Product of quotient map a quotient map when domain is compact Hausdorff? Suppose that $X$ is a compact Hausdorff space and that $q : X \to Y$ is a quotient map. Is it true that the product map $q \times q : X \times X \to Y \times Y$ is also a quotient map? Note I did not assume that the quotient space $Y$ was Hausdorff (which I know to be equivalent to closedness of the quotient map $q$ in this situation). Thanks!
Added: I ask for the following reason. Wikipedia claims that, for a compact Hausdorff space $X$ and a quotient map $q : X \to Y$, the following conditions are equivalent:


*

*$Y$ is Hausdorff.

*$q$ is a closed map.

*The equivalence relation $R = \{ (x,x') : q(x) = q(x') \}$ is closed in $X \times X$. 


I was able to prove that (1) and (2) are equivalent and also that (1) implies (3). However, I do not see how to deduce (1) or (2) from (3). Some googling yields the following ?proof? (the relevant paragraph being the last one) which relies on the claim that $q \times q$ is a quotient map. Thus the question.
 A: It is possible to deduce ($2$) from ($3$) without $q\times q$ being a quotient map.
Let $A\subset X$ be closed. Assume $R=\{(x,x')\in X\times X\mid q(x)=q(x')\}$ is closed, thus compact. Then $A\times X\cap R$ is compact, and so is also its image under the projection $p_2$ onto the second factor. But $p_2(A\times X\cap R)=\{x\in X\mid\exists a\in A:q(a)\sim q(x)\}=q^{-1}(q(A))$. So the saturation of a closed set is compact, and hence closed, which means that $q$ is a closed map.
One could also omit the compactness of $X$ if one assumes directly that $R$ is compact, because the last step only uses the Hausdorff'ness to deduce that a compact set is closed. Note that compactness of $R$ also makes $\{x\}\times X\cap R$ a compact set, so fibers are compact and this makes $q$ a so-called perfect map. These maps preserve many properties of the domain, for example all the separation axioms (except $T_0$)
This doesn't answer the question in the title, which I would really like to know myself. I only know about the product of a quotient map with the Identity $q\times Id:X\times Z\to Y\times Z$, which is a quotient map if $Z$ is locally compact.
Edit: Actually, showing that $q$ is closed requires only the compactness of $X$. Indeed, if $X$ is compact, then the projection $p_2$ is a closed map, so if $R$ is closed, then for every closed $A\subseteq X$, the set $p_2(A\times X\cap R)$ is closed.
