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?
First, assume that $J$ and $G$ are connected. In this case, a lift exists iff $\pi_1(\iota)=0$. This is a sufficient condition due to the homotopy lifting property of covering spaces: $\iota$ has a unique continuous lift $\widetilde{\iota}$ satisfying $\widetilde{\iota}(e)=\widetilde{e}$, and this lift is always a group homomorphism. It is necessary due to fundamental groups: such a $\widetilde{\iota}$ must satisfy $\pi_1(c)\circ\pi_1(\widetilde{\iota})=\pi_1(\iota)$, and sice $\pi_1(\widetilde{G})=0$, this is only possible if $\pi_1(\iota)=0$.
For the general case, let $c:\widetilde{G}\to G$ be the universal covering map, and note that $\widetilde{J}:=c^{-1}(J)$ is a Lie subgroup of $\widetilde{G}$, and the restriction of the universal covering $\hat{c}:\widetilde{J}\to J$ is a covering homomorphism. A necessary condition is that $\iota$ has a lift when restricted to identity compinents; this is the case iff $\pi_1(\iota)=0$ by the argument from the connected case. Another neccesary condition is that $\pi_0(\hat{c})$ has a right inverse. I claim that, when combined, these two conditions are also sufficient. The basic argument is that $\hat{c}$ restricts to a diffeomorphism on each connected component of $\widetilde{J}$, so given a map $\psi:\pi_0(J)\to\pi_0(\widetilde{J})$ with $\pi_0(\hat{c})\circ\psi=\operatorname{id}_{\pi_0(J)}$, we can invert these diffeomorphisms to obtain a lift $\widetilde{\iota}:J\to\widetilde{J}$ with $\pi_0(\widetilde{\iota})=\psi$.