Mathematical Analysis I by Zorich: The diagonal of the Cartesian square of a circle Question #5 of Section 1.2 in Zorich's Mathematical Analysis I defines the set:
$\Delta = \{ (x_1, x_2) \in X^2 \mid x_1 = x_2  \} $
as the diagonal of the Cartesian square $X^2$ of the set $X$. It then proceeds to ask to find geometric representations of the diagonal of the product of two  circles.
What I have figured out is that we can define the circle as
$ S = \{ (x,y) \mid x^2 + y^2 = r^2, r \in \mathbb{R}^+ \} $. 
Now, with $ S \times S = S^2 $ and the parameterization above, I am having trouble visualizing the diagonal of the torus. By definition I know that the diagonal $ \nabla$ will contain all $\{ (s_1, s_2) \in S^2 \mid s_1 = s_2 \}$, but I have no idea how to visualize this using the $(x, y) $ parameterization of the circle. 
Is there a better parameterization or a visualization to this diagonal set? Thank you very much.
 A: Let us "draw the torus" $\Bbb T=I\times I/\sim$, $I$ being the unit interval $[0,1]$, and the equivalence relations identifies $(x,y)\sim (x',y')$ if 


*

*$x,x'\in\{0,1\}$ are (different) end points of the interval, and $y=y'$, and

*$y,y'\in\{0,1\}$ are (different) end points of the interval, and $x=x'$,


and then we will look at some reference points on the diagonal. I will take only four, this is enough i think.
+-----------------------A
|                       |
|                       |
|                 D     |
|                       |
|                       |
|           C           |
|                       |
|                       |
|     B                 |
|                       |
|                       |
A-----------------------+

And now we make / steal / cite a picture of the torus from above...
                  .-"   "-.
                 .'   . ;   `.
                /    : . ' :  \
               |   `  .-. . '  |
               |  :  (   ) ; ` |
               |   :  `-'   :  |
                \   .` ;  :   /
         hjw     `.   . '   .'
                   `-.___.-'

and place the four points on it:
                  .-"   "-.
                 .'   . ;   `.
                /    : . ' :  \
               |   `  (C. . '  |
               |  |) (   ) |B` |
               |   :  `-'   :  |
                \   .` ;  :   /
         hjw     `.   . '   .'
         dan       `-.___.-'
                      /A\

and note that 


*

*the points $\Bbb A=$/A\ and $\Bbb C=$(C are in the plane of symmetry of the donut perpendicular on our view line, which is the radial symmetry line, we see them only tangentially from the $\infty$ point on the line,

*the point $\Bbb B$ is "up", we can see it well,

*the point $\Bbb D=$|D is only seen through a transparent donut, if the donut lies on a table, than $\Bbb D$ is both on the donut and on the table.  


In the process of mapping the diagonal from the $I\times I/\sim$ $2D$ picture to the "$3D$ picture" we are moving around the donut with the same radial velocity as we are moving on the circle of the donut transversal to our place. So we start at 6 o'clock at the sea level, and after a quarter of moving around we are at 3 o'clock, and top on the mountain. Then we move to 12 o'clock, we are again at the sea level, but we are touching the inner lake. Now we buy a submarine...

Picture was taken form http://ascii.co.uk/art/donut
