Lie algebras of lie subgroups

Question

Let $\textbf{p}: \widetilde{G} \to G$ be a covering homomorphism of Lie groups $\widetilde{G}$ and $G$ and $\widetilde{H}$, $p(\widetilde{H})=H$ -- subgroups of $\widetilde{G}$ and $G$ repsectively. It is clear for me that $\textbf{d}_{e}p$ provides the isomorphism of Lie algebras of $\widetilde{G}$ and $G$ just because the kernel of $\textbf{p}$ is dicrete and the fact that any path in $G$ can be lifted.

Then, is it true that $\textbf{Lie}(H)=\textbf{d}_{e}p(\widetilde{H})$? In other words, does it follow that $\textbf{Lie}(H) \subset \textbf{d}_{e}p(\widetilde{H})$?

P.S. One can assume $\widetilde{H}$ to be normal, thus its and $H$ 's tangent spaces are really lie algebras.

Many thanks in advance!

Answer 1

If the kernel of $p$ is discrete, the kernel of the restriction of $p$ to $\widetilde H$ is also discrete. To see this let $g\in Kerp_{\mid \widetilde H}$ since $g\in kerp$, there exists an open subset $U$ of $\widetilde G$ such that $U\cap ker p=\{g\}$. $U\cap \widetilde H$ is an open subset of $\widetilde H$, for the induced topology and $U\cap\widetilde H\subset U\cap \widetilde G=\{g\}$, so if $\widetilde H$ and $H$ are closed subgroups, their Lie algebra are isomorphic.

Answer 2

I will assume that $H$ is a closed subgroup of $G$. Then, given $X\in\operatorname{Lie}(G)$, $X\in\operatorname{Lie}(H)$ if and only if$$(\forall t\in\mathbb{R}):\exp(tX)\in H.$$Now, the goal is to prove that, if $X\in\operatorname{Lie}(H)$, then $X\in D_ep\left(\widetilde X\right)$, for some $\widetilde X\in\operatorname{Lie}\left(\widetilde H\right)$. As you know, $D_ep$ is an isomorphism. So, lt $\widetilde X\in\operatorname{Lie}\left(\widetilde G\right)$ be such that $d_ep\left(\widetilde X\right)=X$. Then$$(\forall t\in\mathbb{R}):\exp\left(t\widetilde X\right)\in\widetilde H,$$since\begin{align}p\left(\exp\left(t\widetilde X\right)\right)&=\exp\left(D_ep\left(t\widetilde X\right)\right)\\&=\exp\left(tD_ep\left(\widetilde X\right)\right)\\&=\exp(tX)\\&\in H.\end{align}