Question on where the assumptions are used in the proof of a corollary of excision theorem (good pair) I want to prove the following theorem, but I have no idea why the assumptions are needed to prove the theorem.
Theorem: Let $A \subset X$ be non-empty and closed. Suppose there exists open $B \subset X$ such that $A \subset B$ and $A$ is a strong deformation retract of $B$. Let $q:(X,A) \to (X/A,*)$ be the quotient map. Then $q_*:H_n(X,A) \to H_n(X/A,*)$ is an isomorphism.
The proof begins with the commutative diagram.

where $\iota_i$ are inclusions. Passing to homology, we get the following diagram.

If we have those isomorphisms above, then $H_n(X,A) \to H_n(X/A,*)$ is also an isomorphism by commutativity of the diagram.
Now, we want to see why the isomorphisms hold.
The isomorphism $\cong_1$: Since $B$ deformation retracts into $A$, the inclusion map $A \to B$ is a homotopy equivalence. Then it induces an isomorphism $H_n(A) \to H_n(B)$. By considering the long exact sequence $$... \to H_n(A) \to H_n(B) \to H_n(B,A) \to ...$$
we get $H_n(B,A)=0$. Then by considering the long exact sequence $$... \to H_n(B,A) \to H_n(X,A) \to H_n(X,B) \to ...$$
we obtain the required isomorphism.
The isomorphism $\cong_2$: I have proved that if $B$ deformation retracts into $A$, then $B/A$ deformation retracts into $*$ using the universal property of quotient map. Then we can repeat the argument above.
Question 1: Here, we only need the fact that $B$ deformation retracts into $A$. But we don't need the strong deformation retract. Can someone explain why strong deformation retract is needed?
The isomorphisms $\cong_3$ and $\cong_4$ comes from excision theorem. (I have shown that if $\overline{A} \subset \text{int}B$, then $\overline{*} \subset \text{int}(B/A)$)
The isomorphism $\cong_5$ comes from the fact that $q|_{X-A}: X-A \to X/A-*$ is a homeomorphism. (I am not sure whether the closedness of $A$ is required here).
Question 2: Can we just replace $A$ non-empty closed and $B$ open by $A \subset X$ such that $\overline{A} \subset \text{int} B$ for some $B \subset X$? The latter conditions are the conditions for the excision theorem to hold.
Thank you very much.
 A: $q \mid_{X \setminus  A} : X \setminus  A \to X/A \setminus \{*\}$ is always a contiunuous bijection and obviously a homeomorphism if $A$ is closed. If $A$ is not closed, it may be a homeomorphism (for example if $X \setminus  A$ is a one-point set), but in general it is not. In fact, assume that there exists an open $U \subset X$ such that neither $U \setminus  A$ nor $U \cup A$ are open in $X$ (an example is $X = I \times I, A = [0,1) \times \{0\}, U = (1/2,1] \times I$). Then $U \setminus  A$ is open in $X \setminus  A$. Assume that $q(U \setminus  A)$ is open in $X/A \setminus \{*\}$. Then there exists an open $V \subset X/A$ such that $V \cap (X/A \setminus \{*\}) = q(U \setminus A)$. Then either $V = q(U \setminus A)$ or $V = q(U \setminus A) \cup \{*\}$. In the first case $q^{-1}(V) = q^{-1}(q(U \setminus  A)) = U \setminus  A$ should be open in $X$, but it is not. In the second case $q^{-1}(V) = q^{-1}(q(U \setminus  A)) \cup q^{-1}(\{*\}) = (U \setminus  A) \cup A = U \cup A$ should be open in $X$, but it is not.
This shows that the assumption "$A$ closed" cannot be dropped if we want that $q \mid_{X \setminus  A}$ induces an isomorphism in homology.
There is another subtle point: Writing $B/A \subset X/A$ requires that $q \mid_B : B \to q(B)$ is a quotient map (only in that case it induces a homeomorphism $q' : B/A \to q(B)$). To verify this, note that $q$ maps an open $U \subset X$ to an open $q(U) \subset X/A$ provided $A \subset U$ or $A \cap U = \emptyset$ (because then $q^{-1}(q(U)) = U$). Now let $U \subset q(B)$ be a set such that $q^{-1}(U)$ is open in $B$. Write $q^{-1}(U) = B \cap W$ with an open $W \subset X$. If $* \in U$, then $A \subset q^{-1}(U) \subset W$. Hence $q(W)$ is open in $X/A$ and $U = q(q^{-1}(U)) = q(B \cap W) = q(B) \cap q(W)$ is open in $q(B)$. If $* \notin U$, then $q^{-1}(U) \cap A = \emptyset$. This implies $W \cap  A = \emptyset$. Hence again $q(W)$ is open in $X/A$ and $U = q(q^{-1}(U)) = q(B \cap W) = q(B) \cap q(W)$ is open in $q(B)$.
This being said, we can answer your questions.
Q2: In fact can work with a closed $A$ and any $B$ such that $A \subset \text{int} B$. In that case also $* \in \text{int} B/A$ because $q(\text{int} B)$ is open in $X/A$.
Q1: The minimal requirements are that $A \hookrightarrow B$ and $* \hookrightarrow B/A$ are homotopy equivalences. This is guaranteed if $A$ is a strong deformation retract of $B$, but may be true also if it is not. However, if we only require that $A$ is a deformation retract of $B$, we cannot be sure that $* \hookrightarrow B/A$ is a homotopy equivalence. In fact, we know that there is a homotopy $H : B \times I \to B$ such that $H_0 = id_B$ and $H_1$ is a retraction $B \to A$, but we do not know that $H_t(A) \subset A$ for all $t$. Only if the latter is satisfied, we can be sure that $H$ induces a homotopy $H' : B/A \times I \to B/A$.
Anyway, you see that "strong deformation retract" can be relaxed a little. A compromise between "strong deformation retract" and "deformation retract" would be to require that there exists a homotopy of pairs $H : (B,A) \times I \to (B,A)$ such that $H_0 = id$ and $H_1$ is a retraction $B \to A$.
