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Using Jordan Curve Thm to Show H_n(R^n) Trivial?

  1. Aug 13, 2011 #1

    WWGD

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    Hi, All:

    I hope I am not missing something obvious: can't we use the Jordan Curve Thm. to show
    that the homology H_n(R^n) of R^n is trivial ? How about showing that Pi_n(R^n) is trivial?

    It seems like the def. of cycles in a space X is geenralized by continuous , injective maps f: S^n -->X . When X=R^n, JCT says that f(S^n) separates R^n into 2 regions, which can be seen as saying that f(S^n) bounds, so that every cycle bounds, and then the homology is trivial.
     
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  3. Aug 13, 2011 #2
    Using the generalized Jordan curve theorem is massively overcomplicating it. All the homology groups of R^n greater than the zeroth are trivial, because R^n is contractible and homology groups remain unchanged under homotopy equivalence. Likewise with the homotopy groups.
     
  4. Aug 13, 2011 #3

    WWGD

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    I understand; I know R^n is contractible, and we can use homotopy equivalence, etc., but I am trying to see if all n-cycles can be represented as images of spheres, and if the JCT actually says that these cycles bound, i.e., the interior region into which the simple-closed curve separates R^n, is the(an) object being bounded.
     
  5. Aug 13, 2011 #4
    Well, obviously not all cycles can be represented as the image of a sphere. Just take the boundary of two disjoint n+1-simplices in the space -- that boundary is not even connected, hence cannot be the image of a sphere.

    As for the other question -- yes, the image of the sphere is always the boundary of the inside, but the inside is not always homeomorphic to a ball. To see this, consider an Alexander horned sphere with the origin on the inside. The outside is not simply connected, but the inside is. Now consider the image of the horned sphere under the map v ↦ v/|v|². This exchanges the inside and the outside, so now the inside is not simply connected and hence is not homeomorphic to a ball.
     
  6. Aug 13, 2011 #5

    WWGD

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    But it seems like you could do away with the first objection by saying that a cycle is the
    union of continuous images of S^n .
     
  7. Aug 13, 2011 #6

    WWGD

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    I guess I am unclear about the geometric definition of a cycle; I guess this would

    depend on the choice of homology we make, but, given that these theories are all

    equivalent ( i.e., they output isomorphic groups for the same space), the choice should

    be independent of choice of homology.
     
  8. Aug 13, 2011 #7
    I think Citan just gave an example of a cycle that is not the image of S^n with

    Alexander's horned sphere, as a space that is bound by a 2-dimensional object

    that is not the image of a sphere.
     
  9. Aug 14, 2011 #8
    And, with respect to cycles, I would say that , in the most general sense, an n- cycle would
    be an n-dimensional subspace that can be oriented (if the subspace is triangulable and can be made into a simplicial complex, then the net boundary should be zero); maybe others here can double-check.
     
  10. Aug 15, 2011 #9
    First, check the dimensions on the jordan curve theorem. Second, maps from the sphere are most definitely not a generalization, since they're a very special case of maps from the n-simplex.
     
  11. Aug 15, 2011 #10
    Third, I'd be interested in seeing a proof of the generalized Jordan curve theorem for continuous maps that doesn't render this circular: I.e. Showing that it separates without knowing anything about the homology of the sphere or Euclidean space.
     
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