Suppose $G$ is a Lie group and $H$ be a connected manifold such that $\phi : H \to G$ is a covering. Suppose $e' \in H $ such that $\phi(e') = e.$ Then our aim is to show that there exist a (unique) Lie group structure on $H$ such that $e'$ is identity and $\phi$ is a map of Lie groups.
Let $m : G \times G \to G$ denote the groups operation. Using map lifting I have constructed $\mu : H \times H \to H$ such that $$\phi \circ \mu = m \circ ( \phi \times \phi ). $$ (where base point for $H$ is $e'.$)
I have shown that $\mu$ defines a group operation on $H$ with identity $e'$ and $\phi$ is a group homomorphism.
I have shown the existence of inverse in $H$ in the following way :
Let $x \in H,$ consider the map $j : H \to H \times H : h \to (h,x).$ We see that the map $F = m \circ \phi \times \phi \circ j : H \to G$ maps $h \to \phi(x)\phi(h)$ which is a covering of $G,$ and $\mu \circ j $ is a unique lift of $F$ thus, $\mu \circ j$ is surjective. So $x$ has a left inverse and similarly a right inverse.
$H$ has a natural manifold structure induced by $G.$ Since $\phi$ is a local diffeomorphism it easily follows that $\mu$ is smooth. Now I want to show that inversion is smooth. So let $i_G : G \to G : g \mapsto g^{-1}$ denote inversion. Then $i_G \circ \phi $ is a covering of $G$ hence lift to a unique map $i_H : H \to H$ such tht $\phi \circ i_H = i_G \circ \phi.$ What I want to show that $\mu(id_H, i_H)$ and $\mu(i_H, id_H)$ both are trivial on diagonal so that $i_H$ is actual inversion of $H.$ Then from $\phi \circ i_H = i_G \circ \phi$ it will follow that $i_H$ is smooth. I am not getting how to show that $\mu(id_H, i_H)$ and $\mu(i_H, id_H)$ both are trivial on diagonal.