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I'm reading about relative homology group but I'm having hard time in understanding this concept.

So I was trying to find $H_1(D^n,S^{n-1})$, but I'm unable to solve this problem. Can someone give some idea for $n=1$ (Then I'll try to generalize it)?

Added By using long exact sequence as suggested in comment I get $H_n(D_n,S^{n-1})$ is isomorphic to $H_{n-1}(S^{n-1})$.How to proceed further?

Dontknowanything
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1 Answers1

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The long exact sequence of the pair (plus the fact $H_1(D^1) = 0$) gives us the exactness of

$0 \to H_1(D^1, S^0) \to H_0(S^0) \to H_0(D^1) \to H_0(D^1, S^0) \to 0$

Hence $H_1(D^1, S^0)$ is just the kernel of the map $H_0(S^0) \to H_0(D^1)$. Recall that (assuming integer coefficients) $H_0(S^0) \cong \mathbb{Z} \oplus \mathbb{Z}$ and $H_0(D^1) \cong \mathbb{Z}$. The application $H_0(S^0) \to H_0(D^1)$ is induced by the inclusion $S^0 \hookrightarrow D^1$. Hence in terms of generators it is the application $\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ given by $(x, y) \mapsto x + y$. Hence its kernel is $\cong \mathbb{Z}$, and thus we have established $H_1(D^1, S^0) \cong \mathbb{Z}$.

Pedro
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  • (+1),Can you please add for general $n$ too?Thank you! – Dontknowanything Sep 01 '15 at 19:34
  • The ideas are the same... what are you having trouble with? – Pedro Sep 01 '15 at 20:05
  • By using long exact sequence I got $H_n(D^n,S^{n-1})$ is isomorphic to $H_{n-1}(S^{n-1})$ then how to proceed?Do we need to use some induction argument? – Dontknowanything Sep 01 '15 at 20:08
  • You can use, for example, the Mayer-Vietoris sequence to show $H_{n-1}(S^{n - 1}) \cong \mathbb{Z}$. This is a well-known calculation which you can find in any algebraic topology textbook – Pedro Sep 01 '15 at 20:13
  • Or the fact that, since $(D^n, S^{n - 1})$ is a good pair, $H_n(D^n, S^{n - 1}) \cong \widetilde{H}_n(D^n / S^{n-1}) \cong \widetilde{H}_n(S^n)$ and so inductively calculate this. – Pedro Sep 01 '15 at 20:16