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

  • Thread starter WWGD
  • Start date
  • #1
WWGD
Science Advisor
Gold Member
5,631
5,063
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.
 

Answers and Replies

Related Threads on S^n not a mapping cylinder. S^n and homeom. subspaces

Replies
2
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
8
Views
3K
Replies
1
Views
812
  • Last Post
Replies
5
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
13
Views
1K
Replies
4
Views
2K
Top