Apparently, $u\overset{M}{\longleftrightarrow}v$ defines an equivalence
relation on $D$.

The following serves as an example.

######  Proposition 7.1.2.

Let $B$ be a subfield of a field $F$. Let $T$ be the operator semigroup
defined in Example 2.1.2.

  1. (a)

If $u\in F$ is algebraic over $B$, then

${{[u]}_{T}}={{[u)}_{T}}=${all roots of $\min(B,u)$},

where $\min(B,u)$ is the minimal polynomial of $u$ over $B$.

  2. (b)

If $u\in F$ is transcendental over $B$, then

${{[u)}_{T}}=F$ and ${{[u]}_{T}}=${$v\in F\,|\,v$ is transcendental over $B$}.

###### Proof.

For (a): Let $u\in F$ be algebraic over $B$.

Let $w\in{{[u)}_{T}}$. By Notation 7.1.1, $\forall f^{*},g^{*}\in T$ (see
Example 2.1.2), $f^{*}(u)=g^{*}(u)$ implies that $f^{*}(w)=g^{*}(w)$. So
$\forall f(x),g(x)\in B[x]$, $f(u)=g(u)$ implies that $f(w)=g(w)$. Thus $w$ is
a root of $\min(B,u)$. Hence

${{[u)}_{T}}\subseteq${all roots of $\min(B,u)$}.

Moreover, let $v\in F$ be a root of $\min(B,u)$. Then $\forall f(x),g(x)\in
B[x]$, $f(u)=g(u)\Leftrightarrow f(v)=g(v)$. It follows that $\forall
f^{*},g^{*}\in T$, $f^{*}(u)=g^{*}(u)\Leftrightarrow f^{*}(v)=g^{*}(v)$; that
is, $u\overset{T}{\longleftrightarrow}v$, and so $v\in{{[u]}_{T}}$. Hence

$({{[u)}_{T}}\subseteq)${all roots of $\min(B,u)$}$\subseteq{{[u]}_{T}}$.

On the other hand, ${{[u)}_{T}}\supseteq{{[u]}_{T}}$ by Notation 7.1.1.
Therefore,

${{[u)}_{T}}=${all roots of $\min(B,u)$}$={{[u]}_{T}}$.

For (b): Let $u\in F$ be transcendental over $B$.

Then $\forall f(x),g(x)\in B[x]$, $f(u)=g(u)$ implies that $f=g$. So $\forall
f^{*},g^{*}\in T$, $f^{*}(u)=g^{*}(u)$ implies that $f^{*}=g^{*}$. Then
$\forall v\in F$, $u\xrightarrow{T}v$, and so ${{[u)}_{T}}=F$.

Hence ${{[u]}_{T}}=\\{v\in F\,|\,v\overset{T}{\longleftrightarrow}u\\}=${$v\in
F\,|\,v\xrightarrow{T}u$}.

$\forall v\in F$, if $v$ is also transcendental over $B$, then as was just
shown for $u$, ${{[v)}_{T}}=F$ and $v\xrightarrow{T}u$. So

${{[u]}_{T}}=\\{v\in F\,|\,v\xrightarrow{T}u\\}\supseteq${$v\in F\,|\,v$ is
transcendental over $B$}.

On the other hand, if $v\in F$ is algebraic over $B$, then by (a),
$u\notin{{[v)}_{T}}$, i.e., the opposite of $v\xrightarrow{T}u$ is true. So

${{[u]}_{T}}=\\{v\in F\,|\,v\xrightarrow{T}u\\}\subseteq${$v\in F\,|\,v$ is
transcendental over $B$}.

Therefore, ${{[u]}_{T}}=${$v\in F\,|\,v$ is transcendental over $B$}. ∎

We will talk more about transcendental elements in Section 23.

Recall that in the classical Galois theory, each element of the Galois group
of a polynomial permutes its roots. By Proposition 7.1.2, we can tell that the
following two propositions are generalizations of this well-known fact.

