Let $G$ be a topological group with a countinuous product $\mu : G \times G \to G$ and $e$ its identity element $( \mu (x,e) = \mu(e,x) = x $ for all $x \in G)$. Let $p: \tilde{G} \to G$ be a covering map and $\tilde{e} \in \tilde{G}$ a point such that $p(\tilde{e}) = e$, i.e. $\tilde{e} \in p^{-1}(e)$. We suppose both spaces ar connected and locally path-connected (thus, path -connected).
I have already proved there exists a unique product in $\tilde{G} \times \tilde{G}$, $\tilde{\mu}$, with identity element $\tilde{e}$, associative, and satisfying $\mu(p(\tilde{x}),p(\tilde{y})) = p(\tilde{\mu}(\tilde{x},\tilde{y}))$, and now I want to define the inverse in $\tilde{G}$ as a countinuous map $\tilde{v}: \tilde{G} \to \tilde{G}$, in order to prove $\tilde{G}$ is also a topological group.
As $G$ has a topological group structure, there is a countinuous map $v : G \to G$ satisfying $v(x) = x^{-1}$ for all $x \in G$. Defining $f : = v \circ p$, I have already showed there exists a lift of $f$ by $p$ via the lifting theorem for countinuous maps, and I have also reached to $p(\tilde{\mu}(\tilde{x},\tilde{v}(\tilde{x}))) = p( \tilde{\mu}(\tilde{v}(\tilde{x}),\tilde{x})) = e$ applying the upper-mentioned property and the fact that $p \circ \tilde{v} = v \circ p$, but I am having a bit of trouble with the following statements:
The lift is unique and satisfies $\tilde{v}(\tilde{e}) = \tilde{e}$. Maybe I have to impose this and apply the uniqueness of lift theorem? Or is it a consequence of something previous?
$\tilde{\mu}(\tilde{x},\tilde{v}(\tilde{x})) = \tilde{\mu}(\tilde{v}(\tilde{x}),\tilde{x}) = \tilde{e}$ for all $\tilde{x} \in \tilde{G}$. Guess I could prove $\tilde{\mu}(\tilde{x},\tilde{v}(\tilde{x})) = \tilde{\mu}(\tilde{v}(\tilde{x}),\tilde{x})$ by having proved $\tilde{v}(\tilde{e}) = \tilde{e}$, as a consequence of uniqueness of lift theorem, but I do not know how that could not equal another point on the fiber of $e$.
I also want to prove that if $G = \mathbb{T}^2$, the torus, and $\tilde{G} = \mathbb{S}^1 \times \mathbb{R}$, and $G$ is abelian, then $\tilde{G} $ is abelian. I am trying to prove this by using the covering map $p(e^{2\pi ix}, y) = (e^{2\pi ix}, e^{2\pi iy})$, with $x \in [0,1)$ and $y \in \mathbb{R}$, but I do not know how to do it. Does this depend on this spaces or is it a more general property that can be deduced from the others?