# How is the category-theoretic product of sets connected to ordered pairs?

Suppose we are working in NBG set theory. A category theoretic product function is a function $f: V \times V \to V$ that satisfies the universal property for products. Is it true that for any product function $f$, there exists an ordered pair function $OP: V \times V \to V$, such that for all sets $S$ and $T$, $f(S,T)$ is the set of all ordered pairs whose first component is in $S$ and whose second component is in $T$?

• The answer to your title question is that if $X \times Y$ is the category-theoretic product of $X$ and $Y$, then morphisms $Z \to X \times Y$ can naturally be identified with ordered pairs of morphisms $Z \to X$ and morphisms $Z \to Y$; this is exactly the universal property of products, and it holds regardless of what the underlying set theory is. What your body question is missing is precisely this notion of two functors being unequal but naturally isomorphic. – Qiaochu Yuan Sep 15 '17 at 19:15

No. First of all, if $f$ is a product function, then the cardinality of $f(S,T)$ must be the product of the cardinalities of $S$ and $T$ (actually, any function with this property serves as a product function).
So if we want to find some function $OP$ with $f(S,T) = \{OP(s,t)\mid s\in S, t\in T\}$ for all $S$ and $T$, $OP$ had better be an injective function. If not, say if $OP(a,b) = OP(c,d)$, where $(a,b)\neq (c,d)$, then taking $S = \{a,c\}$ and $T = \{b,d\}$, the cardinality of $\{OP(s,t)\mid s\in S, t\in T\}$ is less than the product of the cardinalities of $S$ and $T$.
Now the function $f(S,T) = |S|\times |T|$, where $\times$ is multiplication of cardinals, is a product function. If we pick distinct sets $S\neq S'$ of the same cardinality and any nonempty $T$, then since $OP$ is injective, $$\{OP(s,t)\mid s\in S, t\in T\}\neq \{OP(s,t)\mid s\in S', t\in T\}.$$ But since $|S| = |S'|$, we have $f(S,T) = f(S',T)$. So $f$ can't be represented by an ordered pair function.
Of course, the converse is true: if $OP\colon V\times V\to V$ is any injective function, then $f(S,T) = \{OP(s,t)\mid s\in S, t\in T\}$ is a product function.