A relation between product and quotient topology.

I was studying a topic about Algebraic Topology and a question popped into my mind:

Suppose that we have two topological spaces $X$ and $Y$. Let $\sim_X$ and $\sim_Y$ equivalence relations in X and Y. In $X\times Y$, we can define the following equivalence relation:

$$(x,y) \sim (x',y')\ \ \mbox{when}\ \ x\ \sim_X\ x' \ \ \ \mbox{and}\ \ \ y \ \sim_Y \ y'.$$

Then $\frac{X\times Y}{\sim}$ and $\frac{X}{\sim_X}\times \frac{Y}{\sim_Y}$ are homeomorphic?

There's a natural map $f:\frac{X\times Y}{\sim} \longrightarrow \frac{X}{\sim_X}\times \frac{Y}{\sim_Y}$ given by $$f([(x,y)]) = ([x],[y])$$ which is a continuous bijection. But I don't know whether $f^{-1}$ is continuous. Does this result hold? Any hint?

Not an answer, but a remark which is too important (in my opinion) to be just a comment: $\mathsf{Top}$ fails to satisfy several convenient categorical properties, such as the one discussed here that quotient maps don't have to be stable under products. Related to that, $\mathsf{Top}$ is not cartesian closed. However, in convenient categories of spaces all these properties hold (see here). An example is the category of compactly generated (weak) Hausdorff spaces $\mathsf{CGWH}$ (see here and there).

All this indicates that $\mathsf{Top}$ is the "wrong" category for doing topology. A better candidate is $\mathsf{CGWH}$. In practice, there is no additional effort when working in that "correct" category.

Notice that many books and papers in topology just use all these convenient properties although they claim to work with topological spaces. Most authors (seem to) ignore this problem, they even don't let the reader know what their category of "spaces" is (topological spaces? compact spaces? compactly generated spaces? CW complexes? simplicial sets? Almost everything has been called "spaces").

Well, I've found a reference here: http://home.imf.au.dk/marcel/gentop/noter.pdf. Exercise 4.4.7 in those notes describes an example, which is also (I think) referred to here: Products of quotient topology same as quotient of product topology. I'll wave my hands at it a little bit - feel free to correct me if things seem wrong.

Let $X$ be the set of positive rational numbers, and define an equivalence relation on it by declaring any two positive integers to be equivalent. Call that $\sim_{X}$. Let $Y = \mathbb{Q}$, and define the trivial equivalence relation on it, so we identify no points. Then the claim is that $X \times Y/\sim$ is not homeomorphic to $X/\sim_{X} \times Y$.

Define $$ U_{n} = \left\{ (x,y) \in X \times Y \,| \, |x-n| < \textrm{min}(|y - \frac{\sqrt{2}}{n}|,1/2) \right\}.$$ Then set $U = \cup_{n \geq 1} U_{n}$. So each $U_{n}$ is a set around $(n,0)$ in $X \times Y$, where $y$ can be whatever it wants, $x$ is always between $n-1/2$ and $n+1/2$, and if $y$ is close to $\sqrt{2}/n$, then $x$ has to be even closer to $n$. I picture it as sort of an infinitely tall hourglass, where it pinches in near $y = \sqrt{2}/n$.

Then if $q$ is the quotient map from $X \times Y$ to $X \times Y/\sim$, $q^{-1}(q(U)) = U$, so that $q(U)$ is open. This is relatively easy to see - $q$ is only non-injective at points of the form $(m,y)$, where $m$ is an integer. But if $(m,y) \in U$, then every point $(n,y) \in U_{n}$, because $y - \frac{\sqrt{2}}{n}$ can never be $0$, so that $x=n$ is always allowed in $U_{n}$.

Now, if $\epsilon > \sqrt{2}/n$, then there are no $a <n < b$ so that the cylinder $(a,b) \times (-\epsilon, \epsilon) \cap X \times Y$ is contained in $U_{n}$. Essentially, this is because $(a,b) \times (-\epsilon,\epsilon)$ contains the "pinch" of the hourglass, which gets arbitrarily narrow, so no interval $(a,b)$ is narrower than it.

Now, we look at $f(q(U))$. We claim it is not open because it contains no neighborhood of $([1],0)$. Let $\epsilon >0$ be given and assume that $\epsilon < n/ \sqrt{2}$. (Just look at a large enough $n$ to make that true.) If $f(q(U)$ were open, we could find a neighborhood of $([1],0)$ contained in $f(q(U))$ whose $y$-coordinates were all in $(-\epsilon,\epsilon)$. But the previous paragraph essentially proves that can't happen.

Intuitively, $q(U)$ is open because no matter what $n$ is, $q(U)$ contains a neighborhood of $(n,y)$ - the hourglass has SOME width everywhere, even though it's arbitrarily small. But, if you fix $y=0$ and try and put a neighborhood around every $(n,0)$ all at once (i.e. put a neighborhood around $([1],y)$), then for large enough $n$, your small neighborhood will extend around the pinch of the hourglass for that particular $n$.