# Thread: [SOLVED] Algebraic Topology: Regular Cover

1. ## [SOLVED] Algebraic Topology: Regular Cover

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

2. Originally Posted by JJMC89
A covering is said to be regular if for some $\displaystyle \widetilde{x_0} \in \widetilde{X}$ the group $\displaystyle p_{*} \ \pi ( \widetilde{X} , \widetilde{x_0} )$ is a normal subgroup of $\displaystyle \pi (X , x_0 )$. Prove that if $\displaystyle f$ is a closed path in $\displaystyle X$ then either every lifting of $\displaystyle f$ is closed or none is closed.
Lemma: If $\displaystyle \widetilde{X}$ is a regular covering of X, then for every points $\displaystyle e_1, e_2$ of $\displaystyle p^{-1}(b_0)$ for $\displaystyle b_0 \in X$, there is an equivalence $\displaystyle h:\widetilde{X} \rightarrow \widetilde{X}$ with $\displaystyle h(e_1)= h(e_2)$ (link).

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

Not 100% sure though.