# $(X \times X) /{\sim'}\cong (X/{\sim}) \times (X/{\sim})$

The full description of this problem is:

Let $X$ be a topological space. Let $\sim'$ be the equivalence relation on $X\times X$ defined by $(x,y)\sim'(x',y')$ iff $x \sim x'$ and $y \sim y'$

Prove that $(X \times X) /{\sim'}\,\cong (X/{\sim}) \times (X/{\sim})$.

Intuitively it makes so much sense, but I am really stuck on how to write a formal proof. Can anyone give me a outline?

Great thanks!

• What is $X$? Just a set? What does it mean $\cong$? – JonSK Apr 28 '16 at 2:45
• Respectively: a topological space, no, and homeomorphic @JonSK – user228113 Apr 28 '16 at 2:46
• X is a topological space and $\cong$ stands for homeomorphic. I ll edit it now – Xuan Apr 28 '16 at 2:46
• I don't think this is true in general...it's definitely not true in general if you let the two factors be different spaces. – Eric Wofsey Apr 28 '16 at 3:14
• @EricWofsey Shall the quotient map $\pi: X \to X /\sim$ be open? – Xuan Apr 28 '16 at 3:24

Hint

The homeomorphism is given by $$f:(X\times X)/{{\sim}'}\longrightarrow (X/{\sim})\times (X/{\sim})$$ defined by $$f([(x,y)])=([x],[y]).$$

Prove it !

• Protip (that I just searched on TeX.se for because it was driving me nuts!), it turns out enclosing \sim in braces makes it play a bit more nicely with the slash; e.g., $X/{\sim}$ vs $X/\sim$ or $X/_\sim$ – pjs36 Apr 28 '16 at 2:58
• Thanks a lot!. It is clearly bijective, but how to show it is continuous? Is it supposed to use the definition of open sets of quotient space? – Xuan Apr 28 '16 at 2:58

Hint -

Using the universal property of quotient spaces you can get a continuous map $\phi:(X\times X)/ {\sim'}\to X/ {\sim}\times X/ {\sim}$ which in this case you can show to be bijective and open.

• Thanks! Is my idea correct? since $\pi \times \pi: X \times X \to (X/\sim)\times (X/\sim)$ is a quotient map, then by the universal property, there exists such $\phi$? And then show it is bijective and open. – Xuan Apr 28 '16 at 3:02
• @Xuan, absolutely! you are right – R_D Apr 28 '16 at 3:04
• You can't easily show $\phi$ is open, since it isn't in general. – Eric Wofsey Apr 28 '16 at 3:36
• @Eric Wofsey, I didn't mean that it was true in general. I meant it for this problem. I shall remove the word "easily" in my answer at it depends on person to person. – R_D Apr 28 '16 at 3:44