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Local Trivialization in Covering Spaces

  1. Jun 18, 2011 #1
    Hi, All:

    I am trying to understand why covering maps have the

    local triviality condition, i.e., given a cover C:X-->Y, every point y in Y

    has a neighborhood Oy of y with p^-1(Oy)~ Oy x F, where F is the fiber. This

    seems confusing, in that fibers of covering maps are a (discrete) collection of points

    in X (since local diffeomorphisms are local bijections, the preimages are isolated/discrete)

    . Does this statement just mean that the fiber is just a disjoint/discrete

    collection of open sets, indexed by the cardinality of the fiber, or is there more than

    that to it? I am thinking of standard examples like the covering map f:R-->$S^1$,

    given by f(t)=(cost,sint), an infinite-to-one cover; would the Oy here be any 'hood

    (neighborhood) of y that evenly-covers y?

    Also: if this local triviality holds: is every covering map a bundle

    of C over X with singletons as fibers?

    Sorry, I know this is simple, but I haven't seen it in a while, and hopefully someone's

    comments will jog my memory.
     
  2. jcsd
  3. Jun 18, 2011 #2

    quasar987

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    If Oy is evenly covered, then p^{-1}(Oy) is a disjoint union of sets homeomorphic to Oy and there are N of them, where N is the number of sheets of the covering (not necessarily an integer). But it seems indeed that p^{-1}(Oy) can also be seen as a product Oy x S, where S is any set of cardinality N equipped with the discrete topology, thus making a covering space into a locally trivial fiber bundle with discrete fibers.
     
  4. Jun 18, 2011 #3

    mathwonk

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    they have it so you can lift maps of curves into the base.
     
  5. Jun 19, 2011 #4

    WWGD

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    Hmm.. that is interesting; every covering space is (can be made into) a bundle with discrete fiber. Is the opposite also the case, that every bundle is a covering space? It would seem so, but then what is the difference between the two?
     
  6. Jun 19, 2011 #5

    quasar987

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    Do you mean that or "is every bundle with discrete fiber a covering space"?
     
  7. Jun 19, 2011 #6

    lavinia

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    Yes. It just means that there are F disjoint homeomorphic copies of the open set in the covering space.

    I get a better picture of this when I think of the covering space and a group of homeomorphism of it that wrap it up onto itself. The homeomorphisms act properly discontinuously, that is every point has a neighborhood that is mapped to another neighborhood that is disjoint from it. These are the neighborhoods in the preimage of a neighborhood in the quotient space, the same neighborhood you asked about.
     
    Last edited: Jun 19, 2011
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