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I'm just starting topology and am having problems. I have an infinite cylinder $C=${$(x,y,z):x^2+y^2=1$} and want to describe a group action that produces a torus. Intuitively, I imagine bending $C$ and gluing the infinite ends together, so $(x,y,z)$~$(x,y,-z)$. My guess is that the group I want is either $(\mathbb{Z},+)$ or $(\mathbb{Z}/2\mathbb{Z}, *)$, but I'm not sure.

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up vote 2 down vote accepted

Let $G=\Bbb Z$ and $n(x,y,z)=(x,y,n+z)$. Then $C/G$ is the torus $T^2$ with $q:C\to T^2$ given by $q(x,y,z)=\left((x,y),e^{2\pi i z}\right)$. Here $C$ should be thought of as $S^1×\Bbb R$ and $T^2$ is $S^1×S^1$. It is easy to check that $q\circ n=q$ for each $n\in\Bbb Z$, and whenever $p:C\to Y$ is a map satisfying $p\circ n=p$, then there is a unique continuous map $h:T^2\to Y$ such that $h\circ q=p$.

To see that $q$ is an open map, note that $g:z\mapsto e^{2\pi i z}$ is open, since it maps an interval $(a,b),\ a<b<a+1$ to the image of the open and saturated set $[0,1]\cap((a-\lfloor a⌋,b-⌊a\rfloor)\cup(a-⌊a⌋-1,b-⌊a⌋-1))$ under the quotient map $[0,1]\to S^1$. It follows that $q=\text{Id}_{S^1}×g$ is open.

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Let $\mathbb{Z}$ act by translation vertically: $k \cdot (x,y,z) = (x,y,z+k).$ Then the half-open cylinder with $0 \leq z < 1$ is a set of representatives for $C/\mathbb{Z},$ and you can check that you get a torus (with the topology induced by the quotient map). In fact the cylinder is then a covering space for $C/\mathbb{Z},$ since the latter is a quotient under a properly discontinuous action of a discrete group.

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@anon You can go up or down, no? – user126869 Feb 12 '14 at 23:32
@anon: The half-open cylinder is just a set of representatives, the topology is the quotient topology where $(x,y,1-\varepsilon)$ is close to $(x,y,1) = (x,y,0)$. – Berci Feb 12 '14 at 23:32
Ah, OP originally had $x+y^2=1$. Now that it's changed to $x^2+y^2=1$ this is all trivial so my comments can be ignored. Although strangely it was Berci that changed the ^1 (which OP had explicitly typed like that) to a ^2, not the OP. – anon Feb 12 '14 at 23:39
@anon Oh, and yeah of course the quotient of the variety $V(x+y^2-1)\subset\mathbb R^3$ by $\mathbb Z$ in the $z$-direction is topologically a cylinder. – Karl Kronenfeld Feb 12 '14 at 23:42

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