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.