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redbowlover
Apr10-11, 12:40 PM
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
Apr15-11, 01:04 AM
H_{n+1}(S^n\times\mathbb{R},\mathbb{Z}) = 0 because \mathbb{R} is contractible.

lavinia
Apr15-11, 09:50 AM
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
Apr16-11, 10:59 AM
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.