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It is not true in general that the product of two quotient maps is a quotient maps (I don't know any examples though).

Are any weaker statements true? For example, if $X, Y, Z$ are spaces and $f : X \to Y$ is a quotient map, is it true that $ f \times {\rm id} : X \times Z \to Y \times Z$ is a quotient map?

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5 Answers 5

up vote 11 down vote accepted

Your weaker statement is almost true.

If $f: X \to Y$ is a quotient map and $Z$ is locally compact, then $f \times \operatorname{id}$ is a quotient map. I believe that this result is due to Whitehead.

More generally, if $f: X \to Y$ and $g: Z \to W$ are quotient maps and $Y$ and $Z$ are locally compact, then the product $f \times g: X \times Z \to Y \times W$ is a quotient map.

Why? Use the Whitehead theorem twice, since $f \times g = (\operatorname{id} \times g) \circ (f \times \operatorname{id})$.

See Munkres $\S 22$ for counterexamples.

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5  
Here is an example: let $Q$ be the space of rational numbers and $Z$ the subspace of integers. Let $Q/Z$ be $Q$ with $Z$ shrunk to a point, and $p: Q \to Q/Z$ the identification map. Then $p \times 1: Q \times Q \to Q/Z \times Q$ is not an identification map. This is on p.111 of "Topology and Groupoids". Is it the same as in Munkres? This kind of example was my reason for suggesting in 1963 the notion of a "category adequate and convenient for all purposes of topology", at that time Hausdorff k-spaces. –  Ronnie Brown Oct 5 '12 at 10:30
    
It seems that to apply Whitehead's lemma, we need to assume a regular $Z$. A sufficient and necessary condition is given by Day and Kelly, On topological quotient maps preserved by pullbacks or products. Or, maybe locally compact need a stronger definition (than just having a compact neighbourhood). Then it would also be nice to mention the specific definition in use. –  yuning Apr 1 at 23:48
    
Should it be $X$ and $Z$ that are locally compact (to apply the Whitehead thm. twice)? –  hardmath Apr 2 at 1:51
    
No. The second map in the composition is $\operatorname{id}_Y \times g: Y \times Z \to Y \times W$. The factor with the identity map needs to be locally compact (or some slightly weaker condition mentioned in the comments). –  Sammy Black Apr 2 at 18:39

@Ittay: In a cartesian closed category of spaces, a product of identification maps is an identification map. Here is a typical proof essentially from Topology and Groupoids p. 192.

It suffices to prove that if $f:Y \to Z$ is an identification map, then so also is $f \times 1: Y \times X \to Z \times X$ for any $X$.

Let $g:Z \times X \to W$ be a function such that $l=g(f \times 1):Y \times X \to W$ is continuous. By cartesian closedness, we have associated maps

$$ l': Y \to K(X,W), \quad g':Z \to K(X,W)$$

where $K(X,W)$ is the internal hom, and $g'f=l'$. Since $l'$ is continuous, and $f$ is an identification map, then $g'$ is continuous. Hence $g$ is continuous.

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In the category of compactly generated spaces, I think that the product of quotient maps is (always) a quotient map.

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In above cases, the results follow (easily) from the good behavior of compact open topology (related with locally compact topology, and with the category of compactly generated spaces respectively).

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4  
It is not entirely clear how the results follow from what you mention. If you can elaborate a bit that would be helpful. –  Ittay Weiss Nov 3 '12 at 8:23
    
@IttayWeiss - see section 2.4 in Tom Diecks Algebraic Topology. –  user54092 Nov 29 '13 at 1:48

If the quotient maps $p : W \rightarrow X, q:Y \rightarrow Z$ are open, then the product map is also a quotient map:

$U\times V$ open in $X \times Y \implies p^{-1}U\times q^{-1}{V} = (p\times q)^{-1}(U\times V)$ open.

Conversely, $U\times V = (p\times q)(\dots) = p\circ p^{-1}U\times q\circ q^{-1}V$ is open since $p,q$ are open maps.

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