Question on smash product of quotient spaces. 
Let $X$ and $Y$ be two topological spaces and let $A \subseteq X$ and $B \subseteq Y.$ Then $$X/A \wedge Y/B \cong (X \times Y)/ ((X \times B) \cup (A \times Y)).$$

First we need to get a continuous surjective map $f : X \times Y \longrightarrow X/A \coprod Y/B$ and if we compose it with the quotient map $p : X/A \coprod Y/B \longrightarrow X/A \wedge Y/B$ then we get a continuous surjective map $g : X \times Y \longrightarrow X / A \wedge Y / B.$ Now define an equivalence relation on $X \times Y$ as follows $:$ $$(x,y) \sim (x',y') \iff g(x,y) = g(x',y').$$ Then by universal  property  of quotient  topology there exists a continuous
surjective map $\overline g : (X \times Y)/ \sim \longrightarrow X/A \wedge Y/B.$ Furthermore, $\overline g$ is a homeomorphism iff $g$ is a quotient map. This is the so called theory.
Now the questions are $:$
$(1)$ How to define $f$ so that the set of fibres of $g : = (p \circ f)$ is $(A \times Y) \cup (X \times B)\ $?
$(2)$ For such an $f$ how to show that $g$ is a quotient map?
Would anybody please help me in this regard? Thanks in advance.
 A: There is a continuous bijection
$$\phi : (X \times Y)/ ((X \times B) \cup (A \times Y)) \to X/A \wedge Y/B .$$
To see this, consider the the map
$$\psi : X \times Y \stackrel{p_X \times p_Y}{\longrightarrow} X/A \times Y/B \stackrel{q}{\longrightarrow} X/A \wedge Y/B $$
where $p_X, p_Y, q$ are the obvious quotient maps. It is easy to verify that the fibers  $\psi^{-1}(\xi)$ are singletons if $\xi \ne *$ (where $*$ is the point obtained by collapsing $X/A \times \{*\} \cup  \{*\} \times Y/B$) and $\psi^{-1}(*) = X \times B \cup A \times Y$. Hence we get $\phi$.
Now $\phi$ is a homeomorphism if and only if $\psi$ is a quotient map. The problem is that the product of quotients maps $p_X \times p_Y : X \times Y  \to X/A \times Y/B $ is not always a quotient map. Make a search in this forum for "product quotient maps". There are many hits, for example When is the product of two quotient maps a quotient map?
Thus you will need additional assumptions on the pairs $(X,A), (Y,B)$.
Update:
The fibers of $q : X/A \times Y/B \to X/A \wedge Y/B = (X/A \times Y/B)/(X/A \times \{*\} \cup  \{*\} \times Y/B)$ are singletons for $\xi \ne *$ and $q^{-1}(*) = X/A \times \{*\} \cup  \{*\} \times Y/B$.
Clearly
$$(p_X \times p_Y)(X \times B \cup A \times Y) = (p_X \times p_Y)(X \times B) \cup (p_X \times p_Y)(A \times Y) \\= p_X(X) \times p_Y(B) \cup p_X(A) \times p_Y(Y) = X/A  \times \{*\} \cup  \{*\} \times Y/B .$$
Moreover
$$(X \times Y) \setminus (X \times B \cup A \times Y) =  ((X \times Y) \setminus (X \times B)) \cap (X \times Y) \setminus (A \times Y)) \\ = X \times (Y \setminus B) \cap (X \setminus A) \times Y = (X \setminus A) \times (Y \setminus B) .$$
But $p_X \times p_Y$ is injective on $(X \setminus A) \times (Y \setminus B)$ and for $x \in X \setminus A$ resp. $y \in Y \setminus B$ we have $p_X(x) \ne *$ resp. $p_Y(y) \ne *$. Thus
$$(p_X \times p_Y)((X \times Y) \setminus (X \times B \cup A \times Y)) \subset (X/A \times Y/B) \setminus (X \times * \cup * \times Y/B).$$
We conclude that $(p_X \times p_Y)^{-1}(X/A  \times \{*\} \cup  \{*\} \times Y/B) = X \times B \cup A \times Y$ and that for $\eta \notin X/A  \times \{*\} \cup  \{*\} \times Y/B$ the fiber $(p_X \times p_Y)^{-1}(\eta)$ is a singleton.
This shows $\psi^{-1}(*) = X \times B \cup A \times Y$. For $\xi \ne *$ the fiber $q^{-1}(\xi)$ a singleton outside of $X/A  \times \{*\} \cup  \{*\} \times Y/B$, thus $\psi^{-1}(\xi)$ is a singleton.
