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:

1. $Y$ is Hausdorff.
2. $q$ is a closed map.
3. 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.

• – Zhen Lin Aug 3 '12 at 4:56
• @HennoBrandsma: Since you are the author of the original remark, perhaps you can answer the question? – tomasz Dec 26 '15 at 23:39
• It suffices to show $f\times \text{id}_Y:X\times Y \to Y\times Y$ is a quotient map, because we know $\text{id} _X\times f$ is a quotient map by Whitehead's Lemma. We know $X$ and $Y$ are compact, and the Tube Lemma applies to all compact spaces (not necessarily Hausdorff), so I think there is a good chance for this... – Forever Mozart Dec 30 '15 at 0:03
• @ForeverMozart: Could you please let me know what is "Whitehead's Lemma", as you intend it? I suppose the "Tube Lemma" is what I find on wikipedia here... – Mike F Jan 4 '16 at 5:54

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.

• Neat, thanks. I think in the end I proved (3) implies (2) using nets. Let me see if I recall the argument. Suppose (3) holds. We want to show $q$ is closed, so suppose $A \subset X$ is closed. We need to show $q(A)$ is closed in $Y$ i.e. that $B := q^{-1}(q(A))$ is closed in $X$. Let $x \in \overline B$. Then, there is a net $b_i$ in $B$ converging to $x$. Let $a_i \in A$ be chosen arbitrarily subject to $q(a_i) = q(b_i)$ (possible by definition of $B$). – Mike F Mar 30 '13 at 4:26
• But now, $(a_i,b_i)$ is a net in the compact Hausdorff space $R \subset X \times X$, whence has a subnet $(\alpha_j,\beta_j)$ converging to a point in $R$. Now, $\beta_j \to x$ since $\beta_j$ is a subnet of $b_i$ which goes to $x$. Let us say $\alpha_j \to a \in A$ (recall $A$ is closed). Since $(a,x) \in R$, we have $q(a) = q(x)$ which shows $x \in B = q^{-1}(q(A))$. The proof is complete? – Mike F Mar 30 '13 at 4:37
• Your proof looks fine, though it took me a while to see where the Hausdorff property is used. I think without Hausdorff $(\alpha_j,\beta_j)$ could converge to some $(a,y)\in R$ where $y\neq x$ and to $(a,x)\notin R$. There are situations where nets are very suitable, e.g. if a compact group operates on a Hausdorff space, one can show the closedness of $R$ using nets. In this situation I like the my proof better but your proof is correct, too. – Stefan Hamcke Mar 30 '13 at 14:03
• If $X$ is Hausdorff and $R$ is a compact equivalence relation on $X$, then the diagonal $D=\{(x,x)\!:x\in X\}$ is a closed subset of $R$, hence compact. The space $X$ is homeomorphic to $D$, hence compact as well. So the assumptions of the compactness of $X$ and of $R$ have the same strength. – Peter Elias Apr 24 '17 at 13:03