[proofplan] We prove the isomorphism $H_n(X, A) \cong \tilde{H}_n(X/A)$ in three stages. First, we use the good pair hypothesis to find a neighbourhood $U$ of $A$ that deformation-retracts onto $A$, and apply the five lemma to show $H_n(X, A) \cong H_n(X, U)$. Second, we excise $A$ from $(X, U)$ to obtain $H_n(X, U) \cong H_n(X \setminus A, U \setminus A)$. Third, we observe that the quotient map $q: X \to X/A$ identifies $(X, U)$ with $(X/A, U/A)$ and $(X \setminus A, U \setminus A)$ with $(X/A \setminus \{*\}, U/A \setminus \{*\})$, and a symmetric argument on the quotient side gives $H_n(X/A, U/A) \cong H_n(X/A, \{*\}) = \tilde{H}_n(X/A)$. [/proofplan]

[step:Show $H_n(X, A) \cong H_n(X, U)$ using the five lemma]Since $(X, A)$ is a good pair, there exists an open neighbourhood $U$ of $A$ in $X$ such that $A$ is a deformation retract of $U$. Let $r: U \to A$ be the retraction and $H: U \times [0,1] \to U$ the deformation retract homotopy with $H(\cdot, 0) = \operatorname{id}_U$, $H(\cdot, 1) = i \circ r$, and $H(a, t) = a$ for all $a \in A$, $t \in [0,1]$, where $i: A \hookrightarrow U$ is the inclusion. The inclusion of pairs $(X, A) \hookrightarrow (X, U)$ induces a map on long exact sequences. Consider the ladder: \begin{align*} \cdots \to H_n(A) \xrightarrow{i_*} H_n(X) \xrightarrow{} H_n(X, A) \xrightarrow{\partial} H_{n-1}(A) \to \cdots \\ \cdots \to H_n(U) \xrightarrow{j_*} H_n(X) \xrightarrow{} H_n(X, U) \xrightarrow{\partial} H_{n-1}(U) \to \cdots \end{align*} with vertical maps $i_*: H_n(A) \to H_n(U)$ induced by the inclusion $i: A \hookrightarrow U$, the identity $\operatorname{id}: H_n(X) \to H_n(X)$, and the natural map $H_n(X, A) \to H_n(X, U)$. Since $A \hookrightarrow U$ is a homotopy equivalence (with homotopy inverse $r$), the inclusion-induced map $i_*: H_n(A) \to H_n(U)$ is an isomorphism for all $n$. The middle vertical map is the identity, hence an isomorphism. By the [Five Lemma](/theorems/1938), the map $H_n(X, A) \to H_n(X, U)$ is an isomorphism for all $n$.[/step]

[guided]The good pair hypothesis provides a neighbourhood $U$ of $A$ that deformation-retracts onto $A$. Why do we need such a $U$? Because we want to apply excision, but excision requires removing a set whose closure is contained in the interior of the subspace. The subspace $A$ itself may not have this property, but by passing to the larger set $U$ (which contains $\overline{A}$ in its interior), we will be able to excise. Before excising, we must show that replacing $A$ by $U$ does not change the relative homology. The tool is the five lemma applied to the ladder of long exact sequences for the pairs $(X, A)$ and $(X, U)$. The five lemma requires that the maps between corresponding terms (other than the middle term we care about) are isomorphisms. We have: - $H_n(A) \to H_n(U)$: this is an isomorphism because $A \hookrightarrow U$ is a homotopy equivalence (the deformation retraction provides the homotopy inverse). - $H_n(X) \to H_n(X)$: this is the identity map, hence an isomorphism. The [Five Lemma](/theorems/1938) then forces the map $H_n(X, A) \to H_n(X, U)$ to be an isomorphism. (Strictly: the five lemma as stated requires surjectivity of one outer map and injectivity of the other; since all four surrounding maps are isomorphisms, both conditions are satisfied.)[/guided]

[step:Excise $A$ from $(X, U)$ to obtain $H_n(X, U) \cong H_n(X \setminus A, U \setminus A)$]We apply excision with the closed set $Z = A$. The excision hypothesis requires $\overline{A} \subset \operatorname{int}(U)$. Since $U$ is open in $X$, $\operatorname{int}(U) = U$, and $A \subset U$ by construction, so $\overline{A} \subset U = \operatorname{int}(U)$ (using the fact that $A$ deformation-retracts from $U$ and is closed in $X$ for a good pair; in general, the good pair hypothesis ensures $\overline{A} \subset U$ after possibly shrinking $U$). By the [Excision Theorem](/theorems/???), \begin{align*} H_n(X \setminus A, U \setminus A) \xrightarrow{\sim} H_n(X, U) \end{align*} is an isomorphism for all $n$.[/step]

[guided]Excision says that if $Z \subset X$ is a "small enough" subset of the subspace, we can remove it from both spaces without changing relative homology. The precise condition is $\overline{Z} \subset \operatorname{int}(B)$ where $(X, B)$ is the pair and $Z$ is the set being excised. Here we take the pair $(X, U)$ and excise $Z = A$. We need $\overline{A} \subset \operatorname{int}(U) = U$ (since $U$ is open). For a good pair $(X, A)$, the definition guarantees that $A$ is closed in $X$ and there exists an open neighbourhood $U$ of $A$ with a deformation retract. Since $A$ is closed, $\overline{A} = A \subset U$, and the excision hypothesis is satisfied. After excision, we are left with $H_n(X \setminus A, U \setminus A)$. Geometrically, we have removed $A$ entirely, leaving only the "collar" $U \setminus A$ around where $A$ used to be.[/guided]

