# Composition of Mappings is Right Distributive over Pointwise Operation

## 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 $13.11 \ \text{(a)}$