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
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
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $2$