Note: My questions are marked in boldface.
In Devaney's book "An Introduction to Chaotic Dynamical Systems", Westview Press (2003), on p. 190 he defines the $2$-torus by
To describe the torus, let us begin with the plane. We will consider
as identical all points whose coordinates differ by integers. That is to say,
the point $(\alpha, \beta)$ in the plane is to be regarded as the same as the points
$(\alpha + 1, \beta)$, $(\alpha + 5, \beta + 3)$, and, in general, $(\alpha + M, \beta + N)$, where $M$ and $N$ are integers. We let $[\alpha, \beta]$ denote the set of all points equivalent to $(\alpha, \beta)$ under this relation. To be somewhat more formal, the relation $(x, y) \sim (x', y')$ if
and only if $x − x'$ and $y − y'$ are integers gives an equivalence relation on
points in the plane. The torus is thus the set of all equivalence classes under
this relation.
I am able to show that the above relation is an equivalence relation and I can construct examples and counterexamples of points being equivalent or not equivalent. And I understand that an equivalence relation partitions a set.
But I cannot see how that "the torus is the set of all equivalence classes under
this relation".
I also fail to see the relation of geometrically constructing a torus from the unit square as given by the author:
Geometrically, this procedure can be visualized as follows. Consider the
unit square in the plane $0 \leq x, y \leq 1$. Under the above identifications, only
points on the boundary of the square need be considered. Indeed, the top
boundary $y = 1$ should be considered the same as the bottom boundary
$y = 0$, and similarly the left and right boundaries $x = 0$ and $x = 1$ should be
identified. When this occurs, the square becomes first a cylinder and then a
torus
Now, I understand how the torus can be constructed from a square geometrically, by "gluing" together the vertical ends and then the horizontal ends. But how is the geometric construction related to the definition of the torus being the set of all equivalence classes?
From my search on the internet I found that the torus is usually defined as the quotient group
$\mathbb{R}^{2} / \mathbb{Z}^{2}$ of the group $\mathbb{R}^{2}$ by the normal subgroup $\mathbb{Z}^{2}$.
I also found that the torus is homeomorphic to $S^{1} \times S^{1}$, where $S^{1}$ is a circle.
Now, from my intuition with the "gluing" principle, is the $1$-torus then just "gluing" the ends of the unit interval together and hence giving a circle?.
Best Answer
Imagine cutting the plane into $1\times 1$ squares along the lines $x=n$ and $y=n$ for all integers $n$; you get the squares $[m,m+1]\times[n,n+1]$ for all pairs of integers $m$ and $n$. Now stack these squares in a single $1\times 1$ pile on the unit square $[0,1]\times[0,1]$, preserving their orientations; when you’ve done that, all of the lower lefthand corners of the squares will lie above $\langle 0,0\rangle$, all of the upper righthand corners will lie above $\langle 1,1\rangle$, all of the points of the form $\left\langle m+\frac12,n+\frac12\right\rangle$ will lie above $\left\langle\frac12,\frac12\right\rangle$, and so on.
If you examine the equivalence relation $\sim$ carefully, you’ll see that if $x,y\in\Bbb R\setminus\Bbb Z$, the $\sim$-equivalence class of $\langle x,y\rangle$ is exactly the set of points lying on the vertical line through $\langle x,y\rangle$ in this infinite stack of squares. You’ll also see that if $x$ or $y$ is an integer, and $\langle u,v\rangle$ is on the vertical line through $\langle x,y\rangle$, then $\langle x,y\rangle\sim\langle u,v\rangle$. Thus, every point in any of these squares is $\sim$-equivalent to the point directly below it in the square $[0,1]\times[0,1]$ at the bottom of the stack. Thus, the identification of equivalence classes to single points works (in part) by squashing the whole stack vertically down to the unit square $[0,1]\times[0,1]$.
But there’s a little more to it than that. Take the point $\left\langle\frac32,2\right\rangle$ for instance. It’s the midpoint of the upper edge of the square $[1,2]\times[1,2]$, but it’s also the midpoint of the lower edge of the square $[1,2]\times[2,3]$. Thus, when we squash the stack down to $[0,1]\times[0,1]$, it ends up being squashed both to $\left\langle\frac12,1\right\rangle$ and to $\left\langle\frac12,0\right\rangle$, so these points are in the same $\sim$-class and must be identified. The same thing happens with all points of $(\Bbb R\setminus\Bbb Z)\times\Bbb Z$: each of them ends up squashed to a point on the top edge and the matching point on the bottom edge of $[0,1]\times[0,1]$. Identifying these two edges then amounts to rolling the unit square into a cylinder whose axis is parallel to the $x$-axis.
Finally, points like $\left\langle 3,\frac54\right\rangle\in\Bbb Z\times(\Bbb R\setminus\Bbb Z)$ get squashed to corresponding points on the left and right edges of the unit square, in this case $\left\langle 0,\frac14\right\rangle$ and $\left\langle 1,\frac14\right\rangle$. Identifying these amounts to bending the cylinder around into a torus, identifying the left and right circular ends of the cylinder.
The answer to your last question is yes.