Cartesian Product of Projections is Projection on Cartesian Product of Mappings

Theorem
Let $I$ be an indexing set.

Let $\family {S_\alpha}_{\alpha \mathop \in I}$ and $\family {T_\alpha}_{\alpha \mathop \in I}$ be families of sets both indexed by $I$.

For each $\alpha \in I$, let $f_\alpha: S_\alpha \to T_\alpha$ be a mapping.

There exists a unique mapping:
 * $\ds f: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$

such that:
 * $\forall \alpha \in I: \pr_\alpha \circ f = f_\alpha \circ \pr_\alpha$

where:
 * $\circ$ denotes composition of mappings
 * $\pr_\alpha$ denotes the $\alpha$th projection on either $\ds \prod_{\alpha \mathop \in I} S_\alpha$ or $\ds \prod_{\alpha \mathop \in I} T_\alpha$ as appropriate.

Proof of Existence
Let $\mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha$ be arbitrary:
 * $\mathbf x = \family {x_\alpha \in S_\alpha}_{\alpha \mathop \in I}$

Let $\ds f: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$ be defined as:
 * $\forall \mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha: \map f {\mathbf x} = \family {\map {f_\alpha} {x_\alpha} }_{\alpha \mathop \in I}$

We have:

Then:

and it is seen that $f$ is such that:
 * $\forall \alpha \in I: \pr_\alpha \circ f = f_\alpha \circ \pr_\alpha$

as required.

Hence the existence of $f$ as specified.

Proof of Uniqueness
Let $f$ be as defined.

Let $\ds g: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$ also be a mapping such that:
 * $\forall \alpha \in I: \pr_\alpha \circ g = f_\alpha \circ \pr_\alpha$

Let $\mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha$ as before.

Let:

and it is seen that $g = f$.