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The S^infinity sphere

  1. Feb 9, 2007 #1

    I was able to enumerate all the subcomplexes of S^{infinity}, where S^{infinity} has two 0-cells, two 1-cells, two 2-cells, etc. But how do I show that S^{infinity} is contractible?

    Can anyone point me in the right direction? X is contractible if and only if the identity map of X is homotopic to a constant map.

    I guess I don't see what kind of homotopy (a shrinking map) I need to set up between X and a point.

    Thank you!
  2. jcsd
  3. Feb 9, 2007 #2


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    My best guess would be to construct your homotopy piece wise.

    Let P be a point.

    For each of your 0-cells, can you find a homotopy that maps your 0-cell to P?

    For each of your 1-cells, can you find a homotopy that maps your 1-cell to P, that restricts to the aforementioned homotopies on the two endpoints?

    For each of your 2-cells...
  4. Feb 11, 2007 #3
    I worked on it yesterday and I think the following might work: choose an arbitrary point p =(x_1, x_2, x_3,...) in S^infinity. Then show that at time t=1, this point is located at q =(y_1,0,0,0,0,....). This method should work.

    The above advice should work as well (I thought about it before), but remember that we have two 0-cells, two 1-cells, two 2-cells, etc. We know that the unit circle S^1 is not contractible, and neither does the unit sphere. So say we assume that S^infinity = S^2. Then if we look at one cell at a time, sure, we can find a homotopy between a cell and a point, but if we look at S^2 as a whole, it certainly is not contractible.

    So I'm not sure if looking at each of the cells and then find a contracting map works, otherwise, this will imply that S^1, S^2, S^3, etc. are contractible. But this certainly is not true.
  5. Feb 11, 2007 #4

    matt grime

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    Your reasoning about 'cell at a time' is flawed - there is nothing that says that the map must stay within the cell at each point ("in time"). In fact the homotopy *must* map each cell to the point.
  6. Feb 11, 2007 #5


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    what is clear is that each copy of S^n shrinks to a point in S^(n+1).

    what you must do is see whetehr this allows you to write down a global homotopy inductively.

    by the way what is your definition of S(infinity)?
  7. Feb 11, 2007 #6

    matt grime

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    If it's mine then that is trivially the homotopy you need (since it is surely a limit in some appropriate sense).
  8. Feb 12, 2007 #7
    "what is clear is that each copy of S^n shrinks to a point in S^(n+1)."

    That makes sense. However, writing down the global homotopy inductively might be a little difficult, I think.

    The definition of S^{infinity} is the union of S^n, where n=0,1,2,3,..., where each of the subspheres S^k is a subcomplex. It is obtained inductively from the equatorial S^(k-1) by attaching two k-cells, which are the components of S^k - S^(k-1).
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