Homology of S^n x R: Hatcher's Theorem 2B.1

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In summary, 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. The statement of the theorem states that \widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R}) is Z if i=n-1 and 0 otherwise, and this can be shown using a deformation retract argument or the Kunneth formula.
  • #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?
 
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  • #2
quasar987 said:
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}_*(\mathbb{S}^{n-1} \times \mathbb{R})=0[/tex]?

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.
 
  • #3
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.
 
  • #4
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.
 
  • #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.
 
  • #6
That's a lot better an explanation than mine.
 
  • #7
quasar987 said:
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.

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.
 

1. What is the Homology of S^n x R?

The Homology of S^n x R is a mathematical concept that describes the topological space formed by the Cartesian product of the n-dimensional sphere and the real line. It is often denoted as H_*(S^n x R) and is used to study the structure and properties of this space.

2. What is Hatcher's Theorem 2B.1?

Hatcher's Theorem 2B.1 is a theorem in algebraic topology that provides a formula for calculating the homology groups of the product space S^n x R. It states that the homology group H_k(S^n x R) is isomorphic to the direct sum of the homology groups of S^n and R, except for the case of k=n, where there is an additional factor of Z.

3. How is Hatcher's Theorem 2B.1 used?

Hatcher's Theorem 2B.1 is used to simplify the calculation of the homology groups of S^n x R. By breaking down the product space into its individual components and applying the theorem, one can determine the homology groups more easily and efficiently.

4. What are the applications of Homology of S^n x R?

The Homology of S^n x R has various applications in mathematics and science. It is used in algebraic topology to study the structure of topological spaces and to classify them. It also has applications in physics, particularly in the study of gauge theories and string theory.

5. Are there any limitations to Hatcher's Theorem 2B.1?

Like any mathematical theorem, Hatcher's Theorem 2B.1 has its limitations. It only applies to the specific case of the product space S^n x R and cannot be generalized to other spaces. Additionally, it assumes that the homology groups of S^n and R are known, which may not always be the case.

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