Cartensian product of Quotient maps Let $\sim_1 and \sim_2$ be equivalence relations on $X$ and $Y$.
Let $f_1: X \rightarrow X/ {\sim_1}$ and $f_2: Y \rightarrow Y/ {\sim_1}$ be quotient maps. 
Let $f_1 \times f_2: X\times Y \rightarrow X/ {\sim_1} \times Y/{\sim_2}$ be the map defined coordinatewise. Will $f_1 \times f_2$ be a quotient map.
Remark: It is at least true in case $X$ and $Y$ are groups and equivalence relations are defined through a subgroup.
 A: The product of two quotient maps need not be a quotient map.
We can actually find a quotient map $f: X \to Y$ whose square $f \times f$ is not a quotient map. The recipe is as follows: Take any Hausdorff space $X$ with a closed subset $A$ and a point $x \in X$ such that $x$ and $A$ cannot be separated by neighborhoods (so $X$ is not regular). Since $X$ is Hausdorff, the diagonal $\Delta_X$ in $X\times X$ is closed. Since $A$ is closed, so is $A\times A$. Hence $\sim \; = \Delta_X \cup (A\times A)$ is closed. The quotient map $f: X \to X/{\sim}$ is the map identifying $A$ to a single point. Note that $X/{\sim}$ is not Hausdorff since $A$ and $x$ do not have disjoint neighborhoods, so $\Delta_{X/{\sim}}$ is not closed. However, if $f \times f$ were a quotient map, then the diagonal would have to be closed since it has the preimage $\sim$ under $f\times f$.
We can even find a quotient map $f$ and a space $Q$ such that the product of $f$ with the identity on $Q$ (which is obviously a quotient map) fails to be a quotient map. See Spaces in which "$A \cap K$ is closed for all compact $K$" implies "$A$ is closed." for a construction if this map.
