpha \in [K_{2}',U]$. Now, given any $\beta\in [K_{2}',U]\subseteq P(X,y)$, it is clear that $\gamma \cdot \beta (K_{1})=\gamma(K_{1})=\gamma \cdot \alpha(K_{1})\subseteq U$, and $\gamma \cdot \beta (K_{2})=\beta(K_{2}')\subseteq U$, so $\Lambda_{\gamma}(\beta)\in [K,U]$. Therefore, $\Lambda_{\gamma}$ is continuous.

Now, since $\Lambda_{\gamma}$ is compatible with both $\phi_{x}$ and $\phi_{y}$, it passes to the quotient into a continuous map $\Omega_{\gamma}\colon Q(X,y)\to Q(X,x)$, whose inverse is $\Omega_{\overline{\gamma}}$, so $\Omega_{\gamma}$ is a homeomorphism.

Returning to the question at hand: consider a path $\gamma\in P(X,x_{0})$, and a relatively simply connected open set $U\subseteq X$ containing $\gamma(1)=x$. Then:

$$ [\gamma \cdot U]=\{[\gamma\cdot \alpha]\colon \alpha \text{ is a path in } U \text{ such that } \alpha(0)=\gamma(1)=x\} \\ =\{\Omega_{\gamma}([\alpha])\colon \alpha \text{ is a path in } U \text{ such that } \alpha(0)=\gamma(1)=x\}=\Omega_{\gamma}(\phi_{x}([[0,1],U])), $$

So it suffices to prove that $\phi_{x}([[0,1],U])$ is open in $Q(X,x)$, that is, $\phi^{-1}_{x}(\phi_{x}([[0,1],U]))$ is open in $P(X,x)$.I have yet to prove this last bit, but I think it should be possible to conclude from here.