Results 1 to 2 of 2

Math Help - Showing Sn\Sm is homotopy equivalent to Sn-m-1

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    5

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Aug 2011
    Posts
    5

    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....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Verify that R^3\{2 points} is homotopy equivalent to S^2 V S^2
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 22nd 2011, 12:02 PM
  2. Composition of homotopy equivalences is a homotopy equivalence.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 30th 2011, 08:33 PM
  3. showing to statments are equivalent
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 23rd 2011, 02:42 AM
  4. homotopy
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: July 23rd 2010, 05:09 AM
  5. homotopy
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 10th 2010, 05:20 AM

Search Tags


/mathhelpforum @mathhelpforum