Proposition 4.17.
There is a natural bijection
\begin{equation*}
\operatorname{Hom}(\pi_1(X, x_0),\, G) \;\xrightarrow{\ \cong\ }\;
\{G\text{-coverings of }X\text{ with basepoint over }x_0\}\,/\,\text{iso}.
\end{equation*}
Section 4.3 G-Coverings from Homomorphisms
Throughout this section, \(X\) is connected, locally path-connected, and semilocally simply connected, with universal cover \(\tilde{X} \to X\) and basepoints \(x_0 \in X\text{,}\) \(\tilde{x}_0 \in \tilde{X}\text{.}\)
Proof.
From homomorphism to covering. Given \(\rho \colon \pi_1(X, x_0) \to G\text{,}\) let \(\pi_1(X, x_0)\) act on \(\tilde{X} \times G\) by
\begin{equation*}
[\sigma] \cdot (z, g) = \bigl([\sigma] \cdot z,\; g \cdot \rho([\sigma])^{-1}\bigr),
\end{equation*}
where \([\sigma] \cdot z\) is the action from Theorem 4.11. Set \(Y_\rho = (\tilde{X} \times G)/\pi_1(X, x_0)\text{.}\) The map \(P_\rho \colon Y_\rho \to X\) is a \(G\)-covering, where \(G\) acts on \(Y_\rho\) by \(h \cdot \langle z, g\rangle = \langle z, hg \rangle\text{.}\)
From covering to homomorphism. Given a \(G\)-covering \(p \colon Y \to X\) with basepoint \(y_0\) over \(x_0\text{,}\) define \(\rho \colon \pi_1(X, x_0) \to G\) by
\begin{equation*}
\rho([\sigma]) = \text{endpoint of the lift of }\sigma\text{ starting at }y_0,
\end{equation*}
viewing the endpoint as an element of the \(G\)-fibre over \(x_0\text{.}\) These two constructions are inverse to each other.
Exercise 4.18.
Show that, ignoring basepoints, two \(G\)-coverings \(Y_\rho\) and \(Y_{\rho'}\) are isomorphic if and only if \(\rho\) and \(\rho'\) are conjugate in \(G\text{:}\) there exists \(g \in G\) with \(\rho'([\sigma]) = g\,\rho([\sigma])\,g^{-1}\) for all \([\sigma] \in \pi_1(X, x_0)\text{.}\)
