Let \(f: (X, A) \to (Y, B)\) be a map such that both \(f: X \to Y\) and the restriction \(f: A \to B\) are homotopy equivalences.
Show that \(f_*: H_n(X, A) \to H_n(Y, B)\) is an isomorphism for all \(n\).
For the case of the inclusion \(f: (D^n, S^{n-1}) \hookrightarrow (D^n, D^n - \{0\})\), show that \(f\) is not a homotopy equivalence of pairs — there is no \(g: (D^n, D^n - \{0\}) \to (D^n, S^{n-1})\) such that \(fg\) and \(gf\) are homotopic to the identity through maps of pairs. [Observe that a homotopy equivalence of pairs \((X, A) \to (Y, B)\) is also a homotopy equivalence for the pairs obtained by replacing \(A\) and \(B\) by their closures.]
Solution
(a).
1. \(f\) induces a map of long exact sequences of the pairs \((X, A)\) and \((Y, B)\).
functoriality of the long exact sequence of a pair.
2. \(f_* : H_n(X) \to H_n(Y)\) and \(f_* : H_n(A) \to H_n(B)\) are isomorphisms for all \(n\).
hypothesis (both \(f : X \to Y\) and \(f : A \to B\) are homotopy equivalences).
3. By the five lemma applied to the map of long exact sequences, \(f_* : H_n(X, A) \to H_n(Y, B)\) is an isomorphism for all \(n\).
1 and 2 (the five lemma).
(b).
1. \(f : D^n \to D^n\) is a homotopy equivalence (it is the identity), and \(f : S^{n-1} \to D^n - \{0\}\) is a homotopy equivalence (both are homotopy equivalent to \(S^{n-1}\)).
\(D^n\) is contractible, and \(D^n - \{0\}\) deformation retracts onto \(S^{n-1}\).
2. But \(f\) is not a homotopy equivalence of pairs.
suppose \(g : (D^n, D^n - \{0\}) \to (D^n, S^{n-1})\) were a homotopy inverse of pairs.
3. Then \(g\) would restrict to a map \(D^n - \{0\} \to S^{n-1}\) that is a homotopy inverse to the inclusion \(S^{n-1} \hookrightarrow D^n - \{0\}\).
2 (homotopy inverse of pairs restricts to homotopy inverse on the subspaces).
4. This would give a retraction \(D^n - \{0\} \to S^{n-1}\) (or, after the homotopy, a retraction of \(D^n\) onto \(S^{n-1}\)), which is impossible.
\(S^{n-1}\) is not a retract of \(D^n\) (e.g. \(H_{n-1}(S^{n-1}) = \ZZ\) does not split off \(H_{n-1}(D^n) = 0\)).
5. Hence \(f\) is not a homotopy equivalence of pairs.
4.
6. Q.E.D.
3 (a) and 5 (b).