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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.

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    See [this related question](http://math.stackexchange.com/questions/31697/when-is-the-product-of-two-quotient-maps-a-quotient-map) and [this one](http://math.stackexchange.com/questions/91639/x-sim-is-hausdorff-if-and-only-if-sim-is-closed-in-x-times-x).2012-08-03
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    @HennoBrandsma: Since you are the author of the original remark, perhaps you can answer the question?2015-12-26
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    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...2015-12-30
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    @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](https://en.wikipedia.org/wiki/Tube_lemma)...2016-01-04

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