Quotient Space by group action Let $G$ a group and $X$ a topological space. We can define a group action of $G\times G$ over $X\times X$ in the obvious way, and give the quotient topology to $(X\times X)/(G\times G)$. How I can prove that there exists a homeomorphism $(X\times X)/(G \times G)\cong (X/G)\times(X/G)$?
 A: I don’t have a complete answer to the question, but I suspect the claim to be false.
For every set $X$ and every equivalence relation $∼$ on $X$, there exists an action of a group $G$ on $X$ such that $X / {∼} = X / G$.
(One may take $G$ as the group of all permutations of the set $X$ that only permute elements in the same equivalence class.)
Your question can therefore be rephrased as follows:

Let $X$ be a topological space and let $∼$ be an equivalence relation on $X$. Consider on the product $X × X$ the induced equivalence relation
$$
  (x, y) ∼ (x', y')
  \iff
  \text{$x ∼ x'$ and $y ∼ y'$} \,.
$$
Are the spaces $(X × X) / {∼}$ and $(X / ∼) × (X / {∼})$ homeomorphic?
If yes, then how can this be proven?

We always have a continuous bijection given by
$$
  φ
  \colon
  (X × X) / {∼} \longrightarrow (X / {∼}) × (X / {∼}) \,,
  \quad
  [(x, y)] \longmapsto ([x], [y]) \,.
$$
So we may split our question up into two parts:

*

*Is the map $φ$ a homeomorphism?


*If not, are the topological spaces $(X × X) / {∼}$ and $(X / {∼}) × (X / {∼})$ still somehow homeomorphic?
I don’t know a counterexample to the second part, but the first part is known to be false.
There are actually multiple question on this site that deal with the first part (for example, A relation between product and quotient topology. and Products of quotient topology same as quotient of product topology).
However, the provided answers only show something weaker, namely that the map $X × X \to (X / {∼}) × X$ is not a quotient map.
Actual counterexamples can be found in:

*

*Munkres’ Topology (2nd edition, Pearson New International Edition), Chapter 2, Section 22, Exercise 6 (page 145).


*Multiplication is discontinuous in the Hawaiian earring
group (with the quotient topology) by Paul Fabel (http://arxiv.org/abs/0909.3086v2), Theorem 1.
In these counterexamples, simply showing that the map $X × X \to (X / {∼}) × (X / {∼})$ is not a quotient map (which is a reformulation of $φ$ being a homeomorphism) is already tedious, and the counterexamples are also far from obvous:

*

*The first edition on Bourbaki’s General Topology wrongly claimed that $(X × Y) / {∼} ≅ (X / {∼}) × (Y / {∼})$ (according to https://freedommathdance.blogspot.com/2012/12/product-of-two-quotient-spaces-frequent.html). This was corrected in later editions, and the counterexample provided in these later editions is basically the one counterexample that nearly everyone copies.


*I say “nearly” because the counterexample provided in Fabel’s paper seems to be of a different kind.
But this counterexample is also far from obvious, as the popularity of https://mathoverflow.net/questions/26680/fundamental-group-as-topological-group attests.
One of these counterexamples may also work for the second part, but I’m not good enough at point set topology to judge if this is true.
