I'd like to understand the proof of the theorem of correspondace between covering maps and subgroups of fundamental group.
$\textbf{Theorem :}$ Let $X$ connected and semi-locally simply connected. Then $\forall \hspace{0.1cm} H < \pi_{1}(X,x) \hspace{0.1cm} \exists \hspace{0.1cm} p : E \longmapsto X$ covering such that $p_{*}(\pi_{1}(E,\tilde{x})) = H < \pi_{1}(X,x)$, where $\tilde{x} \in p^{-1}(x)$, the covering is unique up to isomorphism.
$\textbf{First part of the proof :}$ Since $X$ is connected and semi-locally simply connected we know the existance of $q : \tilde{X} \longmapsto X$ universal covering of $X$. Since $Aut(\tilde{x},q) \sim \pi_{1}(X,x)$ I denote with $H < Aut(\tilde{x},q)$, the image of $H$ under that isomorphism.
Denoted with $E = \tilde{X}/H$ we have the following $\tilde{X} \overset{\pi}{\longmapsto} \tilde{X}/H = E \overset{p}{\longmapsto} \tilde{X}/\pi_{1}(X,x) = X$
With $p \circ \pi = q$. I'd like to prove that $p$ is well defined and that is a covering map.
I tried to take a evenly covered neighboorhood $U$ of $x \in X$. Since $q$ is a covering map, $q^{-1}(U)$ is a disjoint union of open of $\tilde{X}$, then $\pi$ of this disjoint union should an open in $E$, since the quotient maps induced by groups are always open maps, but I don't how to proceed from here.
Any help would be appreciated.
It is a more general result that if $X \overset{p}{\longmapsto} Y \overset{r}{\longmapsto Z}$ and $q : X \longmapsto Z$ are continuos function such that $r \circ p = q$, with $p,q$ covering maps, $r$ is also a covering map.
In the Theorem cited such $r$ denoted with $p$, certanly exists continuos and gives the commuting diagramm by the fundamental property of identifications, since $q$ is costant on the fibers of $\pi$.