I apologize if this question is naive.
Suppose you know nothing about covering spaces, and they are introduced to you in the context of smooth manifolds, and Lie groups, of which you have a working knowledge.
Now you are given Lie group $G$, and a group $H$ which is a topological space, but not yet shown to be topological group. You have shown that there exists a continuous open surjective homomorphism $\phi:H\rightarrow G$ that has a discrete group as it's kernel, for simplicity say $\mathbb{Z}_2=\{\pm 1\}$. It seems to be true that $H$ is then a double covering of $G$, and that $H$ has the unique structure of Lie group such that $\phi$ is a smooth submersion. Is there a way to show this with no knowledge of covering spaces? Or perhaps with some knowledge of covering spaces that could be readily introduced with a lemma or three?
For context, I have spent a considerable amount of time in the last month working on Clifford algebras with constructing the groups Spin and Pin as motivation. I have shown that these are indeed groups, and that as topological subspaces of the Clifford algebra $\text{Cl}(n)$, there exists a continuous surjective group homomorphism $\phi:\text{Spin}(n)\rightarrow SO(n)$, but the book I am working out then cites as a corollary of this result that $\text{Spin}(n)$ has a unique Lie group structure such that $\phi$ is a smooth double covering of $SO(n)$, which to me does not follow immediately, probably because I do not know much about covering spaces.
Alternatively, if one could be point me to a source on covering spaces that builds up to some result like this, then maybe I could extract a way of doing it myself.
Edit:
Actually, I think I can view covering spaces (in the smooth manifold case) as smooth fibre bundles with discrete fibre i.e. zero dimensional manifolds. Given this, I thought that I could use $\phi$ to create a smoothly compatible formal bundle atlas which covers $G$, which by a theorem I have proven demonstrates $H$ is a smooth fibre bundle. However, I run into issues when doing this.
A standard fact from group theory tells us that if for some $g\in G$, $h\in \phi^{-1}(g)$, then $\phi^{-1}(g)=\{h,-h\}$. Let $U\subset G$, using the aforementioned fact, I thought I could write down a bijection:
$$
\psi_U:\phi^{-1}(U)\longrightarrow U\times\mathbb{Z}_2
$$
by sending $h\rightarrow\phi(h)$ in the first coordinate, and $h\rightarrow \text{sgn}(h)$, but I quickly realized that this is not a good map.
Best Answer
I will show the following: If $G$ is a Lie group, $H$ is a group and a Hausdorff space and $\phi\colon H\rightarrow G$ is continuous, open, surjective and a group homomorphism with kernel of order $2$, then $\phi$ is a $2$-fold covering.
To see some sort of assumption on $H$ is necessary, pick your favorite Lie group $G$ and let $H=G\times(\mathbb{Z}/2\mathbb{Z})_{\text{indiscrete}}$ with the product projection $\phi\colon H\rightarrow G$. Then $\phi$ is clearly continuous, open, surjective and a group homomorphism, but it is not a covering map, because a covering space of a Hausdorff space is Hausdorff, but $H$ is not Hausdorff.
Now, say $\ker(\phi)=\{e,h\}$, where $e\in H$ is the identity and $h\in H$ an element of order $2$. The Hausdorff assumptions means that there are disjoint open $U,V\subseteq H$ such that $e\in U$ and $h\in V$. Replacing $U$ by $U\cap h^{-1}V$ and $V$ by $V\cap hU$, we may assume WLOG that $V=hU$. Let $g\in G$ be arbitrary. Pick a $k\in\phi^{-1}(g)$. Then, $\phi(kU)=\phi(kV)$ is an open neighborhood of $g$ in $G$, $\phi^{-1}(\phi(kU))=kU\cup kV$ is a disjoint union and $\phi$ maps both $kU$ and $kV$ homeomorphically onto $\phi(kU)=\phi(kV)$, because open bijections are homeomorphisms and $kV=khU=hkU$ as $h$ is necessarily central. Thus, $\phi$ is a $2$-fold covering.