# Homologies of the pairs are same but they are not homotopy equivalent as pairs.

Consider the pairs $(\mathbb D^{n},S^{n-1})$ and $(\mathbb D^{n},\mathbb D^{n}-\{0\})$ ,clearly their homologies are same in each dimensions but these pairs are not homotopy equivalent.

Any homotopy equivalence $f:(X,A)\to(Y,B)$ induce a homotopy equivalence $f:(X,\bar A)\to (Y,\bar B)$.

If the pairs in the questions are homotopy equivalent then the pairs $(\mathbb D^{n},S^{n-1})$ and $(\mathbb D^{n},\mathbb D^{n})$ are homotopy equivalent which is certainly not true as their homologies are different at $n$th dimension.

Any comment or discussion will be appreciated.

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But $D^{n}-\{0\}$ is homotopy equivalent to $S^{n-1}$! It's $D^{n}$ which is not homotopy equivalent to $S^{n-1}$...and, naturally, $D^{n}-\{0\}$ is not homotopy equivalent to $D^{n}$. I think this should clear up the matter... – Alex Nelson Dec 23 '12 at 21:19

Sorry yes I misread your problem, indeed your solution is correct. From the LES of the pair $(D^n, S^{n-1})$ we get that $H_n(D^n, S^{n-1}) \cong H_{n-1}(S^{n-1}) \cong \Bbb{Z}$ while in general it is always true that we have $H_n(X,X) = 0$ and so $H_n(D^n,D^n ) = 0$ which means to say that the original pairs that you started out with cannot possibly be homotopy equivalent.
:Since nth relative homology groups of $(D^{n},S^{n−1})$ and $(D^{n},D^{n})$ are different,therefore these two pairs in the question are not homotopy equivalent.I am using here: homotopy equivalent pairs will have same relative homology.Sorry,I did not get your argument. – Shraddha Srivastava Dec 23 '12 at 23:40