We suppose all spaces are connected and locally path connected.
Let $\widetilde{X}\to X$ be an universal covering space. Take the following situation:
Where H,K and G are the respective groups of covering transformations. If all 3 covering spaces are normal, then $K\cong G/H$.
My atempt of a proof is as follows:
Let $p:E\to F$, $q:F\to X$ be the covering spaces.
Because the groups are normal, we have the following
- $$G\cong \pi(X)/(qp)_*(\pi(E))$$ $$H\cong \pi(F)/p_*(\pi(E))$$ $$K\cong \pi(X)/q_*(\pi(F))$$
It follows that:
$$G/H\cong \frac{\pi(X)/(qp)_*(\pi(E)))}{\pi(F)/p_*(\pi(E))}.$$
We also have
2.$$\pi(F)/p_*(\pi(E))=q_*\pi(F)/(qp)_*(\pi(E)).$$
Then by the third isomorphism theorem $$G/H\cong \frac{\pi(X)/(qp)_*(\pi(E)))}{q_*\pi(F)/(qp)_*(\pi(E))}\cong \pi(X)/q_*(\pi(F))\cong K$$
I have doubt whether the set of isomorphisms 1. and 2. are true, as the proof relies on them. If they are not or they are, but I'm missing steps; I would appreciate if anyone could give me a hint on how to do it instead.
Compare the actions of $\pi_1(X, x_0)$ and $K$ on $q^{-1}(x_0)$. If your conventions make one a right action and the other a left action then you need to put in some inverses below or consider antihomomorphisms.
Suppose $[\gamma] \cdot f_0 = f_1$ and let $g \in K$ be the unique element that takes $f_0$ to $f_1$ (the action of $K$ is free by uniqueness of lifts and transitive by definition of a normal cover). Show that:
It follows that image of $q_*$ is normal in $\pi_1(X, x_0)$ and $K \cong \pi_1(X, x_0)/q_*(\pi_1(F, f_0))$.
This is an essential fact that should be proved in any text discussing covering spaces and deck transformations. For instance, it's part of proposition 1.39 in Hatcher's Algebraic Topology.