# Quotient sets and tori

My trek through MacLane and Birkhoff's Algebra has brought me to exercise 1.9.3, which defines the equivalence relation $E$ in $\Bbb R^2$ by $(x, y) E (x', y')$ iff $x-x'$ and $y-y'$ are both integers and asks me to prove that the quotient set $\Bbb R^2/E$ may be described as the set of points on a torus. Sadly, I can't fathom what this question is trying to express, since the quotient set appears to me to be a set of like grids of points that you could connect to make a bunch of $1\times 1$ squares if you wanted. I mean, a torus is three-dimensional last time I checked, so I don't know how you'd go about saying a partition of $\Bbb R^2$ is the same as a subset of $\Bbb R^3$. If someone could help me out in solving this problem, that'd be great. Thanks!

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I imagine the fact that the torus is the unit square with the opposite sides glued together will be useful. – Holdsworth88 Jun 15 '12 at 8:29
The pictures are worth thousands of words, and are really more helpful in handling this question. But if you prefer a more formal way, let's do the following. Map the point $(x,y)$ on the real plane to the point $$f(x,y)=((2+\cos2\pi x)\cos2\pi y,(2+\cos2\pi x)\sin2\pi y, \sin2\pi x).$$ I invite you to verify as an exercise that $f(x,y)=f(x',y')$ if and only if $x-x'$ and $y-y'$ are both integers. Therefore $f$ gives an injective mapping from $\mathbb{R}^2/E$ to $\mathbb{R}^3$. Care to guess what the image of $f$ looks like :-) – Jyrki Lahtonen Jun 15 '12 at 9:40

By torus they mean just the surface of the ‘doughnut’, which is indeed two-dimensional.

Note that the relation actually gives you just one $1\times 1$ square: the points $\langle 0.2,0.3\rangle,\langle 1.2,0.3\rangle$, $\langle 5.2,100.3\rangle,\langle-1.8,23.2\rangle$, etc. all get collapsed to a single point.

Added: Let $S=\{\langle x,y\rangle:0\le x,y<1\}$, the unit square in the first quadrant including the left and bottom edges but not the top or right edge. $S$ contains exactly one point from each $E$-equivalent class. You can think of $E$ as cutting up the plane into squares and melting them all together into the single square $S$. Suppose that you start at $\langle 0.5,0.5\rangle$ in $S$ and travel in the positive $x$-direction, through $\langle 0.6,0.5\rangle$, $\langle 0.7,0.5\rangle$, and so on. You approach the point $\langle 1,0.5\rangle$, which isn’t in $S$. But $$\langle 1,0.5\rangle\;E\;\langle 0,0.5\rangle\;,$$ so $\langle 1,0.5\rangle$ and $\langle 0,0.5\rangle$ are the same point of $\Bbb R^2/E$, and $\langle 0,0.5\rangle$ is in $S$. Thus, if you keep going to the right along the line $y=0.5$, you ‘disappear’ off the righthand edge of $S$ and immediately reappear at $\langle 0,0.5\rangle$ on the lefthand edge. (It’s rather like what happened to torpedoes in some of the early arcade games.) For all practical purposes the square has been rolled into a cylinder with the left and right edges glued together.

The same thing happens when you travel vertically: as you ‘disappear’ off the top edge, you pop back in at the bottom. In effect the vertical cylinder has been wrapped around so that its circular rims can be sewn together to make an inner tube $-$ i.e., a torus.

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Wow!! Math is so beautiful. Thanks for all the help! – Toby Carter Jun 15 '12 at 9:03
@Brian M. Scott : I wish i had a teacher like you :D. – Theorem Jun 15 '12 at 9:32
@Theorem: Thanks! – Brian M. Scott Jun 15 '12 at 9:33

When you factorize points by $x$ coordinate your 2 dimensional plane become a tube.

Now when you try to factorize by $y$ coordinate points of this tube will fold into a torus.

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What is this "factorize" you speak of? Also those pictures look very three-dimensional, but I was assured the torus in question was two-dimensional, though I still am not sure what this "'doughnut'" everyone is talking about is. – Toby Carter Jun 15 '12 at 8:51
this torus is hollow - a two-dimensional surface. Your problem is in understanding of torus. It is a two dimensional surface, not a three dimensional object with interioir – Norbert Jun 15 '12 at 8:54
Yep I thought anything in R^3 was called three-dimensional, but it turns out a 2-manifold in R^3 is 2-dimensional. Pretty soon I'll figure out LaTeX, too. Thanks! – Toby Carter Jun 15 '12 at 9:10
You are wellcome! If you are satisfied with my or with Brian's answer then you can accept it. ( meta.math.stackexchange.com/questions/3286/…). – Norbert Jun 15 '12 at 9:12