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My Algebraic Topology book says the following:

For $\Bbb{S}^n$, $H_p(\Bbb{S}^n)=\Bbb{Z}$ for $p=\{0,n\}$, and $H_p(\Bbb{S}^n)=0$ otherwise.

Also, by Mayer-Vietoris, $H_p(\Bbb{S}^n)\cong H_p(\Bbb{S}^{n-1})$.

How can both be true? Shouldn't $H_{n-1}(\Bbb{S}^n)=0$ and $H_n(\Bbb{S}^n)=\Bbb{Z}$?

Community
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    I'm sure it's just a typo, and should read $H_{p-1}(S^{n-1})$. You get the associated Mayer-Vietoris sequence by considering fattened upper and lower hemispheres of $S^n$. Their intersection is essentially the equator, which is $S^{n-1}$. – Josh Keneda Jun 02 '16 at 01:32
  • @JoshKeneda- You're right. Thanks! –  Jun 02 '16 at 01:33

1 Answers1

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This is a typo. It means to say $H_p(S^n)\cong H_{p-1}(S^{n-1})$. (Actually, this is only true if you used reduced homology, or require $p>1$.)

Eric Wofsey
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  • You're right. Thanks! –  Jun 02 '16 at 01:33
  • @Ayush Khaitan, Please consider accepting this answer if it solved your problem. That way the system won't consider this as an unanswered questions. – R_D Jun 02 '16 at 03:04