######  Proposition 7.1.3.

Let $\sigma$ be a $T$-morphism from a $T$-space $S$ to a $T$-space. Then
$\forall u\in S$, $u\xrightarrow{T}\sigma(u)$ $($and so
$\sigma(u)\in{{[u)}_{T}})$.

###### Proof.

For $u\in S$, let $f,g\in T$ such that $f(u)=g(u)$. Then by Definition 2.3.1,
$f(\sigma(u))=\sigma(f(u))=\sigma(g(u))=g(\sigma(u))$. Hence by Notation
7.1.1, $u\xrightarrow{T}\sigma(u)$. ∎

######  Proposition 7.1.4.

Let $\sigma$ be a $T$-morphism from a $T$-space $S$ to a $T$-space such that
$\sigma$ is injective. Then $\forall u\in S$,
$u\overset{T}{\longleftrightarrow}\sigma(u)$ $($and so
$\sigma(u)\in{{[u]}_{T}})$.

###### Proof.

Let $u\in S$. By Proposition 7.1.3, we only need to show that
$\sigma(u)\xrightarrow{T}u$. Let $f,g\in T$ such that
$f(\sigma(u))=g(\sigma(u))$. Then by Definition 2.3.1,
$\sigma(f(u))=f(\sigma(u))=g(\sigma(u))=\sigma(g(u))$. Hence $f(u)=g(u)$
because $\sigma$ is injective, as desired. ∎

The converse of Proposition 7.1.3 implies a potential way to construct
$T$-morphisms. However, the converse of Proposition 7.1.3 is not always true,
as shown below.

######  Example 7.1.5.

Let $R$ be a differential ring with a derivation $\partial$ such that
$\\{\partial(u)\,|\,u\in R\\}=R$ (it is easy to find such differential ring),
let $T$ be the operator semigroup on $R$ given by
$\\{{{\partial}^{n}}\,|\,n\in{{\mathbb{Z}}^{+}}\\}$ (note that unlike the
operator semigroup defined in Example 2.1.5, now $\operatorname{Id}\notin T$),
let $C$ be the constant ring of $R$, and let a map $\sigma:R\to R$ be given by
$u\mapsto u+c$, where $c\in C$ and $c\neq 0$. Then $\forall f,g\in T$ and
$u\in R$, $f(u)=g(u)$ implies that $f(\sigma(u))=g(\sigma(u))$. Hence $\forall
u\in R$, $u\xrightarrow{T}\sigma(u)$. But obviously $\exists f\in T$ and $u\in
R$ such that $f(\sigma(u))\neq\sigma(f(u))$. So $\sigma$, a map from the
$T$-space ${{\left\langle R\right\rangle}_{T}}=R$ to itself, is not a
$T$-morphism.

This example implies that, to make the converse of Proposition 7.1.3 be true,
we need to impose a stronger condition on $\sigma$. Before doing this, let’s
see a fact.

######  Proposition 7.1.6.

Let $u,v\in D$. Then $u\xrightarrow{T}v$ implies that
$f(u)\xrightarrow{T}f(v)$, $\forall f\in T$.

###### Proof.

Suppose that $u\xrightarrow{T}v$. Let $f,g,h\in T$ such that
$g(f(u))=h(f(u))$. Then by Notation 7.1.1, it suffices to show that
$g(f(v))=h(f(v))$.

By Definition 2.1.1, $g\circ f\in T$ and $h\circ f\in T$. Since
$u\xrightarrow{T}v$ and $g(f(u))=h(f(u))$, by Notation 7.1.1, we have
$g(f(v))=h(f(v))$, as desired. ∎

By Proposition 7.1.6, we can tell that the following theorem contains a
converse of Proposition 7.1.3.

