Pointwise Operation is Composite of Operation with Mapping to Cartesian Product
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Theorem
Let $S$ be a set.
Let $\struct {T, *}$ be an algebraic structure.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let the algebraic structure $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $*$.
Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be mappings.
Let $f \times g : S \to T \times T$ be the mapping from $S$ to the cartesian product $T \times T$ defined by:
- $\forall x \in S : \map {\paren {f \times g}} x = \tuple {\map f x, \map g x}$
Then:
- $f \oplus g = * \circ \paren {f \times g}$
That is, $f \oplus g$ is the composition of the binary operation $*$ with the mapping $f \times g: S \to T \times T$.
Proof
\(\ds \forall x \in S: \, \) | \(\ds \map {\paren {* \circ \paren {f \times g} } } x\) | \(=\) | \(\ds \map * {\map {\paren {f \times g} } x}\) | Definition of Composition of Mappings | ||||||||||
\(\ds \) | \(=\) | \(\ds \map * {\tuple {\map f x, \map g x} }\) | Definition of $f \times g$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f x * \map g x\) | Definition of $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f \oplus g} } x\) | Definition of $f \oplus g$ |
From Equality of Mappings:
- $* \circ \paren {f \times g} = f \oplus g$
$\blacksquare$