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 $13.12 \ \text{(h)}$