Cartesian Product is Set Product/Family of Sets

Theorem
Let $\left\langle{S_i}\right\rangle_{i \in I}$ be a family of sets.

For all $j \in I$, let $\pr_i: \displaystyle \prod_{j \mathop \in I}\left\langle{S_j}\right\rangle \to S_i$ be the $i$th projection from $\displaystyle \prod_{j \mathop \in I} \left\langle{S_j}\right\rangle$ to $S_i$.

Then $\left({\displaystyle \prod_{j \mathop \in I} \left\langle{S_j}\right\rangle, \left\langle{\pr_i}\right\rangle_{i \in I}}\right)$ is a set product.

Proof
Let $i \in I$.

Consider any set $X$ and any indexed family of mappings $\left\langle{f_i: X \to S_i}\right\rangle_{i \in I}$.

Define $h: X \to \displaystyle \prod_{j \mathop \in I} \left\langle{S_j}\right\rangle$ by:
 * $\forall x \in X: h \left({x}\right) = \left\langle{f_i \left({x}\right)}\right\rangle_{i \in I}$

Then for all $x \in X$ and $i \in I$ we have:
 * $\left({\pr_i \circ h}\right) \left({x}\right) = \pr_i \left({\left\langle{f_i \left({x}\right)}\right\rangle_{i \in I}}\right) = f_i \left({x}\right)$

So:
 * $\pr_i \circ h = f_i$

Thus $h$ is shown to exist.

Suppose there exists a mapping $k: X \to \displaystyle \prod_{j \mathop \in I} \left\langle{S_j}\right\rangle$ such that:
 * $\forall i \in I: \pr_1 \circ k = f_i$

Let $x \in X$ and let:
 * $k \left({x}\right) = \left\langle{s_j}\right\rangle_{j \in I}$

for some $\left\langle{s_j}\right\rangle_{j \in I} \in \displaystyle \prod_{j \mathop \in I} \left\langle{S_j}\right\rangle$.

Then:
 * $f_i \left({x}\right) = \left({\pr_i \circ k}\right) \left({x}\right) = \pr_i \left\langle{s_j}\right\rangle_{j \in I} = s_i$

and so:
 * $k \left({x}\right) = \left\langle{s_j}\right\rangle_{j \in I} = \left\langle{f_i \left({x}\right)}\right\rangle_{i \in I} = h \left({x}\right)$

and so $k = h$.

So $h$ is unique.

Hence the result.