Transversal of an equivalence relation Define a relation on $\mathbb{R}^2$ by $(a, b)\sim(c, d)$ if and
only if $(c-a, d- b) \in \mathbb{Z}^{2}$. Prove that $\sim$ is an equivalence relation.
Identifying $\mathbb{R}^2$ with the plane in the usual way, describe the most
natural transversal for $\sim$, which you can find. What, if anything,
has this question to do with doughnuts?
I have already proven the equivalence relation and know the solution of the second part since it is provided, but I have problems with understanding it. I presume it's asking about the equivalence classes?
Solution:

A natural transversal is $I \times I$ where $I = \{r | r \in \mathbb{R}, 0 \leq r < 1\}$. Join the top and bottom of $I$ to form a
tube, and bend the tube round to join the two circles as well. You
have the surface of a doughnut (a torus in mathematical language).

 A: I expect a transversal for an equaivalence relation is a set which intersects each equivalence class in exactly one point.
The set $I\times I$ is a transversal for the relation $\sim$.
When we add to a half-open square $I\times I$ its top and right sides, we obtain a closed unit square $Q$. But each point $(x,1)$ at the top side of $Q$ is equivalent to a point $(x,0)$ at the bottom side of $Q$. So, constructing a model of the quotient space $\Bbb R^2/\sim$, points of each pair $\{(x,1), (x,0)\}$ should be glued together. When we do this, we obtain a tube. Pairs of opposite points at the circles bounding the tube correspond to pairs $\{(0,x), (1,x)\}$ of  the opposite points of the left and the right sides of the square $Q$, which are equivalent, and so they also should be glued together. This two-step construction provides us a torus, as is stated in the proposed solution.
See also Wikipedia:

The torus can also be described as a quotient of the Cartesian plane under the identifications $$(x,y)\sim (x+1,y)\sim (x,y+1),$$
or, equivalently, as the quotient of the unit square by pasting the opposite edges together.

