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Consider the pair of topological space $(\mathbb{D}^n,X)$, where $X \subset \mathbb{D}^n$ is a subspace. We know that there is a long exact sequence of reduced homology groups,

$$\cdots \to \tilde{H}_q(X) \to \tilde{H}_q(\mathbb{D}^n) \to H_q(\mathbb{D}^n,X) \to \tilde{H}_{q-1}(X) \to \tilde{H}_{q-1}(\mathbb{D}^n) \to \cdots.$$

However, since $\mathbb{D}^n$ is contractible, $\tilde{H}_q(\mathbb{D}^n) \cong 0$ for all $q$. So, we have $H_q(\mathbb{D}^n,X) \cong \tilde{H}_{q-1}(X)$ by exactness of the sequence? And this works for all subspace $X$?

Edit: fixed an error where I had $\tilde{H}_{q}(X)$ in places instead of the correct $\tilde{H}_{q-1}(X)$.

Jean Valjean
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You have an index error as Ted mentioned in the comments, and so really the conclusion you should have come to is $\tilde{H}_q(\mathbb{D}^n,X)\cong \tilde{H}_{q-1}(X)$. This works for all subspaces $X$ (in particular for $X=S^{n-1}$ which is a useful result when calculating the homology of a general sphere).

Dan Rust
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  • Yes, that was a typo. Thanks! +1 and accepted. Trying to work through a set of notes on this and was wondering if having it work for all subspaces $X$ is "too powerful" thus wrong. – Jean Valjean Jul 13 '14 at 02:05