Local Trivialization in Covering Spaces

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SUMMARY

The discussion centers on the local triviality condition of covering maps in topology, specifically the requirement that for a covering map \( p: C \to Y \), every point \( y \) in \( Y \) has a neighborhood \( O_y \) such that \( p^{-1}(O_y) \) is homeomorphic to \( O_y \times F \), where \( F \) represents the fiber. The participants clarify that fibers in covering maps are discrete collections of points, and they explore the relationship between covering spaces and fiber bundles. It is established that every covering space can be viewed as a locally trivial fiber bundle with discrete fibers, and the discussion raises the question of whether every bundle with discrete fibers qualifies as a covering space.

PREREQUISITES
  • Understanding of covering maps in topology
  • Familiarity with fiber bundles and their properties
  • Knowledge of homeomorphisms and their role in topology
  • Basic concepts of neighborhoods in topological spaces
NEXT STEPS
  • Study the properties of covering maps in algebraic topology
  • Learn about the relationship between fiber bundles and covering spaces
  • Investigate the role of homeomorphisms in the context of covering spaces
  • Explore examples of covering maps, such as \( f: \mathbb{R} \to S^1 \)
USEFUL FOR

Mathematicians, particularly those specializing in topology, algebraic topology, and differential geometry, will benefit from this discussion, as well as students seeking to deepen their understanding of covering spaces and fiber bundles.

Bacle
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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.
 
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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.
 
they have it so you can lift maps of curves into the base.
 
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?
 
WWGD said:
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"?
 
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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