[step:Identify the excised pair with the corresponding pair in $X/A$ via the quotient map]The quotient map $q: X \to X/A$ collapses $A$ to the basepoint $* = A/A \in X/A$. We observe: - $q$ restricts to a homeomorphism $q|_{X \setminus A}: X \setminus A \xrightarrow{\sim} X/A \setminus \{*\}$, since $q$ is injective on $X \setminus A$ and is a quotient map (hence a homeomorphism onto its image in this case, as a bijective quotient map from an open subset). - Under this homeomorphism, $U \setminus A$ maps to $q(U) \setminus \{*\} = U/A \setminus \{*\}$. Therefore \begin{align*} H_n(X \setminus A, U \setminus A) \cong H_n(X/A \setminus \{*\}, U/A \setminus \{*\}). \end{align*}[/step]

[guided]The quotient map $q: X \to X/A$ identifies all of $A$ to a single point $*$, but is injective on the complement $X \setminus A$. We need to verify that this restriction is actually a homeomorphism onto $X/A \setminus \{*\}$. The map $q|_{X \setminus A}: X \setminus A \to X/A \setminus \{*\}$ is continuous (restriction of a continuous map), bijective (injective because $q$ identifies only points of $A$, surjective because every point of $X/A \setminus \{*\}$ is the image of a unique point of $X \setminus A$). To show it is a homeomorphism, we need to check that it is an open map (or equivalently, that its inverse is continuous). For any open set $V \subset X \setminus A$, the set $q(V)$ is open in $X/A$ if and only if $q^{-1}(q(V))$ is open in $X$. Since $V \cap A = \varnothing$, we have $q^{-1}(q(V)) = V$, which is open in $X$. So $q|_{X \setminus A}$ is an open map and hence a homeomorphism. The image of $U \setminus A$ under this homeomorphism is $q(U \setminus A) = q(U) \setminus \{*\} = U/A \setminus \{*\}$, so the pairs correspond.[/guided]

[step:Apply excision on the quotient side to conclude $\tilde{H}_n(X/A) \cong H_n(X/A, U/A)$]On the quotient side, we apply the same two-step argument in reverse. First, excision on the pair $(X/A, U/A)$ with $Z = \{*\}$: since $\{*\}$ is a point (hence closed) and $\{*\} \subset U/A$ which is open in $X/A$ (because $q^{-1}(U/A) = U$ is open in $X$), we have $\overline{\{*\}} = \{*\} \subset \operatorname{int}(U/A) = U/A$. By excision, \begin{align*} H_n(X/A \setminus \{*\}, U/A \setminus \{*\}) \xrightarrow{\sim} H_n(X/A, U/A). \end{align*} Second, the inclusion $\{*\} \hookrightarrow U/A$ is a homotopy equivalence: the deformation retraction $H: U \times [0,1] \to U$ of $U$ onto $A$ descends to a deformation retraction of $U/A$ onto $\{*\}$ (since $H$ fixes $A$ pointwise, it is compatible with the quotient). By the same five lemma argument as in the first step (applied to the pairs $(X/A, \{*\})$ and $(X/A, U/A)$), we obtain \begin{align*} H_n(X/A, \{*\}) \xrightarrow{\sim} H_n(X/A, U/A). \end{align*} Since $H_n(X/A, \{*\}) = \tilde{H}_n(X/A)$ by definition of reduced homology, and combining all the isomorphisms: \begin{align*} H_n(X, A) \cong H_n(X, U) \cong H_n(X \setminus A, U \setminus A) \cong H_n(X/A \setminus \{*\}, U/A \setminus \{*\}) \cong H_n(X/A, U/A) \cong \tilde{H}_n(X/A). \end{align*}[/step]

[guided]The final step mirrors the first two steps, but applied to the quotient space $X/A$ with the subspace $\{*\}$. The argument is completely symmetric: 1. **Excision on the quotient:** We excise $\{*\}$ from $(X/A, U/A)$, just as we excised $A$ from $(X, U)$. The excision hypothesis $\overline{\{*\}} \subset \operatorname{int}(U/A)$ is satisfied because $U/A$ is open (its preimage under $q$ is $U$, which is open). 2. **Five lemma on the quotient:** The deformation retraction of $U$ onto $A$ descends to a deformation retraction of $U/A$ onto $\{*\}$. Define the descended homotopy $\bar{H}: U/A \times [0,1] \to U/A$ by $\bar{H}(q(u), t) = q(H(u, t))$; this is well-defined because $H(a, t) = a \in A$ for all $a \in A$ and $t \in [0,1]$, so $q(H(a, t)) = * = q(a)$. Thus $\{*\} \hookrightarrow U/A$ is a homotopy equivalence, the five lemma gives $H_n(X/A, \{*\}) \cong H_n(X/A, U/A)$, and the chain of isomorphisms is complete. The full chain is: start with $H_n(X, A)$, enlarge $A$ to $U$ (five lemma), excise $A$ (excision), identify with the quotient (homeomorphism on complements), undo excision on the quotient (excision), and shrink $U/A$ back to $\{*\}$ (five lemma). Each step is an isomorphism, and the composite is the map induced by the quotient $q: (X, A) \to (X/A, \{*\})$.[/guided]