In Hatcher's Algebraic Topology, Theorem 3.18, Hatcher asserts that for CW pairs $(X,A),(Y,B)$, the quotient spaces $(X\times Y)/(A\times Y\cup X\times B)$ and $(X\times (Y/B))/(A\times (Y/B)\cup X \times (B/B))$ are homeomorphic. Intuitively this seems true, but I am having a hard time proving it. I attempted as follows:
Let $q_1:Y \to Y/B, q_2:X \times Y \to (X\times Y)/(A\times Y\cup X\times B)$ denote the quotient maps, and let $1_X$ denote the identity map of $X$. Since the map $f:=1_X \times q_1 : X \times Y \to X \times (Y/B)$ takes $A\times Y\cup X\times B$ into $A\times (Y/B)\cup X \times (B/B)$, $f$ passes to the quotient and induces a map $\bar{f} : (X\times Y)/(A\times Y\cup X\times B) \to (X\times (Y/B))/(A\times (Y/B)\cup X \times (B/B))$. It seems that $\bar{f}$ should be the desired homeomorphism. I think I should find its inverse, but it not works easily because the map $f$ is not general a quotient map.
Hint: Show that $\bar f$ is closed, using that $A$ is closed in $X$ and $B$ is closed in $Y$.