Equivalent Conditions for Entropic Structure/Pointwise Operation of Homomorphisms from External Direct Product is Homomorphism

Theorem
Let $\struct {S, \odot}$ be an algebraic structure.

Let $\struct {S \times S, \otimes}$ denote the external direct product of $\struct {S, \odot}$ with itself:
 * $\forall \tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S: \tuple {x_1, y_1} \otimes \tuple {x_2, y_2} = \tuple {x_1 \odot x_2, y_1 \odot y_2}$

Let $f$ and $g$ be mappings from $\struct {S \times S, \otimes}$ to $\struct {S, \odot}$.

Let $f \odot g$ denote the pointwise operation on $S^{S \times S}$ induced by $\odot$.

Then:
 * If $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism


 * $\struct {S, \odot}$ is an entropic structure.
 * $\struct {S, \odot}$ is an entropic structure.

Sufficient Condition
Let $\struct {S, \odot}$ be such that if $f$ and $g$ are homomorphisms, then $f \odot g$ is also a homomorphism.

So, let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.

Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in S \times S$ be arbitrary.

We have:

Necessary Condition
Let $\struct {S, \odot}$ be an entropic structure.

Let $f: S \times S \to S$ and $g: S \times S \to S$ be arbitrary homomorphisms.

Then:

Hence by definition $f \odot g$ is a homomorphism.