I could not find my mistake in the following argument, though I know it is wrong. This is more like a "Q&A", since there is nothing to "prove" in the positive sense. Here it goes:
For an arbitrary pair (X, A) with A non-empty, we know that for positive $n$
$H_n(X, A) \simeq H_n(X \cup CA - x, CA - x) \simeq H_n(X \cup CA, CA) \simeq H_n(X \cup CA) \simeq H_n(X/A)$
where $CA$ is attached to $X$ by $[(a, 0)] \simeq a \in X$, $x$ is the tip of the cone, and the iso's are by: homotopy invariance, excision, long sequence and again homotopy invariance by the following, which is my concern:
Let $f : X \cup CA \to X/A, x \mapsto x, CA \mapsto [A]$ and $g : X/A \to X \cup CA, x \mapsto x, [A] \mapsto [(a, 1)]$ for any $a \in A$. Then $fg = 1$ and $gf \simeq 1$ by $h_t : X \cup CA \to X \cup CA, x \mapsto x, [(a, 1 - s)] \mapsto [(a, 1 - ts)]$, $0 \le t \le 1$.
What did I do wrong in the above? I would also like to know what is true (Hatcher hints that this leads somewhere, but it is not clear from the phrasing.) This matters in the economy of that chapter.
The point $[(a,1)]\in X\cup CA$ is the top of the attached cone, it has the open neighborhood $$ U = \left\{\, [(a,t)]\, \middle|\, t>\frac 1 2\,\right\}.$$ Now the preimage under $g$ is $g^{-1}(U) = \{[A]\}$, since every $x\notin A$ is mapped outside the cone. Thus, for $g$ to be continuous, you need the one-point set $\{[A]\}$ to be open in $X/A$ which is equivalent to $A$ being open in $X$.