TITLE: Understanding Counter example for "product of two quotient maps is a quotient map" from Ronald Brown
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.