Let $H$ be a group, $G$ be a connected Lie group, and $f: H \rightarrow G$ be a surjective group homomorphism with finite kernel in $Z(H)$. How to show that; $H$ has a unique structure as a Lie group such that $f$ is a homomorphism of Lie groups?
I'm trying to defing a topology on $H$ by defining a basis of $H$ by $f^{-1}(U)$, where $U$ is an open subset of $G$. In this way the topology is well defined, but the open sets in this basis are too big, I can't get a local homeomophism from a $f^{-1}(U)$ to $R^n$.
I have thought about using the kernel(let's denote the cardinality of the kernel by $m$) to cut each $f^{-1}(U)$ by $m$ part (just consider the transitive action of the kernel on any preimage $f^{-1}(g)$, and pick a representative for each $f^{-1}(g)$ to construct open subsets), and then define a topology. Such a topology seems to satisfy what we need, and we have a covering space(of multiplicity $m$) but it is not unique, since we have many ways to pick representatives to construct an open set.
I wonder if it is the right way to construct such a topology? How to understand the uniqueness? Thank you.