I posted a question here some days ago. Here's the question.
Here I have proved that if the covering space $\widetilde X$ is normal then $H_{\widetilde {x_0}}$ is a normal subgroup of $\pi_1 (X,x_0)$. Now Can I say the converse? i.e. if $H_{\widetilde {x_0}}$ is a normal subgroup of $\pi_1 (X,x_0)$ then can we say that the covering space $\widetilde X$ is normal? If so why?
If the answer is "yes" then it is similar to say that for any $\widetilde {x_1} \in p^{-1} (x_0)$ there exists a covering transformation taking $\widetilde {x_0}$ to $\widetilde {x_1}$. How do I find such covering transformation (if it exists at all)? Please help me in this regard.
Thank you very much.
EDIT $:$
Let us assume that the covering space $\widetilde X$ is path connected and the base space $X$ is path connected as well as locally path connected. Then in order to show that there exists a covering transformation taking $\widetilde {x_0}$ to $\widetilde {x_1}$ it is enough to show $p_{*} (\pi_1 (\widetilde X, \widetilde {x_0})) \le p_{*} (\pi_1 (\widetilde X, \widetilde {x_1}))$ i.e. $H_{\widetilde {x_0}} \le H_{\widetilde {x_1}}$.
Since $\widetilde X$ is path connected. We can take a path $\widetilde {\gamma}$ from $\widetilde {x_1}$ to $\widetilde {x_0}$ in $\widetilde X$. Let $\gamma = p \circ {\widetilde {\gamma}}$. Then $\gamma$ is a loop in $X$ based at $x_0$. Let $[f] \in H_{\widetilde {x_0}}$. So $[\gamma] \in \pi_1(X,x_0)$. Then the lift $\widetilde {\gamma * f * \overline {\gamma}}$ of $\gamma * f * \overline {\gamma} \in \pi_1 (X,x_0)$ is same as $\widetilde {\gamma} * \widetilde {f_{\widetilde {x_0}}} * \overline {\widetilde {\gamma}}$ by unique path lifting criterion which begins and ends at $\widetilde {x_1}$ i.e. $[\gamma * f * \overline {\gamma}] \in H_{\widetilde {x_1}}$ where $\widetilde {f_{\widetilde {x_0}}}$ is the unique lift of $f$ beginning at $\widetilde {x_0}$. Since $[f] \in H_{\widetilde {x_0}}$ was completely arbitrarily taken so it follows that $[\gamma] H_{\widetilde {x_0}} [\gamma]^{-1} \le H_{\widetilde {x_1}}$. Since $H_{\widetilde {x_0}}$ is a normal subgroup of $\pi_1(X,x_0)$ so $[\gamma] H_{\widetilde {x_0}} [\gamma]^{-1} = H_{\widetilde {x_0}}$ and hence we have $H_{\widetilde {x_0}} \le H_{\widetilde {x_1}}$. Which proves our claim. Which in turn implies the existence of a covering transformation taking $\widetilde {x_0}$ to $\widetilde {x_1}$ i.e. the covering space $\widetilde X$ is normal.
This completes the proof.
QED
Does my proof hold good? Please verify it.
Best Answer
Let me answer my own question after a prolonged waiting for others reply. I think this is the time to post my answer which I think is logically correct. Here it is $:$
Let us assume that the covering space $\widetilde X$ is path connected and the base space $X$ is path connected as well as locally path connected. Then in order to show that there exists a covering transformation taking $\widetilde {x_0}$ to $\widetilde {x_1}$ it is enough to show $p_{*} (\pi_1 (\widetilde X, \widetilde {x_0})) \le p_{*} (\pi_1 (\widetilde X, \widetilde {x_1}))$ i.e. $H_{\widetilde {x_0}} \le H_{\widetilde {x_1}}$.
Since $\widetilde X$ is path connected. We can take a path $\widetilde {\gamma}$ from $\widetilde {x_1}$ to $\widetilde {x_0}$ in $\widetilde X$. Let $\gamma = p \circ {\widetilde {\gamma}}$. Then $\gamma$ is a loop in $X$ based at $x_0$. Let $[f] \in H_{\widetilde {x_0}}$. So $[\gamma] \in \pi_1(X,x_0)$. Then the lift $\widetilde {\gamma * f * \overline {\gamma}}$ of $\gamma * f * \overline {\gamma} \in \pi_1 (X,x_0)$ is same as $\widetilde {\gamma} * \widetilde {f_{\widetilde {x_0}}} * \overline {\widetilde {\gamma}}$ by unique path lifting criterion which begins and ends at $\widetilde {x_1}$ i.e. $[\gamma * f * \overline {\gamma}] \in H_{\widetilde {x_1}}$ where $\widetilde {f_{\widetilde {x_0}}}$ is the unique lift of $f$ beginning at $\widetilde {x_0}$. Since $[f] \in H_{\widetilde {x_0}}$ was completely arbitrarily taken so it follows that $[\gamma] H_{\widetilde {x_0}} [\gamma]^{-1} \le H_{\widetilde {x_1}}$. Since $H_{\widetilde {x_0}}$ is a normal subgroup of $\pi_1(X,x_0)$ so $[\gamma] H_{\widetilde {x_0}} [\gamma]^{-1} = H_{\widetilde {x_0}}$ and hence we have $H_{\widetilde {x_0}} \le H_{\widetilde {x_1}}$. Which proves our claim. Which in turn implies the existence of a covering transformation taking $\widetilde {x_0}$ to $\widetilde {x_1}$ i.e. the covering space $\widetilde X$ is normal.
This completes the proof.
QED.