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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.

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

Can you offer guidance or do you also need help?

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