Let $G$ and $H$ be Lie groups. Is there an easy way to see why a surjective Lie group homomorphism $\varphi: G \to H\:$ with finite kernel is a covering map?
When assuming $G$ is a Hausdorff space, the only thing I am not sure about is how to prove that $\varphi$ is open, the rest is clear. Do we need such an assumption?
For context, I am interested in proving that the spin group is the double cover of the special orthogonal group if the underlying field is either $\mathbb{R}$ or $\mathbb{C}$.
Theorem: Let $G$ be a Lindelöf and locally compact topological group which acts in a continuous and transitive way on a Baire space $M$. If $m\in M$, then the map$$\begin{array}{ccc}G&\longrightarrow&M\\g&\mapsto&g\cdot m\end{array}$$is an open map.
Proof: It will be enough to prove that if $V$ is a neighborhood of $e$, then $V\cdot m$ is a neighborhood of $m$. Let $W$ be a neighborhood of $e$ such that $W^{-1}\cdot W\subset V$ and suppose that $W\cdot m$ is a neighborhood of some of its points; in other words, suppose that, for some $w_0\in W$, $W\cdot m$ is a neighborhood of $w_0\cdot m$. Then ${w_0}^{-1}\cdot(W\cdot m)$ is a neighborhood of $m$ and therefore $\bigcup_{w\in W}w^{-1}\cdot(W\cdot m)$ is a neighborhood of $m$. But $\bigcup_{w\in W}w^{-1}\cdot(W\cdot m)\subset V\cdot m$, and so $V\cdot m$ is also a neighborhood of $m$.
Therefore, all that remains to be proved is that among all neighborhoods $W$ of $e$ such that $W^{-1}\cdot W\subset V$ there is at least one such that $W\cdot m$ is a neighborhood of some of its points, and this is equivalent to saying that the interior of $W\cdot m$ is not empty. Let $W$ be a compact neighborhood of $e$ such that $W^{-1}\cdot W\subset V$; such a neighborhood exists since we are supposing that $G$ is locally compact. It is clear that the interior of $W\cdot m$ is not empty if and only if, for some $g\in G$, the interior of $g\cdot(W\cdot m)$ is not empty. It follows from the fact that $G$ is a Lindelöf space and from the fact that $\bigcup_{g\in G}g\cdot\mathring W=G$ that there is a sequence $(g_n)_{n\in\mathbb N}$ of elements of $G$ such that $\bigcup_{n\in\mathbb N}g_n\cdot W=G$ and, therefore, such that $\bigcup_{n\in\mathbb N}g_n\cdot(W\cdot m)=M$, since the action of $G$ on $M$ is transitive. For each $n\in\mathbb N$, $g_n\cdot(W\cdot m)$ is a compact set, since $W$ is compact and the action is continuous, and, in particular, each set $g_n\cdot(W\cdot m)$ is a closed set. Since $M$ is a Baire space, there is at least one $n\in\mathbb N$ such that the interior of $g_n\cdot(W\cdot m)$ is not empty and, as it has already been observed, this is equivalent to the assertion that the interior of $W\cdot m$ is not empty. QED
In order to apply this theorem to the spinor groups, it will be enough to prove that these groups are Lindelöf and locally compact. But it is a consequence of the definition of $\mathrm{Spin}(k)$ that this group can be seen as a closed subset of a finite-dimensional real vector space (with the usual topology); therefore, it is both a Lindelöf space and a locally compact space.
Now, let $\varphi\colon G\longrightarrow H$ be a surjective Lie group homomorphism with a finite kernel (or, more generally, a discrete kernel). Then we have an action of $G$ on $H$, defined by$$\begin{array}{ccc}G&\longrightarrow&\operatorname{Aut}(H)\\g&\mapsto&\left(\begin{array}{ccc}H&\longrightarrow&H\\h&\mapsto&\varphi(g)\cdot h\end{array}\right).\end{array}$$So, the previous theorem, taken with $m=e$, tells us that $\varphi$ is an open map.