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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?

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1 Answer 1

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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.

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