$i: (D^n,S^{n-1}) \mapsto (D^n,D^n\setminus \{0\})$ is not a homotopy equivalence of pairs Here $i$ is the inclusion map, $D$ denotes the disk and $S$ denotes the sphere. I would be interested in an answer using homology of pairs.
My idea was to assume that $i$ is a homotopy equivalence of pairs, which implies that the $H_n(D^n,S^{n-1})$ is isomorphic to $H_n(D^n,D^n\setminus \{0\})$. Then by calculating each homology group separately I hope to get a contradiction.
Edit: The original idea was almost the right one. The Homology groups of the pairs are isomorphic. However, the homotopy equivalence of pairs also induces isomorphisms of slightly modified pairs, which then imply the contradiction. I provided a detailled answer.
 A: Assume for contradiction that $i:(D^n, S^{n-1}) \to (D^n,D^n\setminus \{0\})$ is a homotopy equivalence of pairs. Then $i$ also defines a homotopy equialence of the pairs $(D^n, S^{n-1})=(D^n, \overline{S^{n-1}}) \to (D^n,\overline{D^n\setminus \{0\}})=(D^n,D^n),$ (see this question).
 Thus $(D^n, S^{n-1})$ and $(D^n,D^n)$ have isomorphic homology groups.
But this is a contradiction:
$H_n(D^n,D^n)=0$: This follows directly from the definition of relative homology. Alternatively use the long exact sequence of the pair $(D_n,D_n)$
On the other hand, $H_n(D^n,S^{n-1})\neq 0$:
Either use that $H_n(D^n,S^{n-1})=H_n(D^n/S^{n-1})=H^n(S^n)\neq0$ or alternatively use the long exact sequnce of the pair $(D^n,S^{n-1})$ to conclude $H_n(D^n,S^{n-1})=H_{n-1}(S^{n-1})\neq 0$
A: Let me expand on one of the users' comment: this can be seen more easily by noticing that if you have a pair of maps $(D_n,D_n-0)\rightarrow (D_n,S^{n-1})$, then to it is impossible to make this a continuous map because $0\in D_n$ has no where to be mapped to.
