Difference between $[0, 1) \times [0, 1)$ and the 2-torus Let me take the two dimensional case. I have seen a definition of the $2$-torus as
$T^2 = \mathbb{R}^2 / \mathbb{Z}^2$
i.e. the quotient between the real space and the integer space. However, for the equivalence relation
$x \sim y \iff \exists a \in \mathbb{Z}^2 : x = y + a$,
I reckon the quotient space is simply $[0,1) \times [0,1)$. What is the difference between this and the torus?
 A: Nothing, provided you understand $[0,1)\times [0,1)$ to have the quotient topology rather than the subspace topology. In the quotient topology, for example, every open set of a point of the form $(0, t)$ contains points of the form $(1-\epsilon, t)$ for sufficiently small $\epsilon$. This is not the case for the subspace topology.
This is an example where notation can be technically correct yet misleading nonetheless: writing the quotient as "$[0,1)\times [0,1)$" suggests a subspace of $\mathbb{R}^2$ when in fact it is a quotient with a very different topology.
A: As sets, these two spaces are indeed equal, however they differ in their topology. I'll try to give a slightly more intuitive and informal explaination of this instead of just a formal counterexample:
The torus $T^2$ can also be seen to be obtained from the square $[0, 1] \times [0, 1]$ after identifying opposite sides (that is, identifying $(0, t)$ with $(1, t)$ and $(t, 0)$ with $(t, 1)$ for $t \in [0, 1]$), which is just your definition restricted to the square $[0, 1] \times [0, 1]$. This means (informally) that when for example starting in the middle of this new space and walking to the left, when ending up at the left edge, we are in a sense 'simultaneously' on the right edge (since these two points are considered completely the same, not only as points but also in their topology, so in their neighborhoods), so that if we walk left further we come out on the right side.
If we look at $[0, 1) \times [0, 1)$ (with the subspace topology of $\mathbb R^2$ or equivalently the product topology) however, this is not the case, since the neighborhood of a point on the left edge does not include any points to the right, since these points are not related in $\mathbb R^2$ and we did not do any quotienting to change this. Hence you cannot really 'come out' anywhere after 'walking towards the left side'
In summary, while these two spaces contain the same points, the sense in which these points relate, that is, which points are considered close to each other, so ultimately their topology is very different.
