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Topology of punctured plane vs topology of circle?

  1. Apr 12, 2013 #1
    So how does the topology of R^n minus the origin relate to that of the (n-1)-dimensional sphere?

    I would think the topology of the former is equivalent to that of an (n-1)-dimensional sphere with finite thickness, and open edges. But I suppose that is as close as one can get to the (n-1)-dimensional sphere itself? Hm, they seem pretty similar, but not the same... Is there a way in which they are approximately the same topology-wise?
     
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  3. Apr 12, 2013 #2

    micromass

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    Check out homotopy equivalence and deformation retractions.
     
  4. Apr 13, 2013 #3

    Bacle2

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    Yes; a homotopy induces an isomorphism of fundamental groups and homology groups:

    by functoriality of induced maps , if fg=IdX and gf=IdY , then

    (fg)* =f*g* =IdX* , and

    same idea for (gf)*. Then the induced maps are isomorphisms.

    Still, note that the cutsets of the two are different, so that they are not homeomorphic

    (tho maybe the fact that the standard map from R^2\{0| is infinite-to-one may give a clue

    of this..
     
  5. Apr 16, 2013 #4
    Hey, thank you both. Micromass' suggestion made me look up stuff that indirectly made me understand Bacle's post (and more). The homotopy equivalence of spaces is what I was looking for :)
     
  6. May 14, 2013 #5

    mathwonk

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    one has the same topology as the product of the other with the real line.
     
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