Let $\pi:\widetilde{G} \rightarrow G$ be a covering homomorphism of Lie groups and let $\widetilde{H} \leq \widetilde{G}$ and $H \leq G$ be closed subgroups. If $\pi(\widetilde{H}) \leq H$ then the map
$\bar{\pi} :\widetilde{G}/\widetilde{H} \rightarrow G/H$ defined by $\bar{\pi}(\widetilde{g}\widetilde{H})=\pi(\widetilde{g})H$
is well defined and is a covering map.
Here is my attempt:
A standard argument shows that the map above is well defined. We can think of $G/H$ as being the orbit space of the action of $H$ on $G$ given by $h \cdot g = gh^{-1}$ so the quotient map $p:G \rightarrow G/H$ is an open map. Therefore every open set in $G/H$ is of the form $p(U)$ for some open set $U \subset G$. Since $p(U)=\{uH \in G/H : u \in U \}$ with a slight abuse of notation I'll denote this by p(U)=U/H. Therefore the open sets in $G/H$ are of the form $U/H$ for $U \subset G$ open.
I wish to find an evenly covered neighborhood of an arbitrary point $gH \in G/H$.
Since $g \in G$ and $\pi$ is a covering map, there exists an evenly covered neighborhood $V \subset G$ containing $g$. I believe that the open set $V/H \subset G/H$ will serve as an evenly covered neighborhood for $gH$. I am struggling to show this is true, if it is even true.