Let $G$ be a Lie group, $\tilde{G}$ its simply connected universal cover and $\pi: \tilde{G} \longrightarrow G$ the associated covering map.
Then $\tilde{G}$ is also a Lie group and $\pi$ is a Lie group homomorphism.
Because $\pi$ is surjective, it is clear that $\pi$ induces a bijective group homomorphism between $\tilde{G}/\ker$ and $G$. Why is this homomorphism also a homeomorphism?
Best Answer
Let $\pi^\star$ be that map. Since it is a continuous bijection, in order to prove that it is a homeomorphism all that is needed is to prove that $\pi^\star$ is an open map. Let $A$ be an open subset of $\widetilde G/\ker\pi$. Then $\pi^{-1}(A)$ is an open subset of $\widetilde G$. But $\pi^\star(A)=\pi\left(\pi^{-1}(A)\right)$ and, which is an open set, since $\pi$ is an open map (since it is a covering map) and $\pi^{-1}(A)$ is open.