Cartesian Product is Set Product

Theorem
Let $S$ and $T$ be sets.

Let $S \times T$ be the Cartesian product of $S$ and $T$.

Let $\operatorname{pr}_1: S \times T \to S$ and $\operatorname{pr}_2 S \times T \to T$ be the first and second projections respectively on $S \times T$.

Then $\left({S \times T, \operatorname{pr}_1, \operatorname{pr}_2}\right)$ is a set product.

Proof
Consider any set $X$ and mappings $f_1: X \to S$ and $f_2: X \to T$.

Define $h: X \to S \times T$ by:
 * $\forall x \in X: h \left({x}\right) = \left({f_1 \left({x}\right), f_2 \left({x}\right)}\right)$

Then for all $x \in X$ we have:
 * $\left({\operatorname{pr}_1 \circ h}\right) \left({x}\right) = \operatorname{pr}_1 \left({f_1 \left({x}\right), f_2 \left({x}\right)}\right) = f_1 \left({x}\right)$

and
 * $\left({\operatorname{pr}_2 \circ h}\right) \left({x}\right) = \operatorname{pr}_2 \left({f_1 \left({x}\right), f_2 \left({x}\right)}\right) = f_2 \left({x}\right)$

So:
 * $\operatorname{pr}_1 \circ h = f_1$
 * $\operatorname{pr}_2 \circ h = f_2$

Thus $h$ is shown to exist.

Suppose there exists a mapping $k: X \to S \times T$ such that:
 * $\operatorname{pr}_1 \circ k = f_1$
 * $\operatorname{pr}_2 \circ k = f_2$

Let $x \in X$ and let $k \left({x}\right) = \left({s, t}\right)$.

Then:
 * $f_1 \left({x}\right) = \left({\operatorname{pr}_1 \circ k}\right) \left({x}\right) = \operatorname{pr}_1 \left({s, t}\right) = s$
 * $f_2 \left({x}\right) = \left({\operatorname{pr}_2 \circ k}\right) \left({x}\right) = \operatorname{pr}_2 \left({s, t}\right) = t$

and so:
 * $k \left({x}\right) = \left({s, t}\right) = f_1 \left({x}\right), f_2 \left({x}\right) = h \left({x}\right)$

and so $k = h$.

So $h$ is unique.

Hence the result.