For a pair \((X, A)\), let \(X \cup CA\) be \(X\) with a cone on \(A\) attached.
Show that \(X\) is a retract of \(X \cup CA\) iff \(A\) is contractible in \(X\): There is a homotopy \(f_t: A \to X\) with \(f_0\) the inclusion \(A \hookrightarrow X\) and \(f_1\) a constant map.
Show that if \(A\) is contractible in \(X\) then \(H_n(X, A) \approx \tilde{H}_n(X) \oplus \tilde{H}_{n-1}(A)\), using the fact that \((X \cup CA)/X\) is the suspension \(SA\) of \(A\).
Solution
Set \[ Y=X\cup CA. \]
1. If there is a retraction \(r:Y\to X\), then \(A\) is contractible in \(X\).
Parametrize the cone as \[ CA=A\times[0,1]/(A\times\{1\}), \] with \(A=A\times\{0\}\). Restrict \(r\) to the cone and define \[ f_t(a)=r([a,t]). \] Since \(r|_X=\operatorname{id}_X\), one has \[ f_0(a)=a. \] At \(t=1\) all points \([a,1]\) are the cone vertex, so \(f_1\) is constant. Thus \(f_t\) contracts the inclusion \(A\hookrightarrow X\) to a constant map inside \(X\).
2. Conversely, if the inclusion \(A\hookrightarrow X\) is homotopic in \(X\) to a constant map, then \(X\) is a retract of \(Y\).
Let \[ f_t:A\to X, \qquad f_0(a)=a,\quad f_1(a)=x_0, \] be such a homotopy. Define \(r:Y\to X\) by \[ r|_X=\operatorname{id}_X, \qquad r([a,t])=f_t(a) \] on \(CA\). This is well defined at the cone vertex because \(f_1\) is constant, and the two definitions agree along \(A=A\times\{0\}\). Hence \(r\) is a continuous retraction.
This proves part (a).
Assume now that \(A\) is contractible in \(X\), so the retraction exists.
3. The pair \((Y,X)\) has\[ H_n(Y,X)\cong\widetilde H_n(Y/X)\cong\widetilde H_n(SA)\cong\widetilde H_{n-1}(A). \]
Collapsing \(X\) in \(Y=X\cup CA\) leaves \[ CA/A=SA. \] The quotient theorem for relative homology and the suspension isomorphism give the displayed identifications.
4. The retraction splits the long exact sequence of \((Y,X)\), giving\[ \widetilde H_n(Y)\cong \widetilde H_n(X)\oplus\widetilde H_{n-1}(A). \]
The inclusion \(i:X\hookrightarrow Y\) has a left inverse \(r\), so \(i_*\) is injective in every degree. Hence the connecting map \[ H_n(Y,X)\to H_{n-1}(X) \] has zero image, and the long exact sequence breaks into short exact sequences \[ 0\to\widetilde H_n(X)\xrightarrow{i_*}\widetilde H_n(Y) \to H_n(Y,X)\to0. \] The map \(r_*\) splits these sequences. Apply 3.
5. There is a natural isomorphism\[ H_n(X,A)\cong\widetilde H_n(Y). \]
The cone \(CA\) is contractible. Excision identifies \[ H_n(X,A)\cong H_n(Y,CA). \] Since \(CA\) is contractible and nonempty, the long exact sequence of \((Y,CA)\) gives \[ H_n(Y,CA)\cong\widetilde H_n(Y) \] (with the standard reduced interpretation in degree \(0\)).
Combining 4 and 5 yields \[ \boxed{H_n(X,A)\cong\widetilde H_n(X)\oplus\widetilde H_{n-1}(A).} \]