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I have to identify the orbit space of the action of the group $\mathbb{Z_2}$ on the torus $T^2=\left \{ (x,y,z)\in \mathbb{R}^3 |(2-\sqrt{x^2+y^2})^2+z^2 =1 \right \}$ generated by the homeomorphism $f(x,y,z)=(-x,-y,z)$

I think it isn't so hard, but i'm having some trouble understanding quotient spaces, and maybe this example would help me. Any hints?

pedomarco
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    Hint: Start by understanding the quotient of the circle $T^1 = {(x,y) \in \mathbb{R}^2 : x^2+y^2}$ by the homeomorphism $g(x,y) = (-x,-y)$, i.e. reflection through the origin.

    Your example is not so different than this one. You are just working with a "tube" enclosing a circle instead of the circle itself.

     – Mike F Nov 16 '20 at 20:52
  • Okay, I'll try to see that first. So in your example the quotient space is homeomorphic to the sphere, isn't it? – pedomarco Nov 20 '20 at 13:42
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    I guess there's a typo in my comment, should have said $x^2+y^2=1$. Anyway, no the quotient of the circle by the map $g$ I wrote is still a circle. Actually, the quotient of the circle by the equivalence relation generated by rotation by $\frac{2 \pi}{n}$ (for fixed and arbitrary $n$) is still a circle. The case you are dealing with is the case of rotation by $\pi$. – Mike F Nov 20 '20 at 15:45
  • First of all, thank you very much for the answers. Yes, I thought you were talking about a circle. Now I see your example (more less), but I still dont't understand very well my problem, it's really hard for me to think the quotient in 3 dimensions. So if I have a rotation by $\pi$ , is it still true for the torus that the quotient is still the same torus? – pedomarco Nov 20 '20 at 16:40
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    Yes that's correct. You might find things easier if you view the torus as $S^1 \times S^1$, where $S^1$ is the group of complex numbers of length $1$. You could come up with a homeomorphism between this space and your $T^2$. The equivalence relation you are taking the quotient by can be recognized as $(z,w) \sim (-z,w)$. This is also the equivalence relation induced by the surjective map $S^1 \times S^1 \to S^1 \times S^1$ sending $(z,w) \mapsto (z^2,w)$. Continuous surjections of compact Hausdorff spaces are always quotient maps, so you get for free that the quotient is $S^1 \times S^1$. – Mike F Nov 20 '20 at 16:49
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    By the way, once you get more comfortable in topology, you will probably come to view your guess that "the quotient space [of the circle] is homeomorphic to the sphere" as kind of a weird one. The circle is 1-dimensional and the sphere is 2-dimensional. You do not typically expect the dimension of an object to increase when taking a quotient. It is actually possible to get a continuous surjection (hence a quotient map) from the circle to the sphere. Search for "space filling curves". However, it's clear that the relation you are dealing with is nowhere near wild enough to increase dimension. – Mike F Nov 20 '20 at 16:54
  • Again, thank you very much. I will study this immediately and post my resolution. I thought that the quotient "increase" the dimension because i was trying to see this problem like one example I saw about the quotient of a square $[0,1]\times [0,1]$ by a relation that identifies $(0,y)$ with $(1,y)$ wich is homeomorphic to the cylinder $\mathbb{S}^1\times[0,1]$.As I saw that from a square we get a cylinder I thougt that, but yes now thinking better the dimension isn't increasing. – pedomarco Nov 20 '20 at 17:19
  • One more question, reading this post link in the first answer it is told that the quotient of the torus by a rotation of 180 degrees around an horizontal axis is homeomorphic to the sphere. That isn't like my problem because my rotation is around a vertical axis, is it? – pedomarco Nov 20 '20 at 17:58
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    Yes everything you said above is correct. The z-axis is disjoint from your torus $T^2$, and so the 180 degree rotation around the z-axis you consider does not fix any points of $T^2$. For the transformation in the question you linked, however, the horizontal axis will intersect the torus in four points, and those four points will be fixed by rotation about that axis. Those fixed points of the rotation are the "branch points" referred to under that link. – Mike F Nov 20 '20 at 19:25
  • Okey, I understad. Thank you again that's so interesting! – pedomarco Nov 20 '20 at 21:48

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