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Homology of S^n x R

  1. Mar 20, 2009 #1

    quasar987

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    Hi people.

    In the proof of theorem 2B.1 page 169-170 of Hatcher (generalized Jordan curve theorem), item (b) is proved by induction on k and the case k=0 is handled by noticing that S^n - h(S^0) is homeomorphic to S^(n-1) x R. How does he know that [tex]\widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R})[/tex] is Z if i=n-1 and 0 otherwise?
     
    Last edited: Mar 20, 2009
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  3. Mar 20, 2009 #2
    I'm not sure what homology you are using but H(S^n-1xR) is not zero if you mean Z homology. It is zero in dimension,n.
     
  4. Mar 20, 2009 #3

    quasar987

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    Excuse me, I meant

    "How does he know that [tex]\widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R})[/tex] is Z if i=n-1 and 0 otherwise?"
    (as in the statement of the theorem).

    And yes, we are talking about homology with coefficient in Z.
     
  5. Mar 21, 2009 #4

    matt grime

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    What is k? You're inducting on k, but there is no mention of k in the statement of the problem. Actually, have you stated the problem?

    A decent homology theory will turn products into graded tensor products of groups. I.e. the homology of (UxV) in degree n will be the direct sum (over all i) of H_i(U)xH_{n-i}(V) and presumably you've worked out the homology for S^n and R.
     
  6. Mar 21, 2009 #5
    [tex]S^{n-1}\times\mathbb{R}[/tex] is homotopy equivalent to [tex]S^{n-1}[/tex] by a deformation retraction, and the reduced homology of the sphere is already known.
     
  7. Mar 21, 2009 #6

    matt grime

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    That's a lot better an explanation than mine.
     
  8. Mar 21, 2009 #7
    S^n-1 x R has the same homology as S^n-1. the deformation retract argument given in this thread is correct.

    You could also use the Kunneth formula and the knowledge that the homology of R is zero except in dimension zero.
     
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