Composition of Mappings is Right Distributive over Pointwise Operation
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Theorem
Let $A$ and $B$ be sets.
Let $\struct {S, \odot}$ be an algebraic structure.
Let:
- $B^A$ denote the set of mappings from $A$ to $B$
- $S^B$ denote the set of mappings from $B$ to $S$.
Let $f, g \in S^B$ be mappings from $B$ to $S$.
Let $h \in B^A$ be a mapping from $A$ to $B$.
Then:
- $\paren {f \odot g} \circ h = \paren {f \circ h} \odot \paren {g \circ h}$
where:
- $f \odot g$ denotes the pointwise operation on $S^B$ induced by $\odot$
- $g \circ h$ denotes the composition of $g$ with $h$.
Proof
First we establish:
The domain of $h$ is $A$.
The codomain of $h$ is $B$
The domain of both $f$ and $g$ is $B$.
The codomain of both $f$ and $g$ is $S$.
Hence:
- the domain of $\paren {f \odot g} \circ h$ is $A$
- the Domain of $f \odot g$ is $B$
- the codomain of $f \odot g$ is $S$
- the codomain of $\paren {f \odot g} \circ h$ is $S$.
Then:
- the domain of $f \circ h$ is $A$
- the domain of $g \circ h$ is $A$
- the codomain of $f \circ h$ is $S$
- the codomain of $g \circ h$ is $S$
- the domain of $\paren {f \circ h} \odot \paren {g \circ h}$ is $A$
- the codomain of $\paren {f \circ h} \odot \paren {g \circ h}$ is $S$.
Hence both $\paren {f \odot g} \circ h$ and $\paren {f \circ h} \odot \paren {g \circ h}$ have the same domain and codomain.
Let $a \in A$ be arbitrary.
Let $b \in B$ be such that $b = \map h a$.
We have:
\(\ds \map {\paren {\paren {f \odot g} \circ h} } a\) | \(=\) | \(\ds \map {\paren {f \odot g} } {\map h a}\) | Definition of Composite Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f \odot g} } b\) | Definition of $b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f b \odot \map g b\) | Definition of Pointwise Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f {\map h a} \odot \map g {\map h a}\) | Definition of $b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f \circ h} } a \odot \map {\paren {g \circ h} } a\) | Definition of Composite Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {\paren {f \circ h} \odot \paren {g \circ h} } } a\) | Definition of Pointwise Operation |
Hence the result by Equality of Mappings.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(a)}$