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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?

6 Answers 6

26

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.

  • 0
    By the way, for a proof of Whitehead's theorem see Lemma 4 [here](http://www.math.ku.dk/~moller/e02/3gt/opg/S29.pdf)2015-01-19
7

@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.

4

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

3

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.

2

There is another example that I just learnt from exercise 16, section 2.2 of Algebraic Topology by Tammo Tom Dieck. This is :

Let $X$ be a topological space and $A$ be a compact subspace of $X$. Denote the quotient map $p:X \to X/A$. Then $ p \times id_Y : X \times Y \to X/A \times Y$ is a quotient map for any arbitrary space $Y$.

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    Do you know how to prove it? I would be very helpful if you could give a hint.2017-09-13
-1

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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    @IttayWeiss - see section 2.4 in Tom Diecks Algebraic Topology.2013-11-29