# The S^infinity sphere

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!

## Answers and Replies

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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...

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.

matt grime
Science Advisor
Homework Helper
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.

mathwonk
Science Advisor
Homework Helper
2020 Award
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)?

matt grime
Science Advisor
Homework Helper
If it's mine then that is trivially the homotopy you need (since it is surely a limit in some appropriate sense).

"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).