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Thread: [SOLVED] Algebraic Topology: Regular Cover

  1. #1
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    [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.
    Last edited by JJMC89; Apr 25th 2010 at 07:02 PM.
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  2. #2
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    Quote Originally Posted by JJMC89 View Post
    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.
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