# Thread: Showing Sn\Sm is homotopy equivalent to Sn-m-1

1. ## Showing Sn\Sm is homotopy equivalent to Sn-m-1

Consider $S^m$ embedded in $S^n (m < n)$ as the subspace $\{(x_1,x_2,...,x_{m+1},0,....,0) | {\sum x_i^2 = 1}\}$. Show $S^n$ \ $S^m$ is homotopy equivalent to $S^{n-m-1}$.

If we consider m=1, n=2, then according to the theorem we're trying to prove, our quotient space should be homotopy equivalent to a point. However from our definition, $S^1$ embedded in $S^2$ is $\{(x_1,x_2,0) | {\sum x_i^2 = 1}\}$, i.e. the circle on the x-y plane. I thought the quotient space meant "shrinking all the points in the same equivalence class into a point", so our quotient space ends up more like $S^2 \wedge S^2$? So my first question is, is the question giving the right definition of embedding for us to prove the proposition?

In the case of m=1, n=2, if we use an embedding definition based on rotation about the z-axis (e.g. using spherical polars, points are in the same equivalence class if they have the same radius and elevation angle from x-y plane), we can see very clearly that $S^2$ \ $S^1$ is [-1,1]. Even if the embedding definition is correct in question, maybe there is an easier one to work with?

I probably have completely misunderstood the question, would love to know my mistake. Any comments are welcome.

2. ## Re: Showing Sn\Sm is homotopy equivalent to Sn-m-1

Wow, it took me so long to find out what is wrong with this - I thought I was taking a quotient space, but I actually need to remove the points of the embedding, and then can explicitly construct a homotopy. Well that was dumb of me....