Lie-group existence on universal covering manifold

Let $X$ be an n-dimensional smooth manifold with Lie group $G$ acting transitively on $X$, i.e. $X$ is a homogeneous space. Let $\tilde{X}$ be the associated universal covering space. To what extent is it known that there exists a Lie group $\tilde{G}$ on $\tilde{X}$ and a surjective group homomorphism $\psi:\tilde{G} \rightarrow G$ such that the covering map $\varphi$ is equivariant with respect to this homomorphism, i.e. that \begin{equation} \varphi(\tilde{g}(\tilde{x})) = \psi (\tilde{g}) \varphi (\tilde{x}), \end{equation} where $\tilde{x}\in \tilde{X}$?

• In case $G$ is connected: So there's always a universal covering group $\psi :\tilde{G}\rightarrow G$. Then we have a map $$\tilde{G}\times \tilde{X}\rightarrow X, (\tilde{g},\tilde{x})\mapsto \psi(\tilde{g})\phi(\tilde{x})$$ which lifts to a map $\tilde{G}\times \tilde{X} \rightarrow \tilde{X}$, which should give you the action you want. Feb 19 '16 at 0:22
• In case $G_0\subset G\rightarrow G/G_0$ is split, you can just multiply $\tilde{G_0}$ with $G/G_0$ To get $\tilde{G}$ ($G_0$ is the component of the identity). Feb 19 '16 at 0:31