# [SOLVED] Algebraic Topology: Regular Cover

• Apr 25th 2010, 01:55 PM
JJMC89
[SOLVED] Algebraic Topology: Regular Cover
A covering is said to be regular if for some $\widetilde{x_0} \in \widetilde{X}$ the group $p_{*} \ \pi ( \widetilde{X} , \widetilde{x_0} )$ is a normal subgroup of $\pi (X , x_0 )$. Prove that if $f$ is a closed path in $X$ then either every lifting of $f$ is closed or none is closed.
• Apr 25th 2010, 10:24 PM
aliceinwonderland
Quote:

Originally Posted by JJMC89
A covering is said to be regular if for some $\widetilde{x_0} \in \widetilde{X}$ the group $p_{*} \ \pi ( \widetilde{X} , \widetilde{x_0} )$ is a normal subgroup of $\pi (X , x_0 )$. Prove that if $f$ is a closed path in $X$ then either every lifting of $f$ is closed or none is closed.

Lemma: If $\widetilde{X}$ is a regular covering of X, then for every points $e_1, e_2$ of $p^{-1}(b_0)$ for $b_0 \in X$, there is an equivalence $h:\widetilde{X} \rightarrow \widetilde{X}$ with $h(e_1)= h(e_2)$ (link).

If $\widetilde{X}$ is path-connected and locally path-connected, then there exists a lifting of f which is a closed path by the lifting property (link). Then by the above lemma, every lifting of f should be closed.

I think if $\widetilde{X}$ is not path-connected, the criteria of "none is closed" is applied along with the above lemma.

Not 100% sure though.