Attaching a cell Could you help me to explain this argument:
Let $f: S^{n-1} \rightarrow A$ for $n \ge 1$, form
$$X= C(f) := \dfrac{A\coprod D^n}{f(x) \sim x, \forall x \in S^{n-1}}$$
"$(D^n,S^{n-1}) \rightarrow (X,A)$ induces isomorphisms in $H_q(D^n,S^{n-1}) \rightarrow H_q(X,A)$"
 A: Let $g : (D^n, S^{n-1}) \to (X, A)$ be the characteristic map of the cell $e = X - A$.  There are homotopy equivalences $(D^n, S^{n-1}) \to (D^n, D^n - \{0\})$ and $(X, A) \to (X, X - \{g(0)\})$, which induce isomorphisms of relative homology groups.  By excising the boundary $\partial D^n$, one sees $((D^n)^\circ, (D^n)^\circ - \{0\}) \hookrightarrow (D^n, D^n - \{0\})$ induces an isomorphism in homology, and the same for $(e, e - \{g(0)\}) \hookrightarrow (X, X - \{g(0)\})$ by excising $A$.  But $g$ restricts to a homeomorphism $(D^n)^\circ \stackrel{\sim}{\to} e$, hence it induces an isomorphism $H_*((D^n)^\circ, (D^n)^\circ - \{0\}) \stackrel{\sim}{\to} H_*(e, e - \{g(0)\})$.  It follows then that $g$ also induces an isomorphism $g_* : H_*(D^n, S^{n-1}) \stackrel{\sim}{\to} H_*(X, A)$.
A: Note $H_q(D^n, S^{n-1}) = H_q(D^n/S^{n-1}) = H_q(S^n)$ since $D^n/S^{n-1}$ is homeomorphic to $S^n$.  On the other hand $H_q(X, A) = H_q(X/A) = H_q(S^n)$ since $X/A$ is also homeomorphic to $S^n$.
Let me know if you don't understand any of these isomorphisms.
