# The universal cover of a path-connected, locally path-connected space $X$ covers any other covering space

I'm currently reading Hatcher's Algebraic topology book. In page 68 he says:

A consequence of the lifting criterion is that a simply-connected covering space of a path-connected, locally path-connected space $X$ is a covering space of every other path-connected covering space of $X$ A simply-connected covering space of $X$ is therefore called a universal cover. It is unique up to isomorphism, so one is justified in calling it the universal covering.

Let's precise that a little bit. Let $X$ be a path-connected, locally path-connected space and $x_0\in X$. Let $p_1:(\tilde{X}_1,\tilde{x}_1)\to (X,x_0)$ be a simply-connected covering space and $p_2:(\tilde{X}_2,\tilde{x}_2)\to (X,x_0)$ be any other path-connected covering space. Using the fact that $\tilde{X}_1$ is simply connected and the lifting criterion one can find a map $\tilde{p}_1:(\tilde{X}_1,\tilde{x}_1)\to(\tilde{X}_2,\tilde{x}_2)$ such that $p_2\tilde{p}_1=p_1$ and $\tilde{p}_1$ must be the desired covering map. Note that this is where one uses the hypothesis $X$ path connected and locally path-connected and $\tilde X_1$ and $\tilde X_2$ both path-connected

However I'm having some trouble proving it. Let $x\in \tilde{X}_2$ then $p_2(x)\in X$ and there is a nbd $U$ of $p_2(x)$ such that $p^{-1}_1(U)=\cup_i V_i$ where the $V_i$ are open and disjoint subsets of $\tilde{X}_1$ and $p|V_i:V_i\to U$ is an homeomorphism. The desired nbd of $x$ that makes $\tilde{p}_1$ into a covering map should be $p^{-1}_2(U)$. We have $\tilde{p}_1^{-1}(p^{-1}_2(U))=p^{-1}_1(U)$ and then one expects that $\tilde{p}_1|V_i:V_i\to p_2^{-1}(U)$ is an homeomorphism to make this work. But I can't prove that last statement, I'm not even sure it's surjective (Hatcher doesn't assume covering spaces to be surjective).

Is this the right approach? Or maybe there is a simpler way to do it.

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:

1. $q$ 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 $p_1(x_0) = x_0'$ and $q(x)$. 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.

2. $q$ 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 $q^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ 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.