# The topology on $X/{\sim}\times X/{\sim}$ is not induced by $\pi\times\pi$.

I'm looking for an example of a topological space $X$ together with an equivalence relation $\sim$ where the product topology on $X/{\sim}\times X/{\sim}$ is not induced by $\pi\times\pi$ as a final topology, where $\pi:X\to X/{\sim}$ is the canonical projection.

I know that the two topologies coincide when $\pi$ is open, for example $X=G$ a topological group and $X/{\sim}=G/H$ for a normal subgroup $H$.

Equip $\Bbb R$ with the topology $\tau=\{U\cup(V\setminus K)\mid U,V\text{ open in the Euclidean topology}\}$, where $$K=\left\{1,\tfrac12,\tfrac13,\dots\right\}$$ This is the coarsest topology containing the Euclidean one and having $K$ as a closed set, and it is generated by the sets $(a,b)$ and $(a,b)\setminus K$ as a basis, where $a$ and $b$ range over the reals. Call this space $X$.
Let $q:X\to Y$ be the quotient map collapsing $K$ to a point, and denote by $Y\times_q Y$ the product with the final topology of $q\times q$. We'll show that there is a closed set $C$ in $Y\times_q Y$ which is not closed in the product space $Y\times Y$. Let $$\textstyle C = (K\times K)\cup \Delta_X$$ This set is closed, since $K$ is closed and $X$ is Hausdorff. Note that $C$ is the equivalence relation induced by $q$, and it is easy to show that for any map $q$ the induced relation is saturated with respect to $q\times q$. So $C$ is closed and saturated, and it follows that $(q\times q)(C) = \Delta_Y$ is closed in the quotient space $Y \times_q Y$. Now for $\Delta_Y$ to be closed in $Y\times Y$, this space had to be Hausdorff. However, we can show that $\{0\}$ and $K$ cannot be separated by neighborhoods in $Y$, and we do this by showing that any open sets $U$ and $V$ around $0$ and $K$, respectively, intersect.
So let $\varepsilon$ and $\varepsilon_n$ be positive numbers such that $$0 \in (\varepsilon,\varepsilon)\setminus K \subseteq U \\ \textstyle K \subset \coprod_{n=1}^\infty\left(\tfrac1n-\varepsilon_n,\;\tfrac1n+\varepsilon_n\right) \subseteq V$$ Choose natural numbers $k>\tfrac1\varepsilon$ and $m > \tfrac1{\varepsilon_k}+k(k+1)$. Then $1/k - 1/m$ is an element of $\left(\tfrac1k- \varepsilon_k, \, \tfrac1k+\varepsilon_k\right) \cap (\varepsilon,\varepsilon)\setminus K$, hence $U$ and $V$ intersect. This means that $\Delta_Y$ is not closed in $Y \times Y$, and therefore $Y \times Y$ does not have the final topology with respect to $q\times q$.
The general idea here (which was suggested by t.b. in the comments here, although I had not seen this comment when I wrote my answer) is that we have a non-regular Hausdorff space $X$ with a closed subset $A$ which cannot be separated from a point $x$ by neighborhoods. Then $\{x\}$ cannot be separated from $A$ in $X/A$, so this quotient space is not Hausdorff. Then $\Delta_{X/A}$ is not closed although its preimage $\Delta_X \cup A\times A$ under $q \times q$ is closed in $X \times X$. This means that $q \times q$ is not a quotient map.
There is an example on p. 111 of Topology and Groupoids. Let $h:\mathbb Q \to \mathbb Q / \mathbb Z$ be obtained by identifying the integers in the rationals to a single point. Then it is shown there that $h \times 1_{\mathbb Q}$ is not an identification map.