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I am having some trouble with the following problem:

Define a relation on the square $[0,1]\times[0,1]$ as follows: for every $x,y\in[0,1]$, let $(x,y)\sim(x,y)$, $(0,y)\sim(1,y)$, and $(x,0)\sim(x,1)$. Show that the quotient map $\pi:[0,1]\times[0,1]\to([0,1]\times[0,1])/\sim$ is not an open map.

My knowledge of the problem: I know that a mapping $f:(X,\tau_{X})\to (Y,\tau_{Y})$ is open if and only if for any open set $\mathcal{O}\in \tau_{X}$, we have $f(\mathcal{O})\in\tau_{Y}$. I also know that the map $\pi$ is just the map which sends $x\mapsto [x]$, where $[x]$ is the equivalence class of $x$ under $\sim$. Finally, I think (not entirely sure) that this quotient should give us the torus.

Where I'm Stuck: I am having difficulty figuring out how to show that a set is not open in the quotient space. I think I might have to start with an open ball in $[0,1]^{2}$ and then show that its image is not necessarily open in the quotient space. I'm not sure if this is right though.

Thanks in advance for any suggestions!

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The set $\pi([0,1/2)\times [0,1/2))$ is not open in the quotient space (which is indeed a torus).

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    $\begingroup$ Okay! But because it's open in the original space, we can conclude that the map isn't open? $\endgroup$ – Sir_Math_Cat Sep 14 '17 at 3:29
  • $\begingroup$ @WarTurtle Right. $\endgroup$ – Alex Ravsky Sep 14 '17 at 3:30
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    $\begingroup$ Thank you very much! I think this makes more sense now. $\endgroup$ – Sir_Math_Cat Sep 14 '17 at 3:31
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    $\begingroup$ @AlexRavsky Could you elaborate as to why it is not open in the quotient space? $\endgroup$ – Perturbative Sep 14 '17 at 18:28
  • $\begingroup$ @WarTurtle This can be seen directly at a torus. :-) For a formal proof we may note that a sequence $\{\pi(1-\frac 1n, 1-\frac 1n)\}$ from its complement converges to its point $\pi(0,0)$, which is impossible for open sets. $\endgroup$ – Alex Ravsky Sep 14 '17 at 18:34

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