[proofplan]
We combine two ingredients: the [Subcomplex Pairs Are Good](/theorems/2259) theorem (which guarantees that CW pairs $(X, A)$ are good pairs) and the general quotient isomorphism for good pairs (excision applied to the quotient). For a CW complex $X$ with subcomplex $A$, the pair $(X, A)$ is good, and the quotient map $q: (X, A) \to (X/A, A/A)$ induces an isomorphism $H_n(X, A) \cong H_n(X/A, A/A) \cong \tilde{H}_n(X/A)$ for all $n$, where $\tilde{H}_n$ denotes reduced homology.
[/proofplan]
[step:Verify that $(X, A)$ is a good pair]
By the [Subcomplex Pairs Are Good](/theorems/2259) theorem, if $X$ is a CW complex and $A \subseteq X$ is a subcomplex, then $A$ is a neighbourhood deformation retract of some open neighbourhood $V \supseteq A$ in $X$. This means there exists an open set $V$ with $A \subseteq V \subseteq X$ and a homotopy $H: [0,1] \times V \to X$ with $H(0, x) = x$, $H(1, x) \in A$, and $H(t, a) = a$ for all $a \in A$ and $t \in [0,1]$. In particular, the inclusion $A \hookrightarrow V$ is a homotopy equivalence, which is the defining condition for $(X, A)$ to be a good pair.
[guided]
The condition for $(X, A)$ to be a good pair is that $A$ possesses an open neighbourhood $V$ in $X$ that deformation retracts onto $A$. This is stronger than merely requiring $A$ to be closed: it demands that the "collar" around $A$ in $X$ can be continuously collapsed back onto $A$.
For CW pairs, this deformation retract is constructed by using the product structure of the cells near $A$: each $n$-cell $e_\alpha$ whose closure meets $A$ has a characteristic map $\Phi_\alpha: D^n \to X$, and the attaching region $\Phi_\alpha(\partial D^n) \cap A$ can be "pushed outward" using the radial deformation of $D^n$ near its boundary. The union of these collar neighbourhoods, one for each cell adjacent to $A$, gives the open set $V$. The proof of this construction is given in [Subcomplex Pairs Are Good](/theorems/2259).
[/guided]
[/step]
[step:Apply excision to the quotient map]
Consider the quotient map $q: X \to X/A$ that collapses $A$ to a point $[A] \in X/A$. This induces a map of pairs $q: (X, A) \to (X/A, [A])$. Since $(X, A)$ is a good pair, the inclusion $A \hookrightarrow V$ is a homotopy equivalence for some open $V$. The map $q$ restricts to a homeomorphism $X \setminus A \xrightarrow{\sim} (X/A) \setminus [A]$.
We apply excision in the following form. Let $Z = A$ and observe that $\overline{A} = A \subseteq V = A^\circ \cup (V \setminus \partial A)$ (here $A$ is closed in $X$ since subcomplexes of CW complexes are closed, and $A \subseteq V^\circ$ since $V$ is open and contains $A$). Since the pair $(X, A)$ is good, the long exact sequences for the pairs $(X, A)$ and $(X/A, [A])$ are related by the quotient map $q$, and excision gives
\begin{align*}
q_*: H_n(X, A) \xrightarrow{\cong} H_n(X/A, [A])
\end{align*}
for all $n$.
[guided]
The precise mechanism is as follows. Since $(X, A)$ is a good pair, the homotopy equivalence $A \simeq V$ implies that $i_*: H_n(A) \xrightarrow{\cong} H_n(V)$ is an isomorphism. The five-lemma applied to the map of long exact sequences for $(X, A)$ and $(X/A, [A])$ then shows that $q_*: H_n(X, A) \to H_n(X/A, [A])$ is an isomorphism. Alternatively, one can use the excision theorem directly: since $V$ deformation retracts onto $A$, the inclusion $(X \setminus A, V \setminus A) \hookrightarrow (X, V)$ induces isomorphisms on relative homology by excision, and composing these isomorphisms with the deformation retract isomorphism yields the result.
The point is that collapsing a subcomplex to a point does not change the relative homology, provided the subcomplex has a sufficiently nice neighbourhood in $X$.
[/guided]
[/step]
[step:Identify $H_n(X/A, [A])$ with the reduced homology $\tilde{H}_n(X/A)$]
The point $[A] \in X/A$ has trivial homology: $H_n([A]) = 0$ for $n \geq 1$ and $H_0([A]) \cong \mathbb{Z}$. The long exact sequence of the pair $(X/A, [A])$ gives, for $n \geq 1$:
\begin{align*}
H_n([A]) \to H_n(X/A) \to H_n(X/A, [A]) \to H_{n-1}([A]).
\end{align*}
For $n \geq 2$, both $H_n([A]) = 0$ and $H_{n-1}([A]) = 0$, so $H_n(X/A, [A]) \cong H_n(X/A) = \tilde{H}_n(X/A)$.
For $n = 1$, we have $H_1([A]) = 0 \to H_1(X/A) \to H_1(X/A, [A]) \to H_0([A]) \cong \mathbb{Z}$. The map $H_0([A]) \to H_0(X/A)$ sends the generator to the path component of $[A]$. Since $X/A$ is path-connected (assuming $X$ is — otherwise work component-by-component), this map is surjective, and its kernel is zero if $X/A$ has one path component. The connecting map $H_1(X/A, [A]) \to H_0([A])$ is therefore the zero map, so $H_1(X/A, [A]) \cong H_1(X/A) = \tilde{H}_1(X/A)$.
For $n = 0$, the long exact sequence gives $H_0([A]) \xrightarrow{j_*} H_0(X/A) \to H_0(X/A, [A]) \to 0$. The cokernel of $j_*$ is the reduced homology $\tilde{H}_0(X/A)$, so $H_0(X/A, [A]) \cong \tilde{H}_0(X/A)$.
In all degrees, $H_n(X/A, [A]) \cong \tilde{H}_n(X/A)$.
[/step]
[step:Combine to obtain the quotient isomorphism]
Composing the isomorphism $q_*: H_n(X, A) \xrightarrow{\cong} H_n(X/A, [A])$ from the excision step with the identification $H_n(X/A, [A]) \cong \tilde{H}_n(X/A)$ from the long exact sequence analysis, we obtain
\begin{align*}
H_n(X, A) \cong \tilde{H}_n(X/A)
\end{align*}
for all $n \geq 0$. This isomorphism is natural with respect to maps of CW pairs.
[/step]
