I am trying to prove the following fact:
Let $(G, \cdot)$ be a topological group and let $E$ be a universal covering space of $G$. If $G$ and $E$ are both locally path-connected, then for any choice of element $e$ in the fiber over $1_G \in G$, there exists a unique topological group structure on $E$, with $e$ as the identity, for which the covering map $p: E \to G$ is a homomorphism.
Following the idea found on Wikipedia:
The construction is as follows. Let $a$ and $b$ be elements of $E$ and let $f$ and $g$ be paths in $E$ starting at $e$ and terminating at $a$ and $b$ respectively. Define a path $h : I \to G$ by $h(t) = p(f(t))p(g(t))$. By the path-lifting property of covering spaces there is a unique lift of $h$ to $E$ with initial point $e$. The product $ab$ is defined as the endpoint of this path. By construction we have $p(ab) = p(a)p(b)$. One must show that this definition is independent of the choice of paths $f$ and $g$, and also that the group operations are continuous.
I have already checked the well-definition and the group axioms, but I am stuck on the proof of the continuity of the group operations.
P.S. I am aware that there exists another construction to prove the fact, namely the one used in the Greenberg Harper book, but I don't want to use the Lifting Criterion, just the homotopy lifting property.
Best Answer
I will skip the tedious proof of well-definition and group axioms since you have already got them. Hence I will only focus on the proof of the continuity of the map
$$\star\colon E\times E\to E, \qquad (a,b)\mapsto a\star b^{-1}$$
From the construction and the fact that $p$ is a group homomorphism, we have a natural commutative diagram
$$\require{AMScd} \begin{CD} E\times E @>{\star}>> E\\ @VVV @VVV \\ G\times G @>{\cdot}>> G \end{CD}$$
where the vertical lines are the product map $p\times p$ and $p\colon E\to G$ respectively, and the horizontal arrow is the continuous map $\cdot \colon G\times G\to G$ sending $(g_1,g_2)\mapsto g_1\cdot g_2^{-1}$.
$\textbf{Definition}$ Given an open subset $U\subseteq E$, I say that $U$ is ${\it elementary}$ with respect to $p$ if $p_{|U} \colon U\to p(U)$ is an homeomorphism.
$\textbf{Lemma:}$ It there exists a topological basis of $E$ given by elementary sets with respect to $p$.
$\textit{proof:}$ Let $U$ be an open set of $E$ and $x\in U$. It there exists an open neighbourhood $V$ of $p(x)$ such that $p^{-1}(V)=\sqcup_{\alpha}U_\alpha$ and $p_{|U_\alpha}\colon U_\alpha\to V$ is an homeomorphism. Let $U_x:=U_{\alpha(x)}$ be the unique open elementary set of the family containing the point $x$. Then $U_x\cap U\subset U$ is elementary with respect to $p$. By the arbitrary choice of $x\in U$, we get $U=\cup_{x\in U}U_x\cap U$.
$\textbf{Remark:}$ The definition of a open elementary set together with the above lemma works for any covering $p\colon X\to Y$.
$\textbf{Proposition:}$ The above map $\star$ is continuos.
$\textit{proof:}$ For simplicity of notation, let us define $f:=\cdot \circ (p\times p)$. From the previous lemma, it is sufficient to prove the continuity for any open elementary set $U$ of $E$ with respect $p$. Hence we only need to prove that $\star^{-1}(U)$ is open. Given $x\in \star^{-1}(U)$, since $E\times E$ is locally path connected, then it there exists a open path connected neighbourhood $V$ of $x$. Moreover, this can be chosen such that $f(V)\subseteq p(U)$ (remember that $p(U)$ is open in $G$). Given a point $y=(y_1,y_2)\in V$, we choose a path $\gamma$ from $(e, e)$ to $x$ and a path $\eta$ from $x$ to $y$ all contained in $V$. Then $ f_*(\gamma\eta)$ is a path from $1_G$ to $p(y_1)p(y_2^{-1})$, which is homotopic to the path $p(f(t))p(g(t))$ that you defined above. Thus their lifting via $p$ starting at $e$ are homotopic and in particular have the same ending point $\star(y)$. By construction, since $U$ is elementary, we observe that the lifting of $ f_*(\gamma\eta)$ is simply $p^{-1}\circ f(\gamma\eta)$, so that the end point $\star(y)$ falls into $U$. Hence $y\in \star^{-1}(U)$ and from the arbitrary choice of $y\in V$ we have $V\subset \star^{-1}(U)$. This proves $\star^{-1}(U)$ is open.
Please let me know if you agree with me or if you have some doubts and questions :).