Suppose the Lie group $G$ contains the Lie group $J$ as a subgroup, so
$$
G \supset J.
$$


If $G$ has a nontrivial first homotopy group $\pi\_1(G) \neq 0$.


If $G$ has a universal cover $\widetilde{G}$, so $\pi\_1(\widetilde{G}) = 0$.



> 
> Question: What are the necessary and sufficient conditions to derive that
> $$
> \widetilde{G} \supset J \quad (?)
> $$
> is also true via a lifting?
> $$
> \begin{array}{ccc}
>  & & \widetilde{G}\\
>  &\nearrow & \downarrow\\
>  J & \longrightarrow & G
> \end{array}.
> $$
> 
> 
> 


My thought:


1. Sufficient condition may be $\pi\_1(J)=0$, due to the property $\pi\_1(\widetilde{G})=0$.
2. Necessary condition may be that
$\pi\_1(J) \to \pi\_1({G})$ is a trivial group homomorphism, which maps all $\pi\_1(J)$ to 0,
because this map
can be decomposed by $\pi\_1(J) \to \pi\_1(\widetilde{G})=0$ and $\pi\_1(\widetilde{G})=0 \to \pi\_1({G})$.
3. The above conditions are for universal covering $\widetilde{G}$, what would be the conditions for the (non-universal) covering space $\widetilde{G}$?



Necessary and sufficient conditions for the Lie group embedding $G \supset J$ can be lifted to $G$'s covering space
Lifting to covers is completely understood; the necessary and sufficient requirement is that the image of the fundamental group lies in the subgroup associated to the cover. So, for example, you are correct that to lift to the universal cover the map on fundamental groups must be trivial. For a reference, see chapter 1 of Hatcher proposition 1.33.


