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

by WWGD
Tags: curve, hnrn, jordan, trivial
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WWGD
#1
Aug13-11, 12:05 PM
P: 578
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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Citan Uzuki
#2
Aug13-11, 12:44 PM
P: 274
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.
WWGD
#3
Aug13-11, 12:46 PM
P: 578
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.

Citan Uzuki
#4
Aug13-11, 01:07 PM
P: 274
Using Jordan Curve Thm to Show H_n(R^n) Trivial?

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|. This exchanges the inside and the outside, so now the inside is not simply connected and hence is not homeomorphic to a ball.
WWGD
#5
Aug13-11, 01:24 PM
P: 578
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 .
WWGD
#6
Aug13-11, 01:32 PM
P: 578
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.
Bacle
#7
Aug13-11, 01:38 PM
P: 662
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.
Bacle
#8
Aug14-11, 12:10 PM
P: 662
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.
zhentil
#9
Aug15-11, 06:50 PM
P: 491
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.
zhentil
#10
Aug15-11, 07:15 PM
P: 491
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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