Reduced homology of sphere cross reals?

[redbowlover]

tex doesn't seem to be working right...sry for the notation.

Working through of a proof of the generalized jordan curve theorem. Keep getting stuck on calculating the reduced homology of S^n by R, (ie n-sphere cross the real line).


My book (hatcher) seems to imply its 0 except the n^th homology is Z.

But doesn't S^n cross R have dimension n+1? And shouldn't this imply the (n+1)th homology group is Z? Or is this only true of closed manifolds?

Any thoughts would be appreciated.

 

[fzero]

$$H_{n+1}(S^n\times\mathbb{R},\mathbb{Z}) = 0$$ because $$\mathbb{R}$$ is contractible.

 

[lavinia]

tex doesn't seem to be working right...sry for the notation.

Working through of a proof of the generalized jordan curve theorem. Keep getting stuck on calculating the reduced homology of S^n by R, (ie n-sphere cross the real line).


My book (hatcher) seems to imply its 0 except the n^th homology is Z.

But doesn't S^n cross R have dimension n+1? And shouldn't this imply the (n+1)th homology group is Z? Or is this only true of closed manifolds?

Any thoughts would be appreciated.

the homotopy of S^n x R into itself defined by ((s,r),t) -> (s,rt) deforms S^n x R onto S^n
keeping S^n fixed.

 

[Tinyboss]

But doesn't S^n cross R have dimension n+1? And shouldn't this imply the (n+1)th homology group is Z? Or is this only true of closed manifolds?

Orientable and closed, in the manifold case.

 

[blue_raver22]

It is possible that there may be an issue with the notation or calculations in your proof. However, it is also possible that the reduced homology of S^n cross R is indeed 0 for all dimensions except n, as suggested by Hatcher. This may be due to the fact that the n-sphere is contractible to a point, and therefore the cross product with the real line does not add any additional homology. This may only be true for closed manifolds, as you mentioned, since the open manifolds may have additional homology due to the boundary. It would be helpful to double check your calculations and potentially consult with other mathematicians or resources for further clarification.