Product of quotient topology not homeomorphic to the quotient of product topology How can I construct a counter-example for the following statement: if $\sim_X$ is an equivalence relation on $X$, $\sim_Y$ is an equivalence relation on $Y$, then $X/{\sim_X} \times Y/{\sim_Y}$ is homeomorphic to $(X \times Y)/{\sim_{XY}}$ where $(x_1,y_1) \sim_{XY} (x_2,y_2)$ iff $x_1 \sim_X x_2$ and $y_1 \sim_Y y_2$?
There is a related question on this, Products of quotient topology same as quotient of product topology , which says that it is false that if $p : X \to Z_1$ and $q : Y \to Z_2$ are quotient maps, then $p \times q$ is necessarily a quotient map. However, that this particular map is not a quotient map doesn't imply that an homeomorphism cannot be constructed.
 A: This answer is incorrect, and does not contribute to answering the posted question. Don't read it, and if you do, keep in mind it only illustrates how one could make a mistake.
Let $X=Y=[0,1]$ with $0\sim1$. Then $X/\sim$ and $Y/\sim$ each is homeomorphic to a circle, so their product is a torus. On the other hand $(X\times Y)/\sim$
is a square with the four corners identified, which is not homeomorphic to a torus.
Edit to address the comment below.
Take a handkerchief and lift the four corners, bending it, so they touch. (But, apart from the corners meeting at a single point, the edges do not touch.) This is $(X\times Y)/\sim$.
Here is a more formal justification that there is no homeomorphism between $(X/\sim)\times(Y/\sim)$ (a torus) and $(X\times Y)/\sim$.
Removing one point from $(X\times Y)/\sim$ - the point where the corners of the handkerchief touch - leaves a simply connected space (a closed square with the four corners removed). There is no point with this property on the torus.
