Cartesian Product of Projections is Projection on Cartesian Product of Mappings

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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

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:

\(\ds \forall \mathbf x \in \prod_{\alpha \mathop \in I} S_\alpha: \, \) \(\ds \map {\paren {f_\alpha \circ \pr_\alpha} } {\mathbf x}\) \(=\) \(\ds \map {f_\alpha} {\map {\pr_\alpha} {\mathbf x} }\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {f_\alpha} {x_\alpha}\) Definition of $\alpha$th Projection on $\ds \prod_{\alpha \mathop \in I} S_\alpha$


Then:

\(\ds \forall \mathbf x \in \prod_{\alpha \mathop \in I} S_\alpha: \, \) \(\ds \map {\paren {\pr_\alpha \circ f} } {\mathbf x}\) \(=\) \(\ds \map {\pr_\alpha} {\family {\map {f_\alpha} {x_\alpha} }_{\alpha \mathop \in I} }\) Definition of $f$
\(\ds \) \(=\) \(\ds \map {f_\alpha} {x_\alpha}\) Definition of $\alpha$th Projection on $\ds \prod_{\alpha \mathop \in I} T_\alpha$


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.

$\Box$


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:

\(\ds \map {\paren {\pr_\alpha \circ g} } {\mathbf x}\) \(=\) \(\ds \map {\pr_\alpha} {\map g {\mathbf x} }\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {f_\alpha} {x_\alpha}\) by definition
\(\ds \leadsto \ \ \) \(\ds \forall \mathbf x \in \prod_{\alpha \mathop \in I} S_\alpha: \, \) \(\ds \map g {\mathbf x}\) \(=\) \(\ds \family {\map {f_\alpha} {x_\alpha} }_{\alpha \mathop \in I}\)

and it is seen that $g = f$.

$\blacksquare$


Sources

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