Let $\mathbb{T}^2$ be the torus and have $\mathbb{Z}_2$ action on the torus by permuting the coordinates. I am trying to prove the orbit space is congruent to the mobius strip $\mathbb{T}^2/\mathbb{Z}^2 \cong M$.

I am writing $M\cong I^2/\alpha$, where $I=[0,1]$, $\alpha(x)=1-x$ and we identify the points $(x,0)\sim (\alpha(x),1)$. I know it is useful for this problem to identify the torus as the quotient $I^2/\sim_0$ with left glued to right and top glued to bottom (referring to sides) and then the action on the torus corresponds to switching coordinates on the square, which is reflecting over the diagonal $(x,x)$. What I don't understand is how to combine these relations of gluing and permuting to show this space is homeomorphic to the quotient above with $\alpha$.

I think that with both relations we will identify the edges together since $(x,0)\sim_0(x,1)\sim(1,x) \sim_0(0,x)$. Then points not on the edges will either be identified with themselves if they are on the diagonal or just their permutation. But I can't figure out how to map this to the Mobius strip...

How do you proceed formally with proving these spaces are homeomorphic? Obviously I'm not looking for all the details but I'm stuck both trying to visualize it (i get to folding the square over the diagonal $(x,x)$ but then I cant picture how to get the mobius strip from there on). Any help is appreciative.

Also a different generalization than the natural one: instead of permuting the coordinates in $\mathbb{T}^k$ and taking the action by $S_n$ consider the action by reversing the coordinates, which can still be done by using $\mathbb{Z}_2$. What is this orbit space homeomorphic to? Would it be $M^k$ when $k$ is even and something else for odd $k$?

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    $\begingroup$ Just observe that the quotient us nonorientable and compute it's Euler characteristic (zero) and number if boundary components (one) $\endgroup$ Aug 4, 2016 at 19:04
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    $\begingroup$ @studiosus ah! Excellent, but i havent studied homology proper yet. I imagine its more tedious but is there a way to do this with more basic tools like gluing and properties of quotient maps? (I will start studying homology now anyway) $\endgroup$ Aug 4, 2016 at 22:05

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From your other question it seems that you've figured out how to show that the quotient is a Mobius strip. For sake of completeness, here's the argument in pictorial form (I apologize for the crude drawings.): enter image description here

Regarding the last step, maybe this (even more poorly-drawn) picture would help: enter image description here

  • $\begingroup$ Tbh i actually havent been able to figure out (in formal detail) the last step for the picture: from the triangle back to the rectangle with sides "twisted". I moved on without the formal proof because i had spent quite some time on this without getting it. If you could outline that as well i'd be very thankful! I will also have to take sometime to digest your comment on my new question (and im at work). Again, thank you! $\endgroup$ Aug 31, 2016 at 14:05

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