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 $$(\forall x \in S)(\forall y \in S)(\exists ! z\in S) x \sim_1 z \wedge y \sim_2 z$$ $$(\forall z\in S)(\exists ! x\in S)(\exists ! y\in S) x \sim_1 z \wedge y \sim_2 z$$

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:wherecan 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?

togetherwith its projection maps to the factors, could be characterized by a suitable universal mapping property. Then that construction is unique up to suitable isomorphism as all objects satisfying universal properties are. This can let you relax a little about the fact that there may be more than one way to make the construction of a product of sets (and its projections to the factors). $\endgroup$ – KConrad Oct 9 '11 at 13:08