1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

S^n not a mapping cylinder. S^n and homeom. subspaces

  1. Nov 21, 2008 #1


    User Avatar
    Science Advisor
    Gold Member

    Hi, everyone:

    I have been trying to show this using the following:

    Given f: Y-->X

    IF S^n ~ Y_f(x) , then S^n deformation-retracts to Y , and ( not sure of this)

    also is homeomorphic to Y (I know Y_f(x) is homotopic to Y ) . But ( so I am

    branching out into more sub-problems) S^n is not homeomorphic to any of its

    subspaces : if f:S^n -->Z , Z<S^n is a homeomorphism, then Z is compact,

    and, by Invariance of Domain, Z is also open. Then , by connectedness, Z=S^n.

    Anyway. I also have --tho I am not sure if this helps -- that , I think that

    the only mapping cylinder that is a manifold is the identity i: M-->M , with

    M a manifold. If this is true, this means that f:X-->Y as above has X=Y =S^n

    Any Other Ideas?

  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: S^n not a mapping cylinder. S^n and homeom. subspaces
  1. Homology of S^n x R (Replies: 6)

  2. Deg 1 map S^3 -> RP(3) ? (Replies: 20)