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.
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.
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.