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

[WWGD]

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.

 

[pat_connell]

Yes, it helps to know that the only mapping cylinder that is a manifold is the identity map i: M-->M with M a manifold. This means that for f:X-->Y, X and Y both equal S^n. Now, if S^n deformation-retracts to Y, then it also follows that S^n is homeomorphic to Y. Recall that a deformation retract is a continuous map which contracts a topological space X onto a subspace A of X such that the inclusion map from A to X is homotopic to the identity map on X. In this case, since S^n deformation-retracts to Y, it follows that S^n is homeomorphic to Y. As for the second part of your question, it is true that if f:S^n-->Z, where Z<S^n is a homeomorphism, then Z is compact and open, and thus, by connectedness, Z=S^n.I hope this helps!