Suppose that $p:\tilde{X} \rightarrow X$ is a covering obtained from a properly discontinuous action of G on $\tilde{X}.$ Prove that $p:\tilde{X} \rightarrow X$ is regular.
my attempt: Since p is a covering obtained from a properly discontinuous action of G on X, I know that $p:\tilde{X} \rightarrow \tilde{X}/G$ . Now to show that p is regular, I have to show that for some $\tilde{x_0}\in \tilde{X}$, the group $p_*\pi(\tilde{X}, \tilde{x_0})$ is a normal subgroup of $\pi(X, x_0)$. But then, I have no idea. Please help. thanks
(a) changing the basepoint $\bar x_0 \in p^{-1}(x_0)$ to $\bar x_1 \in p^{-1}(x_0)$ corresponds to conjugating $p_*\pi(X, x_0)$ by an element of $\pi(X, x_0)$.
Let $\gamma$ be a loop (representing $g \in \pi(X, x_0)$) which lifts to a path $\bar\gamma$ from $\bar x_0$ to $\bar x_1$. Let $H_0=p_*\pi(X, x_0)$ and $H_1=p_*\pi(X, x_0)$. Now you can prove that $g^{-1}H_0g \subset H_1$ and $gH_1g^{-1} \subset H_0$. Therefore $g^{-1}H_0g = H_1$. The converse is easy to show.
So $g$ is in the normalizer $N(H_0)$ iff $p_*\pi(X, x_0)=p_*\pi(X, x_0)$
(b) $p_*\pi(X, x_0)=p_*\pi(X, x_0)$ iff there exists a deck transformation taking $\bar x_0$ to $\bar x_1$
Suppose $p_*\pi(X, x_0)=p_*\pi(X, x_0)$, then lift $p$ to a map $\bar {p_1}:(\bar X,x_0) \rightarrow (\bar X,x_1)$ with $p \bar {p_1} =p$. Similarly we get $\bar {p_1}:(\bar X,x_1) \rightarrow (\bar X,x_0)$ with $p \bar {p_2} =p$. By the unique lifting property, $\bar {p_1} \bar {p_2}=1$, $\bar {p_2} \bar {p_1}=1$. So $\bar{p_1}$ $\bar{p_2} $ are inverse isomorphisms. The converse is easy to show.
(c) All the Deck transformations (homeomorphisms $\bar X \rightarrow \bar X $ which preserve the projection) form a group $G$ which is a properly discontinuous action on $X$.
Easy to show.
(d) $N(H)=\pi(X, x_0)$
$g_2g_1^{-1}$ takes $g_1(U)$ to $g_2(U)$