4
$\begingroup$

One can prove that the product of two open quotient maps is a quotient map. Ronald Brown gives a counter example for the fact that this is in general not true for arbitrary quotient maps, in his book Topology and Groupoids on page 111. The counter example is:

$$p \times id: \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q} / \mathbb{Z} \times \mathbb{Q},$$ where $p$ is the quotient map from $\mathbb{Q}$ to $\mathbb{Q} / \mathbb{Z}$.

But since $\mathbb{Q}$ is a topological group and $\mathbb{Z}$ a subgroup it follows that $p$ is open. The $id$ is clearly open, too. Now $p \times id$ is the product of two open quotient maps and therefore a quotient map.

This is clearly contradicting. Where is my mistake?

$\endgroup$
3
$\begingroup$

More a misunderstanding than a mistake. The $\mathbb{Q}/\mathbb{Z}$ that Brown talks about is not the quotient group, but the topological space that you get by identifying all of $\mathbb{Z}$ to a single point, leaving all nonintegers equivalent only to themselves. Formally,

$$x\sim y :\iff (x = y)\lor (x \in \mathbb{Z}\land y \in \mathbb{Z})$$

and the map is

$$p\times \operatorname{id} \colon \mathbb{Q}\times\mathbb{Q} \to (\mathbb{Q}/\sim)\times\mathbb{Q}.$$

$p\colon\mathbb{Q}\to\mathbb{Q}/\sim$ is not open.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.