Let $p:E\to X$ be the universal covering of $X$; let $e\in E $ and $p(e)=x $. Given a homomorphism $\rho :\pi_1 (X,x)\to G$, let $Y_\rho =E×G/\pi_1 (X,x) $ with the quotient topology, and let $p_\rho:Y_\rho\to X$ be the projection over $E/\pi_1 (X,x) =X$. Then $p_\rho:Y_\rho\to X$ is a $G $-covering of $X $, where the action of $G $ is given by $h [e,g]=[e,hg]$.
There are a few things that I don't understand in the proof of this theorem: first off, I read that $p_\rho $ is continuous because it is induced by a map $E×G\to X$, which is invariant respect to the action of $\pi_1 (X,x)$. However I don't have clear this situation; I presume that the map is the projection $E×G\to E $, but I don't see how the action of $\pi_1 (X,x)$ is defined.
Then, in order to prove that $Y_\rho $ is a $G $-covering, one has to prove that the action of$G $ over $Y_\rho $ is properly discontinuous and that the fibers of $p_\rho $ are $G $-orbits. Let $N $ be a evenly covered open for the universal cover; then $p^{-1}(N)×G \cong N×\pi_1 (X,x)×G$. The map $$N×\pi_1 (X,x)×G\to N×G, \ \ \ \ (z,[\gamma],g)\mapsto (z,g\rho [(\gamma^{-1}]))$$ is $\pi_1 (X,x)$ invariant and induces an homomorphism, so this should prove our thesis. I really don't understand what all this procedure means: basically, I don't understand why the map above is $\pi_1 (X,x)$ invariant; moreover, I presume that the choice of that map should be in some way "natural", but I can't realize its intuitive meaning. Thanks a lot for any clarify
I believe the intention is that $G$ is a group, and that one uses $\rho$ to define an action of $\pi_1(X,x)$ on $G$ by left multiplication: $$[\gamma] \cdot h = \rho[\gamma] \, h $$ where the expression $\rho[\gamma] \, h$ simply means the product of $\rho[\gamma] \in G$ with $h \in G$ using the given group operation on $G$. One then has the diagonal action of $\pi_1(X,x)$ on the product $E \times G$ which is used to define the quotient $Y_\rho$. The rest should be straightforward, but feel free to ask if there's still anything confusing.