EDIT: Why is $J$ an embedding: It is a continuous bijection with is image $[B]$ (brackets to denote class). If $C \subseteq B$ is closed, we note that $[X \sqcup C]$ is a closed, since prior to quotient $q:X \sqcup B \rightarrow Y$, it is a closed saturated set. Then $[X \sqcup C] \cap [B] = [C]$. So $J$ is an embedding. $\square$
Why is $\bar{F}$ a homeomorphism: $X \rightarrow Y/B$ induces a map $\bar{F} :X/A \rightarrow Y/B$ from commutativity. The map is given by $[x] \mapsto [F(x)] = [x]$, which is bijective. We show explicitly, the inverse map $Y/B \rightarrow X/A$ is continuous.
- If $[U]_{X/A} \subseteq X/A$ is open, then in $X$ it is an open set containg $A$.
- The preimage of $[U]_{Y/B}$ is $[U]_Y \cup [B]_Y$ which is open since its preimage in $X \sqcup B$ is $U \sqcup B$. $\square$
Questios:
- I wonder if my proof is correct
- I am hoping to see a formal proof rather than point set topology. Since this is part of a derivation of cofibre sequence. The author states, "the derivation uses only formal properties".
That $J$ is an embedding does not follow from formal properties of the pushout. In any category a pushout diagram as the left square has the following property: If $j$ is a monomorphism, then so is $J$. In the category $\mathbf{Top}$ the monomorphisms are precisely the injective maps, so this does not help you. Thus you need a down-to-earth-proof. This is given in Proposition (1.8.1) of
tom Dieck, General topology.
Unfortunately your proof is not correct because in general $X \sqcup C$ is not saturated. As an example take $A = B = X = [0,1]$ and $f = j = id$. Then $Y = [0,1]$ and $F = J = id$. We have $p^{-1}(p([0,1] \sqcup \{ 0 \})) = [0,1] \sqcup [0,1]$.
Similarly I do not see how to prove that $\overline{F}$ is a homeomorphism just by using universal properties of pushouts and quotients. But your proof is correct.