Universal cover of homogeneous space $G/H$. Let $G$ be a compact connected Lie group with finite fundamental group and $H \leq G$ be a closed subgroup. Consider the homogeneous space $G/H$. Let $H'=\pi^{-1}(H)$ and $\bar{H}=(H')_0$ be the identity component of the closed subgroup $H'$. I am trying to prove that $\widetilde{G}/\bar{H}$ is the universal cover of $G/H$ where $\pi:\widetilde{G} \rightarrow G$ is the universal cover of $G$. 
I have already proved that $G/H$ is simply connected. Therefore I need only show that $\widetilde{G}/H$ is a covering space for $G/H$.
Let $K=H'/\bar{H}$ be the group of components of $H'$. I have shown that $K$ is finite and therefore it suffices to construct a free action of $K$ on $\widetilde{G}/\bar{H}$ such that the corresponding orbit space is $G/H$. This is where I am stuck. I am having trouble constructing any action of $K$ on $G/H$, let alone one that is free and gives the proper orbit space. 
 A: Here are two different approches.
First, one can define an action of $H'$ on $\tilde{G}/\overline{H}$ as follows.  Given $g\in \tilde{G}$ and $h\in H'$, set $h\ast g\overline{H} = (gh^{-1})\overline{H}$.
In general, this kind of "right multiplication" isn't well defined.  But I claim that is is in this case, owing to the fact that $\overline{H}$ is a normal subgroup of $H'$.  I'll leave it to you to show this, and also that this action induces a free action by $K$.
For an alternative solution, a closed subgroup $L\subset G$ always gives rise to a principal $L$-fiber bundle $L\rightarrow G\rightarrow G/L$.  If you have a favorite $L$-action on a manifold $M$, this allows you to create something called an "associated bundle" where the base is still $G/L$, but the fiber turns into $M$.  The total space is $(M\times G)/L$ with $L$ acting diagonally on both pieces.  This action is free (because it's free on the second factor) so $(M\times G)/L$ really is a manifold.  (I often denote this $M\times_L G$.)  The projection map $(M\times G)/L\rightarrow G/L$ maps $[m,g]$ to $[g]$
A particular instance of this occurs when one has a triple of subgroups:  $L_1\subseteq L_2\subseteq G$.  Then $L_2$ acts on the homogeneous space $L_2/L_1$, so one gets the fiber bundle $L_2/L_1\rightarrow (L_2/L_1 \times G)/L_2\rightarrow G/L_2$.
It's not too hard to show that there is a really nice diffeomorphism between the total space $(L_2/L_1\times G)/L_2$ and $G/L_1$.  Really nice means that in the associated bundle $L_2/L_1\rightarrow G/L_1\rightarrow G/L_2$, the fiber inclusion is literal inclusion, and the projection is just "quotient by even more".
Applying this with $(L_1,L_2,G) = (\overline{H},H',\tilde{G})$ gives the result you want.
