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

  1. Nov 21, 2008 #1

    WWGD

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    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?

    Thanks.
     
  2. jcsd
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