# Homology of S^n x R

Homework Helper
Gold Member
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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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.

Homework Helper
Gold Member
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.

matt grime
Homework Helper
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.

matt grime
Homework Helper
That's a lot better an explanation than mine.

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.