Topology Group action 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.
 A: 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.
A: 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.
