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:

Hence the result by Equality of Mappings.