# $E \overset{p}{\longrightarrow}X$ covering space, connected and locally path-connected. Obtain homomorphism from $N$ onto $G$

I'm doing exercise 5.9 of Greenberg and Harper which is the following: Let $E \overset{p}{\longrightarrow}X$ covering space, connected and locally path-connected. Obtain homomorphism from $N$ normalizer of $p_\ast \left(\pi_1(E,e_0) \right)$ onto $G$ the group of covering transformations with kernel $p_\ast\left( \pi_1(E,e_0) \right)$.

So far I have done the following:

Let $e_0, e_1 \in p^{-1}(x_0)$. Since $E$ is connected and locally path-connected it is also connected. Then let $\tilde{\alpha} : I \longrightarrow E$ be the path that connects $e_0$ and $e_1$. Thus, $p \circ \tilde{\alpha} = \alpha$ with $\alpha$ a loop in $X$ which we represent as the element $n\in \pi_1(X,x_0)$. $E$ is path connected, so $\pi_1(X,x_0)$ acts transitively on $E$ and we have $p_\ast\left( \pi_1(E,e_0)\right) = n \cdot p_\ast\left( \pi_1(E,e_1)\right) \cdot n^{-1}$. Additionally, if $n \in N \Longleftrightarrow p_\ast\left( \pi_1(E,e_0)\right) = p_\ast\left( \pi_1(E,e_1)\right)$.

At this point, my idea was to prove that if we have $p_\ast\left(\pi_1(E,e_0)\right) = p_\ast\left( \pi_1(E,e_1)\right)$, then there exists a homeomorphism $\phi \in G$ s.t. $\phi(e_0)=e_1$, however I don't know how to proceed so some hints or advice would be greatly appreciated.

Some remarks:

• You are right that $p_\star(\pi_1(E, e_0)) = \alpha . p_\star(\pi_1(E, e_1)) . \alpha^{-1}$, where $\alpha = p \circ \widetilde \alpha$ and $\widetilde \alpha$ is any path in $E$ from $e_0$ to $e_1$.

• You are right that, if $p_\star(\pi_1(E, e_0)) = p_\star(\pi_1(E,e_1))$, then there exists a unique covering transformation sending $e_0$ to $e_1$. To prove this, you could use the general lifting criterion (Hatcher pages 61 and 62):

If $p: (E, e) \to (X,x)$ is a covering map and $f : (Y, y) \to (X, x)$ is another map, with $Y$ path connected and locally path connected, then there exists a lift $\widetilde f : (Y,y) \to (E, e)$ such that $p \circ \widetilde f = f$ iff $f_\star(\pi_1(Y,y)) \subset p_\star (\pi_1(E, e))$. If these conditions are satisfied, then this lift is unique.
• I presume you want to define your map $N \to G$ in the following way. Taking a homotopy class $n \in N \subset \pi_1(X,x_0)$, pick any loop $\alpha$ in this class, and lift $\alpha$ to a path $\widetilde \alpha$ on $E$ such that $\alpha(0) = e_0$ and $p \circ \widetilde \alpha = \alpha$. Define $e_1$ to be the endpoint $\alpha(1)$, and define the covering transformation $\tau \in G$ associated to $n$ to be the unique covering transformation sending $e_0$ to $e_1$. This works, provided that you check that the covering transformation you get is independent of your choice of representative $\alpha$ for the homotopy class $n$!

• You need to check this your map from $N$ to $G$ is actually a group homomorphism. Try to use the "uniqueness of lifts" theorem to show that, if the homotopy classes $n_1$ and $n_2 \in N$ are represented by loops $\alpha_1$ and $\alpha_2$ lifting to the paths $\widetilde \alpha_1$ and $\widetilde \alpha_2$ starting at $e_0$ in $E$, and if $n_1$ and $n_2$ give rise to the covering transformations $\tau_1$ and $\tau_2 \in G$, then the loop $n_1 . n_2$ must lift to the path $\widetilde \alpha_1 . \tau_1(\widetilde \alpha_2)$. Try to deduce that the endpoint of the lift of $n_1 . n_2$ is $\tau_1(\tau_2(e_0))$, so the covering transformation associated to $n_1 . n_2$ maps $e_0$ to $\tau_1(\tau_2(e_0))$, and hence, is equal to $\tau_1 \circ \tau_2$.

• Just for fun, you might like to show that the kernel of your homomorphism $N \to G$ is $p_\star(\pi_1(E, e_0))$. (Of course, your homomorphism $N \to G$ is surjective.)

• The lifting criterion in Greenberg and Harper is only introduced in the following chapter, so I guess I shouldn't rely on it for the proof. Do you recall another method of proving the existence of the unique covering transformation? – peter19 Mar 28 '17 at 15:23
• Hmm. Off the top of my head, I can't think of another way! Perhaps somebody else can. But you could always prove the general lifting criterion! It's quite a fun proof! :-) – Kenny Wong Mar 28 '17 at 16:14