Equivalent Conditions for Entropic Structure/Pointwise Operation is Homomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Let $\struct {T, \circledast}$ be an arbitrary algebraic structure.

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

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


Then:

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

if and only if:

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


Proof

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

Let $a, b \in T$ be arbitrary.

Because $T$ is arbitrary, and $f$ and $g$ are arbitrary, it follows that:

\(\ds \forall w, x, y, z \in S: \exists a, b \in T: \, \) \(\ds \map f a\) \(=\) \(\ds w\)
\(\, \ds \land \, \) \(\ds \map f b\) \(=\) \(\ds x\)
\(\ds \map g a\) \(=\) \(\ds y\)
\(\, \ds \land \, \) \(\ds \map g b\) \(=\) \(\ds z\)


Thus we have:

\(\ds \paren {w \odot x} \odot \paren {y \odot z}\) \(=\) \(\ds \paren {\map f a \odot \map f b} \odot \paren {\map g a \odot \map g b}\) a priori
\(\ds \) \(=\) \(\ds \paren {\map f {a \circledast b} }\odot \paren {\map g {a \circledast b} }\) as $f$ and $g$ are both homomorphisms
\(\ds \) \(=\) \(\ds \map {\paren {f \odot g} } {a \circledast b}\) Definition of Pointwise Operation
\(\ds \) \(=\) \(\ds \map {\paren {f \odot g} } a \odot \map {\paren {f \odot g} } b\) as $f \odot g$ is a homomorphism
\(\ds \) \(=\) \(\ds \paren {\map f a \odot \map g a} \odot \paren {\map f b \odot \map g b}\) Definition of Pointwise Operation
\(\ds \) \(=\) \(\ds \paren {w \odot y} \odot \paren {x \odot z}\) Definition of Pointwise Operation

As $w$, $x$, $y$ and $z$ are arbitrary, $\struct {S, \odot}$ is an entropic structure.

$\Box$


Necessary Condition

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

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

Then:

\(\ds \forall x, y \in T: \, \) \(\ds \map {\paren {f \odot g} } {x \circledast y}\) \(=\) \(\ds \map f {x \circledast y} \odot \map g {x \circledast y}\) Definition of Pointwise Operation
\(\ds \) \(=\) \(\ds \paren {\map f x \odot \map f y} \odot \paren {\map g x \odot \map g y}\) as $f$ and $g$ are both homomorphisms
\(\ds \) \(=\) \(\ds \paren {\map f x \odot \map g x} \odot \paren {\map f y \odot \map g y}\) Definition of Entropic Structure
\(\ds \) \(=\) \(\ds \paren {\map {\paren {f \odot g} } x} \odot \paren {\map {\paren {f \odot g} } y}\) Definition of Pointwise Operation

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalent_Conditions_for_Entropic_Structure/Pointwise_Operation_is_Homomorphism&oldid=686457"