# Torus defined by relation

If constructing a Torus from a square $$[0, 1]^2$$ in $$R^2$$, identified by the relation $$p$$ as follows:

$$(x,y), (x',y') \in [0, 1]^2\ ; \ \ (x,y) \ p \ (x',y') \iff$$\left\{ \begin{array}{@{}ll@{}} (x,y) = (x', y') \ \text{or} \\ (\{x, x'\} = \{0,1\} \land y = y') \ \text{or} \\ (\{y,y'\} = \{0,1\} \land x=x') \ \text{or} \\ \{x, x', y,y'\} = \{0,1\} \end{array}\right.$$$$

What I don't understand is the importance/meaning of the last condition. The first one makes sure the equivalence between the same points is preserved, the second that we can fold the square horizontaly, the third that we can then fold it vertically.

But If all three are now included in the defintion of $$(x,y) \ p \ (x',y')$$, then we already have our Torus shape. What is the significance of the last one? It seems to show that the points on the initial square that stand oppposite to each other on the diagonals are corresp. through $$p$$ equal to each other too. But why do we need that if that follows from the 2. and 3. conditions by transitivity?

• Without a definition of $x',y'$ in terms of $x,y$ it's hard to see what the first one means. Commented Aug 14, 2020 at 18:38
• my bad, i edited it
– Aelx
Commented Aug 14, 2020 at 18:43

When constructing a quotient space of a topological space $$X$$, one is given a decomposition of $$X$$, which means a collection of pairwise disjoint subsets of $$X$$ whose union is $$X$$.

From set theory, we learn that there is a one-to-one correspondence as follows: $$\{\text{decompositions of X}\} \leftrightarrow \{\text{equivalence relations on X}\}$$ So, the relation $$p$$ in your post must be an equivalence relation (... and it is ...) in order to be able to use it to obtain a decomposition and thus a quotient space.

If you had instead defined $$p$$ without the last requirement, then $$p$$ would not be an equivalence relation.

However, there is a shortcut we often take. Another theorem of set theory says that for any set $$X$$ and for any relation $$R$$ on $$X$$, there exists a unique "smallest" equivalence relation that contains $$R$$, namely the "reflexive, symmetric, transitive closure" of $$R$$. One calls this the equivalence relation generated by $$R$$.

One can then define a quotient space using any relation $$R$$ on any topological space $$X$$: use the decomposition of $$X$$ that corresponds to the equivalence relation generated by $$R$$.

So, if in your example you had left out the last condition, you would obtain a relation $$R$$ which is not an equivalence relation; but then by taking the transitive closure you would be putting the last condition back in, and the result would be the equivalence relation $$p$$ written in your post.

• I see now. A definition without the last condition would break the transitivity requirement for $p$. In this case using $p$ one could see the validity of $(0,0) \ p \ (0,1)$, then $(0,0) \ p \ (1,0)$ but without the last condition there would not exist an implication after that, that $(0,1) \ p \ (1,0)$ and thus $p$ would not be usable for defining a quotient space as not being an equivalence relation.
– Aelx
Commented Aug 14, 2020 at 19:34