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

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SUMMARY

The discussion centers on the properties of the n-sphere, S^n, particularly its relationship with mapping cylinders and homeomorphic subspaces. It is established that if S^n is homeomorphic to a space Y_f(x), then S^n deformation-retracts to Y, indicating a strong topological connection. The participant asserts that S^n is not homeomorphic to any of its subspaces, emphasizing that if a mapping f: S^n --> Z results in Z being a homeomorphism, then Z must be compact and open, leading to the conclusion that Z equals S^n. Additionally, it is proposed that the only mapping cylinder that qualifies as a manifold is the identity mapping.

PREREQUISITES
  • Understanding of topological spaces and homeomorphisms
  • Familiarity with the concepts of deformation retracts and homotopy
  • Knowledge of the Invariance of Domain theorem
  • Basic principles of manifold theory
NEXT STEPS
  • Explore the properties of mapping cylinders in topology
  • Study the Invariance of Domain theorem in detail
  • Investigate the concept of deformation retracts and their implications
  • Learn about the classification of manifolds and their homeomorphic properties
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Mathematicians, topologists, and students studying algebraic topology, particularly those interested in the properties of spheres and manifold theory.

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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.
 

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