Covering space of a topological group is itself a topological group Let $p:\widetilde{G}\to G$ be a covering, where $G$ is a (locally path-connected and path-connected) topological group with neutral element $e$. Let $\widetilde{e}\in p^{-1}(\{e\})$. I have to show that $\widetilde{G}$ can be made to a topological group with neutral element $\widetilde{e}$ in a unique way.
I allready have show using the lifting property, that there exists a unique continuous map $\widetilde{\mu}:\widetilde{G}\times \widetilde{G}\to \widetilde{G}$ satisfying $\widetilde{\mu}(x,\widetilde{e})=\mu(\widetilde{e},x)=x$ for all $x\in\widetilde{G}$, as well as $p(\widetilde{\mu}(x,y))=\mu (p(x),p(y))$ for all $x,y\in \widetilde{G}$, where $\mu$ denoted the multiplication of $G$. But how to show associativity?
Using the associativity of $\mu$ and the lifting property, I was also able to show that
$$p(\widetilde{\mu}(x,\widetilde{\mu}(y,z)))=p(\widetilde{\mu}(\widetilde{\mu}(x,y),z))$$
for all $x,y,z\in\widetilde{G}$, but how can I conclude associativity?
 A: Consider the maps $\alpha,\beta\colon\widetilde G^3\longrightarrow\widetilde G$ defined by$$\alpha(g_1,g_2,g_3)=\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\quad\text{and}\quad\beta(g_1,g_2,g_3)=\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right).$$You want to prove that they are equal. Note that, if $g_1,g_2,g_3\in\widetilde G$, then\begin{align}p\left(\alpha(g_1,g_2,g_3)\right)&=p\left(\widetilde\mu\left(\widetilde\mu(g_1,g_2),g_3\right)\right)\\&=p\left(\widetilde\mu(g_1,g_2)\right)\cdot p(g_3)\\&=p(g_1)\cdot p(g_2)\cdot p(g_3)\\&=p(g_1)\cdot p\left(\widetilde\mu(g_2,g_3)\right)\\&=p\left(\widetilde\mu\left(g_1,\widetilde\mu(g_2,g_3)\right)\right)\\&=p\left(\beta(g_1,g_2,g_3)\right).\end{align}In other words, $p\circ\alpha=p\circ\beta$. Since, furthermore, $\alpha\left(\widetilde e,\widetilde e,\widetilde e\right)=\beta\left(\widetilde e,\widetilde e,\widetilde e\right)=\widetilde e$, $\alpha=\beta$.
A: Since you know that $p(\widetilde{\mu}(x,\widetilde{\mu}(y,z)))=p(\widetilde{\mu}(\widetilde{\mu}(x,y),z))$ holds, the underlying functions differ only by an automorphism of the cover, i. e. $$\widetilde{\mu}(\cdot,\widetilde{\mu}(\cdot,\cdot))=\widetilde{\mu}(\widetilde{\mu}(\cdot,\cdot),\cdot) \circ \phi,$$
where $\phi$ is an automorphism on the cover $\widetilde{G}$.
However since they agree on the identity element $\widetilde{e}$, said automorphism has to be the identity and we get $$\widetilde{\mu}(x,\widetilde{\mu}(y,z))=\widetilde{\mu}(\widetilde{\mu}(x,y),z)$$
