(b) Let $p : A→B$ and $q : C→D$ be quotient maps, and let $B$ and $C$ be locally compact Hausdorff. Show $p×q$ is a quotient map.
Proof: (a) It is clear that $π$ is continuous, so it suffices to show $A⊆Y×Z$ is open when $π^{-1}(A)⊆X×Z$ is open. For any given $y×z∈A$, when $x×z$ is such that $π(x×z)=y×z$ (i.e. $x∈p^{-1}(y)$) we find a neighborhood $U$ of $x$ saturated with respect to $p$ and a neighborhood $V$ of $z$ such that $U×V⊆π^{-1}(A)$, so that $π(U×V)=p(U)×V$ will be a neighborhood of $y×z$ contained in $A$. In this fashion, let $U_1×V$ be a basis element in $π^{-1}(A)$ containing $x×z$ since $π^{-1}(A)$ is open. Since $Z$ is locally compact Hausdorff, we may assume $V$ is such that $\overline{V}$ is compact and $U×\overline{V}⊆π^{-1}(A)$.
Given open $U_i$ such that $x×z∈U_i×\overline{V}⊆π^{-1}(A)$, we construct $U_{i+1}$ such that $x×z∈U_{i+1}×\overline{V}⊆π^{-1}(A)$ and $p^{-1}(p(U_i))⊆U_{i+1}$, and we shall set $U=∪U_i$. To see this construction, note that for each $u∈p^{-1}(p(U_i))$ we see $u×\overline{V}$ is compact, so that by the tube lemma we may choose a neighborhood $V_u$ of $u$ such that $V_u×\overline{V}⊆π^{-1}(A)$. Letting $U_{i+1}=∪V_u$, we see each of the two desired properties are fulfilled.
We see $p^{-1}(p(U))=U$ by the second desired property of each $U_i$, so that $U$ is saturated. Since also $U×V⊆U×\overline{V}⊆π^{-1}(A)$ we have demonstrated the proper $U$ and $V$, so that $π$ is a quotient map.
(b) This is evident by considering the composition of quotient maps $p×q=(1×q)∘(p×1)$.$~\square$
