Let \(CA\) be the cone on \(A\), show that \(\tilde H_*(X \union CA) \cong \tilde H_*(X, A)\).
Solution
Goal: Let \((X, A)\) be a topological pair. Let \(CA = (A \times [0, 1]) / (A \times \{0\})\) be the cone on \(A\), attached to \(X\) along \(A \times \{1\} \equiv A \subseteq X\), so \(X \cup CA = X \cup_A CA\). Prove that \(\widetilde{H}_n(X \cup CA) \cong H_n(X, A)\) for all \(n \ge 0\).
1. Apply the long exact sequence of the pair \((X \cup CA, CA)\).
1.1. For the pair \((Y, B) = (X \cup CA, CA)\), the long exact sequence in reduced homology is: \[\cdots \to \widetilde{H}_n(CA) \xrightarrow{i_*} \widetilde{H}_n(X \cup CA) \xrightarrow{j_*} H_n(X \cup CA, CA) \xrightarrow{\partial} \widetilde{H}_{n-1}(CA) \to \cdots\]
1.2. The cone \(CA\) is contractible to the cone vertex \(v = [A \times \{0\}]\) via \(H([(a, t)], s) = [(a, (1-s)t)]\).
1.3. Since \(CA\) is contractible, its reduced homology vanishes in all degrees: \(\widetilde{H}_n(CA) = 0\) for all \(n \ge 0\).
1.4. The exact sequence becomes \(0 \to \widetilde{H}_n(X \cup CA) \xrightarrow{j_*} H_n(X \cup CA, CA) \to 0\), which gives an isomorphism: \[j_* \colon \widetilde{H}_n(X \cup CA) \xrightarrow{\cong} H_n(X \cup CA, CA) \quad \text{for all } n \ge 0.\]
1.5. Substituting \(\widetilde{H}_n(CA) = 0\) and \(\widetilde{H}_{n-1}(CA) = 0\) into the long exact sequence leaves the short exact sequence \(0 \to \widetilde{H}_n(X \cup CA) \to H_n(X \cup CA, CA) \to 0\), whose middle map is an isomorphism.
2. Apply excision to relate \(H_n(X \cup CA, CA)\) to \(H_n(X, A)\).
2.1. Let \(q:A\times[0,1]\to CA\) be the cone quotient, with \(q(A\times\{0\})\) the cone vertex and \(q(a,1)\) identified with \(a\in A\subseteq X\). Set\[ U=q\bigl(A\times[0,1/2)\bigr)\subseteq CA\subseteq X\cup_A CA. \]
2.2. The set \(U\) is open in \(X\cup_A CA\), and its closure is contained in\[ q\bigl(A\times[0,1/2]\bigr), \]which lies in the interior of \(CA\) inside \(X\cup_A CA\) because it is disjoint from the attaching locus \(q(A\times\{1\})=A\).
The preimage \(A\times[0,1/2)\) is open in \(A\times[0,1]\) and is saturated for the cone quotient, hence its image is open in \(CA\); since it is disjoint from the attaching locus, it is also open in the adjunction space. Likewise \(q(A\times(1/2,1])\) is open, so the closure of \(U\) is contained in \(q(A\times[0,1/2])\). Every point with cone coordinate at most \(1/2\) has a neighborhood contained in the cone part and away from the attaching locus.
2.3. Excision therefore gives an isomorphism\[ H_n\bigl((X\cup_A CA)\setminus U,\,CA\setminus U\bigr) \xrightarrow{\cong} H_n(X\cup_A CA,CA). \]
The closure of the excised open set \(U\) is contained in the interior of the subspace \(CA\), exactly the hypothesis of the excision theorem for the pair \((X\cup_A CA,CA)\).
2.4. The complementary pair is\[ \bigl((X\cup_A CA)\setminus U,\,CA\setminus U\bigr) = \Bigl(X\cup_A q\bigl(A\times[1/2,1]\bigr),\,q\bigl(A\times[1/2,1]\bigr)\Bigr). \]
2.5. This pair deformation retracts onto \((X,A)\) by moving the cone coordinate linearly from \(t\in[1/2,1]\) to \(1\) while fixing \(X\) pointwise.
For \(s\in[0,1]\), send \(q(a,t)\) to \[ q\bigl(a,(1-s)t+s\bigr) \] and fix every point of \(X\). At \(t=1\) this agrees with the attaching identification \(q(a,1)=a\), so the homotopy is well-defined on the adjunction space. It preserves the cone-collar subspace and at \(s=1\) sends that subspace onto \(A\) while fixing \((X,A)\) pointwise.
2.6. Hence\[ H_n(X,A)\xrightarrow{\cong} H_n\bigl((X\cup_A CA)\setminus U,\,CA\setminus U\bigr) \xrightarrow{\cong} H_n(X\cup_A CA,CA). \]
The first isomorphism is induced by the deformation retraction in 5 and the second by excision in 3.
3. Combine isomorphisms.
3.1. From 1 and 2: \[\widetilde{H}_n(X \cup CA) \cong H_n(X \cup CA, CA) \cong H_n(X, A) \quad \text{for all } n \ge 0.\]
3.2. The isomorphism of 1 and the isomorphism of 2 compose to give \(\widetilde{H}_n(X \cup CA) \cong H_n(X, A)\).
4. Q.E.D.
4.1. Steps <1>1–<1>3 establish \(\widetilde{H}_*(X \cup CA) \cong H_*(X, A)\).