# Set of Endomorphisms on Entropic Structure is Closed in Induced Structure on Set of Self-Maps/Converse

## Theorem

Let $\struct {S, \odot}$ be a magma.

Let $S^S$ be the set of all mappings from $S$ to itself.

Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.

Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.

Let $\struct {T, \oplus_T}$ be closed in $\struct {S^S, \oplus}$.

Then it is not necessarily the case that $\struct {S, \odot}$ is an entropic structure.

## Proof

Let $S = \set {a, b}$.

Let $\odot$ be the operation on $S$ defined by Cayley table as:

$\begin{array} {c|cc} \odot & a & b \\ \hline a & b & b \\ b & a & b \end{array}$

First we note that:

 $\ds \paren {a \odot a} \odot \paren {b \odot a}$ $=$ $\ds b \odot a$ $\ds = a$ $\ds \paren {a \odot b} \odot \paren {a \odot a}$ $=$ $\ds a \odot b$ $\ds = b$

demonstrating that $\struct {S, \odot}$ is specifically not an entropic structure.

Let $T \subseteq S^S$ denote the set of endomorphisms on $\struct {S, \odot}$.

Let $f, g \in T$ be arbitrary.

That is, let $f: S \to S$, $g: S \to S$ be endomorphisms on $\struct {S, \odot}$.

Let $x, y \in S$ be arbitrary.

We are to show that:

$\map {\paren {f \oplus g} } {x \odot y} = \paren {\map {\paren {f \oplus g} } x} \odot \paren {\map {\paren {f \oplus g} } y}$

Let us investigate all the mappings from $S$ to $S$.

From Cardinality of Set of All Mappings, there are exactly $4$ of these, so inspecting them all is not onerous.

Let $I_S$ denote the identity mapping:

 $\ds \map {I_S} a$ $=$ $\ds a$ $\ds \map {I_S} b$ $=$ $\ds b$

From Identity Mapping is Automorphism, $I_S$ is an automorphism.

Hence $I_S$ is a fortiori an endomorphism on $\struct {S, \odot}$.

That is:

$I_S \in T$

Let $f_a$ and $f_b$denote the constant mappings:

 $\ds \map {f_a} a$ $=$ $\ds a$ $\ds \map {f_a} b$ $=$ $\ds a$
 $\ds \map {f_b} a$ $=$ $\ds b$ $\ds \map {f_b} b$ $=$ $\ds b$

We have that:

 $\ds \map {f_a} {a \odot a}$ $=$ $\ds \map {f_a} b$ $\ds$ $=$ $\ds a$
 $\ds \map {f_a} a \odot \map {f_a} a$ $=$ $\ds a \odot a$ $\ds$ $=$ $\ds b$

and so $f_a$ is not a homomorphism, and so not an endomorphism on $\struct {S, \odot}$.

Hence:

$f_a \notin T$

We have that:

 $\ds \forall x, y \in S: \,$ $\ds \map {f_b} {x \odot y}$ $=$ $\ds b$ $\ds \map {f_b} x \odot \map {f_b} y$ $=$ $\ds b \odot b$ $\ds$ $=$ $\ds b$

and so $f_b$ is a homomorphism, and so an endomorphism on $\struct {S, \odot}$.

Hence:

$f_b \in T$

Now let $h \in S^S$ be defined as:

 $\ds \map h a$ $=$ $\ds b$ $\ds \map h b$ $=$ $\ds a$

We have:

 $\ds \map h {a \odot a}$ $=$ $\ds \map h b$ $\ds$ $=$ $\ds a$ $\ds \map h a \odot \map h a$ $=$ $\ds b \odot b$ $\ds$ $=$ $\ds b$

and so $h$ is not a homomorphism, and so not an endomorphism on $\struct {S, \odot}$.

Hence:

$h \notin T$

So we are left with:

$T = \set {I_S, f_b}$

From Composite of Endomorphisms is Endomorphism, each of $I_S \circ I_S$, $I_S \circ f_b$, $f_b \circ I_S$ and $f_b \circ f_b$ are endomorphisms.

By definition of identity mapping:

$I_S \circ I_S = I_S$
$I_S \circ f_b = f_b$
$f_b \circ I_S = f_b$

and from Composite with Constant Mapping is Constant Mapping: $f_b \circ f_b = f_b$

Hence we have that $\struct {T, \oplus}$ is closed in $\struct {S^S, \oplus}$.

But as we have seen, $\struct {S, \odot}$ is not an entropic structure.

$\blacksquare$

## 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 $13.12 \ \text{(h)}$