# Is the converse of this is true?

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.

• I have just proved it. Can I add it? Oct 9, 2018 at 17:02
• Sure, as long as it's correct, people will upvote anyway.
– Laz
Oct 9, 2018 at 18:49

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.