I am reading the section on covering actions from Glen Bredon's Tranformation groups.
Let $G$ be a Lie group (not necessarily connected) acting effectively/faithfully on a connected, locally path connected, semi-locally simply connected space $X$, not necessarily with fixed points. Let $p:\tilde{X}\to X$ be the universal covering of $X$.
For any $g\in G$, $\theta_g:X\to X$ is the map given by $x\mapsto g\cdot x$. Now he makes the following statement -
$\theta_g$ can be covered by a homeomorphism of $\tilde{X}$ since $\tilde{X}$ is simply connected, and any two such liftings differ by a deck transformation. Clearly, all such liftings for all $g$ form a subgroup $G'$ of $\operatorname{Homeo}(\tilde{X})$.
My question is how do we get such a homeomorphism of $\tilde{X}$?
I think it should be by general lifting theorem. So for a choice of base point $x_0\in X$ and $x'_0\in\tilde{X}$ such that $p(x'_0)=g\cdot x_0$ there will be a unique homeomorphism from $\tilde{X}\to\tilde{X}$ sending $x'_0$ to itself. So why does he talk about any two such liftings? Does he mean for different choice of base points? Also why do they differ by a deck transformation?
Thank you.
First, you ask how to get lifts of $\theta_g$ to $\tilde X$. Here is the construction, an application of the lifting theorem.
Consider the original base point $x_0$, and consider the alternate base point $x'_0 = \theta_g(x_0)$. Fix a lift $\tilde x_0 \in \tilde X$ of $x_0$.
For each choice of a lift $\tilde x'_0$ of $x'_0$, let's consider the following lifting problem: how to lift the map $\theta_g : (X,x_0) \to (X,x'_0)$ to a map $\tilde \theta_g : (\tilde X,\tilde x_0) \to (\tilde X,\tilde x'_0)$. By the lifting lemma, to determine whether this lift $\tilde\theta_g$ exists, we must consider the following two subgroups of $\pi_1(X,x'_0)$, namely: $$(\tilde\theta_g)_*(\pi_1(\tilde X,\tilde x_0)) $$ and $$p_*(\pi_1(\tilde X,\tilde x'_0)) $$ The necessary and sufficient condition for $\tilde\theta_g$ to exist is that the first subgroup is contained in the second, which is obvious because both subgroups are trivial since $\tilde X$ is simply connected.
You also asked "Why does he talk about two such liftings?" Notice, the construction of $\tilde \theta_g$ has a choice, namely: a choice of lift $\tilde x'_0$ of $x'_0$. Two different choice of $\tilde x'_0$ would yield two different lifts of $\theta_g$.
From another point of view, suppose that $\theta_{g,1}, \theta_{g,2} : \tilde X \to \tilde X$ are two lifts of $g : X \to X$. To say that these two lifts "differ by a covering transformation" means that their difference $\theta_{g,1}^{-1} \circ \theta_{g,2} : \tilde X \to \tilde X$ is a covering transformation of the universal covering map $p : \tilde X \to X$.
To prove this, first notice that $\theta_{g,1}^{-1} \circ \theta_{g,2}$ is a lift of $\theta_g^{-1} \circ \theta_g$ which equals the identity map on $X$. Almost by their very definition, the deck transformations are the maps $\tilde X \mapsto \tilde X$ which are lifts of the identity map on $X$. So, $\theta_{g,1}$ and $\theta_{g,2}$ differ by a deck transformation.