At first sight there is no abstract (= structural) definition of "product" in set theory. E.g. the Cartesian product of sets $A$ and $B$ is defined as the set of all ordered pairs $(x,y)$, $x \in A$, $y \in B$, and thus depends on the definition of "ordered pair" which is notoriously arbitrary.

I wonder if the following can count as an abstract (= structural) definition of "product" in the context of set theory.

Consider a set $S$ with two equivalence relations $\sim_1$ and $\sim_2$.

> **Definition**: $(S,\sim_1,\sim_2)$ is a product iff
> 
> $$(\forall x)(\forall y)(\exists ! z)
> x \sim_1 z \wedge y \sim_2 z$$
> 
> $$(\forall z)(\exists ! x)(\exists !
> y) x \sim_1 z \wedge y \sim_2 z$$

If $(S,\sim_1,\sim_2)$ is a product

- $S$ can be understood as $S/_{\sim_1} \times S/_{\sim_2}$

- the relations $\sim_i$ can be read as *has the same $i$-th component*

- the canonical projection map $\pi_i (x) = [x]_{\sim_i}$ can be understood as the *$i$-th component* 

> **Question:** Isn't this definition somehow on par - concerning
> structuralness - with the definition
> of category theory? If so, why is it
> so rarely found, or rather: *where* can
> I find it (in which textbook, e.g.)?

Considering the product of a set with itself, i.e. $S = X \times X$, one relation $\sim$ does suffice, which does not have to be an equivalence relation, not even symmetric, but from which two equivalence relations can be defined:

$$ x \sim_1 y :\equiv (\exists z) x \sim z \wedge y \sim z $$
$$ x \sim_2 y :\equiv (\exists z) z \sim x \wedge z \sim y $$

If $(S,\sim_1,\sim_2)$ is a product the relation $x \sim y$ can be read as *the first component of $x$ equals the second component of $y$*.

> **Question**: Are there conditions on a relation $\sim$ such that $\sim_1$,
> $\sim_2$ as defined above make
> $(S,\sim_1,\sim_2)$ automatically a
> product?