######  Theorem 7.1.7.

Let $u,v\in D$. Then

  1. (a)

$u\xrightarrow{T}v$ if and only if there exists a map $\sigma:{{\left\langle
u\right\rangle}_{T}}\to{{\left\langle v\right\rangle}_{T}}$ given by
$f(u)\mapsto f(v),\forall f\in T$.

  2. (b)

The above map $\sigma$ is a $T$-morphism.

###### Proof.

For (a):

$u\xrightarrow{T}v$;

$\Leftrightarrow$ $\forall f,g\in T$, $f(u)=g(u)$ implies that $f(v)=g(v)$;

$\Leftrightarrow$ $\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle
v\right\rangle}_{T}}$ given by $f(u)\mapsto f(v),\forall f\in T$, is a well-
defined map.

For (b): Let $a\in{{\left\langle u\right\rangle}_{T}}$. Then $\exists f\in T$
such that $f(u)=a$. Let $g\in T$. Then $\sigma(g(a))=\sigma(g(f(u)))$. By
Definition 2.1.1, $g\circ f\in T$, and so by the definition of $\sigma$,
$\sigma(g(f(u)))=g(f(v))$. Besides, $g(\sigma(a))=g(\sigma(f(u)))=g(f(v))$
also by the definition of $\sigma$. Combining the above equations, we have
$g(\sigma(a))=\sigma(g(a))$. Therefore, $\sigma$ is a $T$-morphism. ∎

For $T$-isomorphisms (Definition 2.4.1), we have the following straightforward
result of Theorem 7.1.7.

######  Corollary 7.1.8.

Let $u,v\in D$.

  1. (a)

$u\overset{T}{\longleftrightarrow}v$ if and only if there exists a bijective
map $\sigma:{{\left\langle u\right\rangle}_{T}}\to{{\left\langle
v\right\rangle}_{T}}$ given by $f(u)\mapsto f(v),\forall f\in T$.

  2. (b)

The above bijective map $\sigma$ is a $T$-isomorphism.

The converse of statement (b) in Theorem 7.1.7 raises a question as follows.

######  Question 7.1.9.

For $u\in D$ and a $T$-morphism $\sigma$ from ${{\left\langle
u\right\rangle}_{T}}$ to a $T$-space, does there exist any $v\in D$ such that
$\sigma(f(u))=f(v),\forall f\in T$?

The significance of this question is that, in the cases where the answer is
yes, we can construct $\sigma$ from $u$, $v$, and $T$. Theorem 7.1.7 implies
that there are cases where the answer to the question is positive, and we will
talk about these cases in Subsection 7.2. It seems not very easy to find a
case which gives negative answer to the question, but we found one as follows.

######  Example 7.1.10.

Define $h:\mathbb{R}\to\mathbb{R}$ by $x\mapsto x+1$ and
$g:\mathbb{R}\to\mathbb{R}$ by $x\mapsto\pi x$. Let
$T=\left\langle\\{h,g\\}\right\rangle$ (Definition 2.1.6) and
$S={{\left\langle 1\right\rangle}_{T}}$. Then each $f\in T$ can be expressed
as a finite composite of alternating “powers” of $h$ or $g$ as follows, where
each ${{h}^{{{k}_{i}}}}$ or ${{g}^{{{k}_{i}}}}$ denotes the composite of
${{k}_{i}}$ copies of $h$ or $g$, respectively.

(7.1) $f={h^{{k_{n}}}}{g^{{k_{n-1}}}}\cdots{h^{{k_{2}}}}{g^{{k_{1}}}},$ (7.2)
$f={g^{{k_{n}}}}{h^{{k_{n-1}}}}\cdots{h^{{k_{2}}}}{g^{{k_{1}}}},$ (7.3)
$f={h^{{k_{n}}}}g^{{k_{n-1}}}\cdots{g^{{k_{2}}}}h^{{k_{1}}},$

or

(7.4) $f={g^{{k_{n}}}}h^{{k_{n-1}}}\cdots{g^{{k_{2}}}}h^{{k_{1}}},$

where $n$ is an even number for $($7.1$)$ and $($7.4$)$, $n$ is an odd number
for $($7.2$)$ and $($7.3$)$, and ${{k}_{i}}\in{{\mathbb{Z}}^{+}},\forall
i=1,\cdots,n$. (In particular, $h$ has the form (7.3), where $n=1$, and $g$
has the form (7.2), where $n=1$.) And correspondingly, $\forall a\in
S={{\left\langle 1\right\rangle}_{T}}$, $a=f(1)$ has the form

