Let $\pi: \widetilde{G} \rightarrow G$ be the universal covering group of a Lie group $G$. Let $H$ be a closed subgroup of $G$ and let $\widetilde{H}$ be a closed subgroup of $\widetilde{G}$ that has the property that $\pi(\widetilde{H}) \subset H$.
Let $\widetilde{G}/\widetilde{H}$ and $G/H$ be the corresponding coset spaces. We think of the coset space $G/H$ as being the orbit space under the left action of $H$ on $G$ by $h \cdot g = gh^{-1}$.
Claim: The map $\bar{\pi}: \widetilde{G}/\widetilde{H} \rightarrow G/H$ defined by $\bar{\pi}(\widetilde{g}\widetilde{H})=\pi(g)H$ is well defined.
I would like verification that this proof is correct or incorrect.
I think this is well defined because if $\widetilde{g}_1\widetilde{H}=\widetilde{g}_2\widetilde{H}$ then $\widetilde{g}_2=\widetilde{g}_1\widetilde{h}^{-1}$ for some $\widetilde{h} \in \widetilde{H}$ by the definition of the action. Therefore $\bar{\pi}({\widetilde{g}}_2\widetilde{H})=\bar{\pi}(\widetilde{g}_1\widetilde{h}^{-1}\widetilde{H})=\pi(\widetilde{g}_1\widetilde{h}^{-1})\widetilde{H}=\pi(\widetilde{g}_1)\pi(\widetilde{h})^{-1}H=\pi(\widetilde{g}_1)H=\bar{\pi}({\widetilde{g}}_1\widetilde{H})$
where the second from last equality follows from the fact that $\pi(\widetilde{h})^{-1} \in H$ and the definition of the orbit space for the action of $H$ on $G$.
I feel this proof is not difficult but I've been told that this claim is not true because $\pi(\widetilde{H})$ is a subset of $H$ rather than equal to $H$. However, I've spent a long time thinking about it and still feel that it is, so I would like verification from others.