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

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:
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\odot$ be the operation on $S$ defined by Cayley table as:
:<nowiki>$\begin{array} {c|cc} \odot & a & b \\
\hline
a & b & b \\
b & a & b
\end{array}$</nowiki>
First we note that:
{{begin-eqn}}
{{eqn | l = \paren {a \odot a} \odot \paren {b \odot a}
      | r = b \odot a
      | rr= = a
}}
{{eqn | l = \paren {a \odot b} \odot \paren {a \odot a}
      | r = a \odot b
      | rr= = b
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map {I_S} a
      | r = a
}}
{{eqn | l = \map {I_S} b
      | r = b
}}
{{end-eqn}}
From Identity Mapping is Automorphism, $I_S$ is an automorphism.
Hence $I_S$ is {{afortiori}} an endomorphism on $\struct {S, \odot}$.
That is:
:$I_S \in T$
Let $f_a$ and $f_b$ denote the constant mappings:
{{begin-eqn}}
{{eqn | l = \map {f_a} a
      | r = a
}}
{{eqn | l = \map {f_a} b
      | r = a
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {f_b} a
      | r = b
}}
{{eqn | l = \map {f_b} b
      | r = b
}}
{{end-eqn}}
We have that:
{{begin-eqn}}
{{eqn | l = \map {f_a} {a \odot a}
      | r = \map {f_a} b
}}
{{eqn | r = a
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {f_a} a \odot \map {f_a} a
      | r = a \odot a
}}
{{eqn | r = b
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = \map {f_b} {x \odot y}
      | r = b
}}
{{eqn | l = \map {f_b} x \odot \map {f_b} y
      | r = b \odot b
}}
{{eqn | r = b
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map h a
      | r = b
}}
{{eqn | l = \map h b
      | r = a
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \map h {a \odot a}
      | r = \map h b
}}
{{eqn | r = a
}}
{{eqn | l = \map h a \odot \map h a
      | r = b \odot b
}}
{{eqn | r = b
}}
{{end-eqn}}
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.
{{qed}}