I have come across two definitions which are as follows $:$
Definition $:$
Covering Transformation $:$
Let $(\widetilde X , p)$ be a covering space. A homeomorphism $f : \widetilde X \longrightarrow \widetilde X$ is said to be a covering transformation if $p \circ f = p$ holds ; i.e. if $f$ is a lift of $p$.
Normal Covering Space $:$
A covering space $\widetilde X$ together with the covering map $p : \widetilde X \longrightarrow X$ is said to be a normal covering space if for any $x_0 \in X$ and for any pair of points $\widetilde {x_0},\widetilde {x_1} \in p^{-1} (x_0)$ there exists a covering transformation $f$ sending $\widetilde {x_0}$ to $\widetilde {x_1}$.
Also I know that the fundamental group $\pi_1 (X,x_0)$ acts on the fibres of $p$. Now let $\widetilde {x_0} \in p^{-1} (x_0)$. Let $H_{\widetilde {x_0}}$ be the stabilizer of $\widetilde {x_0}$ under the action of $\pi_1 (X,x_0)$ on $p^{-1} (x_0)$. Then I have seen that $H_{\widetilde {x_0}} = p_{*} (\pi_1 (\widetilde X , \widetilde {x_0})),$ where $p_{*} : \pi_1 (\widetilde X , \widetilde {x_0}) \longrightarrow \pi_1 (X,x_0)$ is the homomorphism between the fundamental groups induced by the covering map $p$.
Now suppose that the underlying covering space is normal then I am trying to show that $H=H_{\widetilde {x_0}}$ is a normal subgroup of the fundamental group $\pi_1 (X,x_0)$. Which I find hard to prove. Please help me in this regard.
Thank you very much.
EDIT $:$
I have tried it in the following way $:$
Let $[\gamma] \in \pi_1 (X,x_0)$. Then $\gamma$ is a loop in $X$ based at $x_0$. Consider the unique lift $\widetilde {\gamma}$ of $\gamma$ beginning at $\widetilde {x_0}$. Let $\widetilde {\gamma}$ ends at $\widetilde {x_1}$. Then $[\gamma]. H_{\widetilde {x_1}}. [\gamma]^{-1} = H_{\widetilde {x_0}} = H$. Since the underlying covering space is normal so there exists a covering transformation $f$ sending $\widetilde {x_0}$ to $\widetilde {x_1}$. But then by applying the lifting criterion for $f$ and $f^{-1}$ we get $p_{*} (\pi_1(\widetilde X,\widetilde {x_0})) = p_{*} (\pi_1 (\widetilde X,\widetilde {x_1}))$ i.e. $H_{\widetilde {x_1}} = H$. So $[\gamma] \in N(H),$ where $N(H)$ denotes the normalizer of $H$. Since $[\gamma] \in \pi_1 (X,x_0)$ is taken in a completely arbitrary fashion so it follows that $N(H) = \pi_1 (X,x_0)$. In other words $H$ is a normal subgroup of $\pi_1 (X,x_0)$.
QED
Is that correct at all? Please verify it.
Best Answer
Let $p:(X,x_0) \to (Y,y)$.
If there is a deck transformation $\varphi:X \to X$ carrying $x_0$ to $x_1$ in $p^{-1}(y)$, then since $p \circ \varphi=p$, so $p_*(\pi_1(X,x_0)) \subset p_*(\pi_1(X,x_1))$, and interchanging the roles of basepoint gives equality.
On the other hand, fix some $x_0 \in p^{-1}(y)$ and let $\alpha \in \pi_1(Y)$. I claim that there is some $x_1 \in X$ so that $p_*(\pi_1(X,x_1))=\alpha^{-1}p_*(\pi_1(X,x_0)) \alpha $.
Consider the lift $\tilde{\alpha}:x_0 \mapsto x_1$ and take that to be the correct point.
It is evident from these two considerations together that you have the right idea!