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

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Discussion Overview

The discussion revolves around the application of Hatcher's Theorem 2B.1, specifically regarding the homology of the space \( S^n \times \mathbb{R} \) and the implications of the generalized Jordan curve theorem. Participants explore the homology groups of \( \mathbb{S}^{n-1} \times \mathbb{R} \) and the reasoning behind certain claims made in the theorem's proof.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question how Hatcher concludes that \( \widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R}) \) is \( \mathbb{Z} \) if \( i=n-1 \) and 0 otherwise.
  • Others argue that the homology of \( S^{n-1} \times \mathbb{R} \) is not zero in dimension \( n \) and clarify that they are discussing homology with coefficients in \( \mathbb{Z} \).
  • A participant points out that \( S^{n-1} \times \mathbb{R} \) is homotopy equivalent to \( S^{n-1} \) due to a deformation retraction, which supports the claim about its homology.
  • Another participant suggests using the Künneth formula and the known homology of \( \mathbb{R} \) to further understand the homology groups involved.
  • There is a request for clarification regarding the variable \( k \) mentioned in the induction process, as it is not defined in the problem statement.

Areas of Agreement / Disagreement

Participants express differing views on the homology of \( S^{n-1} \times \mathbb{R} \), with some asserting it is zero in certain dimensions while others provide arguments for its equivalence to \( S^{n-1} \). The discussion remains unresolved regarding the specific claims about the homology groups.

Contextual Notes

There is uncertainty regarding the definitions and assumptions related to the homology theories being discussed, particularly in relation to the dimensions and coefficients used. The role of the variable \( k \) in the induction process is also unclear.

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 \widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R}) is Z if i=n-1 and 0 otherwise?
 
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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 \widetilde{H}_*(\mathbb{S}^{n-1} \times \mathbb{R})=0?

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.
 
Excuse me, I meant

"How does he know that \widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R}) 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.
 
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.
 
S^{n-1}\times\mathbb{R} is homotopy equivalent to S^{n-1} by a deformation retraction, and the reduced homology of the sphere is already known.
 
That's a lot better an explanation than mine.
 
quasar987 said:
Excuse me, I meant

"How does he know that \widetilde{H}_i(\mathbb{S}^{n-1} \times \mathbb{R}) 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.
 

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