Composition of Mappings is Left Distributive over Homomorphism of Pointwise Operation

Theorem
Let $A$ be a set.

Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be algebraic structures.

Let:
 * $S^A$ denote the set of mappings from $A$ to $S$.

Let $f$ be a homomorphism from $S$ to $T$.

Let $g, h \in S^A$ be mappings from $A$ to $S$.

Then:
 * $f \circ \paren {g \odot h} = \paren {f \circ g} \otimes \paren {f \circ h}$

where:
 * $g \odot h$ denotes the pointwise operation on $S^A$ induced by $\odot$
 * $f \circ h$ denotes the composition of $f$ with $h$.

Proof
First we establish:

The domain of $g$ and $h$ is $A$.

The codomain of $g$ and $h$ is $S$.

The domain of $f$ is $S$.

The codomain of $f$ is $T$.

Hence:
 * the Domain of $g \odot h$ is $A$
 * the codomain of $g \odot h$ is $S$
 * the domain of $f \circ \paren {g \odot h}$ is $A$
 * the codomain of $f \circ \paren {g \odot h}$ is $T$.

Then:
 * the domain of $f \circ g$ is $A$
 * the domain of $f \circ h$ is $A$
 * the codomain of $f \circ g$ is $S$
 * the codomain of $f \circ h$ is $S$
 * the domain of $\paren {f \circ g} \otimes \paren {f \circ h}$ is $A$
 * the codomain of $\paren {f \circ g} \otimes \paren {f \circ h}$ is $T$.

Hence both $f \circ \paren {g \odot h}$ and $\paren {f \circ g} \otimes \paren {f \circ h}$ have the same domain and codomain.

Let $a \in A$ be arbitrary.

Let $b \in B$ be such that $b = \map g a$.

Let $c \in B$ be such that $c = \map h a$.

Let $d \in T$ such that $d = b \odot c$.

We have:

Hence the result by Equality of Mappings.