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!

  • $\begingroup$ What is $X$? Just a set? What does it mean $\cong$? $\endgroup$ – JonSK Apr 28 '16 at 2:45
  • $\begingroup$ Respectively: a topological space, no, and homeomorphic @JonSK $\endgroup$ – user228113 Apr 28 '16 at 2:46
  • $\begingroup$ X is a topological space and $\cong$ stands for homeomorphic. I ll edit it now $\endgroup$ – Xuan Apr 28 '16 at 2:46
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Apr 28 '16 at 3:14
  • $\begingroup$ @EricWofsey Shall the quotient map $\pi: X \to X /\sim $ be open? $\endgroup$ – Xuan Apr 28 '16 at 3:24


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 !

  • $\begingroup$ 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 $X/_\sim$ $\endgroup$ – pjs36 Apr 28 '16 at 2:58
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Xuan Apr 28 '16 at 3:02
  • $\begingroup$ @Xuan, absolutely! you are right $\endgroup$ – R_D Apr 28 '16 at 3:04
  • $\begingroup$ You can't easily show $\phi$ is open, since it isn't in general. $\endgroup$ – Eric Wofsey Apr 28 '16 at 3:36
  • $\begingroup$ @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. $\endgroup$ – R_D Apr 28 '16 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.