Local Trivialization in Covering Spaces

  • Thread starter Bacle
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  • #1
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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.
 

Answers and Replies

  • #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.
 
  • #3
mathwonk
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they have it so you can lift maps of curves into the base.
 
  • #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?
 
  • #5
quasar987
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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?
Do you mean that or "is every bundle with discrete fiber a covering space"?
 
  • #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.
 
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