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Math Help - [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 \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.
    Last edited by JJMC89; April 25th 2010 at 08:02 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 \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.
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