The S^infinity sphere

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Hi,

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!
 

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  • #2
Hurkyl
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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...
 
  • #3
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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.
 
  • #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.
 
  • #5
mathwonk
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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)?
 
  • #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).
 
  • #7
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"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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