# Is a covering space of a topological group a fiber bundle with the structure group the fundamental group of the topological space?

Question. Is a covering space of a topological group a fiber bundle with the structure group the fundamental group of the topological space?

Let $p:E\rightarrow X$ be a covering space of X. I know how $\pi_1(X,x_0)$ acts on $p^{-1}(x_0)$. And if the space is path-wise connected all the $\pi_1(X,x_0)$ are all isomorphic for all $x_0\in X$. Thus $\pi_1(X)$ acts on the fibers.

Am I correct or am I missing something?

Yes. In fact, any regular covering space $p\colon E\to B$ is a princpal bundle with structure group given by $\pi_1(B)/p_*(\pi_1(E))$. So if $E$ is a universal cover (which is always regular), then the structure group is just $\pi_1(B)$.
• I understand how $\pi_1(B)$ acts on the fibers. But I don't understand how does this action factors through $\pi_1(B)/p_*(\pi_1(E))$. Apr 17, 2015 at 14:23
• @Babai The monodromy action of $\pi_1(B, x)$ on $p^{-1}(x)$ is defined by lifting a loop $\gamma$ based at $x$ to $E$ and looking at the endpoint of $\tilde{\gamma}$. $p_*(\pi_1(E)) \subset \pi_1(B)$ stabilizes $x$ as lift of any element of this subgroup is always a loop, i.e., the endpoints are the same. Apr 17, 2015 at 14:33
• Consider a simple example like the double cover $p$ of the circle $S^1$ over itself. Sure $\mathbb{Z}$ acts on the two-point fiber by either the trivial permutation if $n$ is even, or the non-trivial permutation if $n$ is odd, but we might as well mod out by the trivial permutations, which are precisely the image of $p_*$ on $\pi_1(S^1)$, the even loops. So we get a free action of $\mathbb{Z}/2\mathbb{Z}$ acting on the fiber $p^{-1}(x_0)$. Apr 17, 2015 at 14:34