(7.5)
$a=(\cdots({\pi^{{k_{1}}}}+{k_{2}}){\pi^{{k_{3}}}}+\cdots+{k_{n-2}}){\pi^{{k_{n-1}}}}+{k_{n}},$
(7.6)
$a=(\cdots({\pi^{{k_{1}}}}+{k_{2}}){\pi^{{k_{3}}}}+\cdots+{k_{n-1}}){\pi^{{k_{n}}}},$
(7.7)
$a=(\cdots((1+{k_{1}}){\pi^{{k_{2}}}}+{k_{3}}){\pi^{{k_{4}}}}+\cdots+{k_{n-2}}){\pi^{{k_{n-1}}}}+{k_{n}},$

or

(7.8)
$a=(\cdots((1+{k_{1}}){\pi^{{k_{2}}}}+{k_{3}}){\pi^{{k_{4}}}}+\cdots+{k_{n-1}}){\pi^{{k_{n}}}},$

where $n$ is an even number for $($7.5$)$ and $($7.8$)$, $n$ is an odd number
for $($7.6$)$ and $($7.7$)$, and ${{k}_{i}}\in{{\mathbb{Z}}^{+}},\forall
i=1,\cdots,n$.

Let $a\in S$. Since $\pi$ is transcendental over $\mathbb{Q}$, the expression
for $a$ in the above forms is unique. It follows that there is only one $f\in
T$ in the form of $($7.1$)$, $($7.2$)$, $($7.3$)$, or $($7.4$)$ such that
$a=f(1)$. Hence we can define a map $\sigma:S\to S$ as follows.

Let $a\in S$. If $f\in T$ has the form $($7.1$)$ or $($7.2$)$ such that
$a=f(1)$, let $\sigma(a)=f(\pi)$, which clearly belongs to $S$. If $f\in T$
has the form $($7.3$)$ or $($7.4$)$ such that $a=f(1)$, let $\sigma(a)=a$.

Now let’s show that $\sigma\in\operatorname{End}_{T}(S)$. If $a=f(1)$ has the
form $($7.5$)$ or $($7.6$)$, then $f$ has the form $($7.1$)$ or $($7.2$)$, and
so $\forall l\in T$, $l\circ f$ also has the form $($7.1$)$ or $($7.2$)$.
Hence by the definition of $\sigma$,
$\sigma(l(a))=\sigma(l(f(1)))=l(f(\pi))=l(\sigma(a))$. If $a=f(1)$ has the
form $($7.7$)$ or $($7.8$)$, then $f$ has the form $($7.3$)$ or $($7.4$)$, and
so $\forall l\in T$, $l\circ f$ also has the form $($7.3$)$ or $($7.4$)$.
Hence by the definition of $\sigma$,
$\sigma(l(a))=\sigma(l(f(1)))=l(f(1))=l(a)=l(\sigma(a))$. Therefore in both
cases, $\sigma(l(a))=l(\sigma(a))$, $\forall l\in T$ and $a\in S$. So
$\sigma\in\operatorname{End}_{T}(S)$.

Assume that there exists $v\in\mathbb{R}$ such that $\sigma(f(1))=f(v),\forall
f\in T$. Recall the definitions of $h$ and $g$ at the beginning of the
example. Since $h$ has the form (7.3), $h(v)=\sigma(h(1))=h(1)=2$, and so
$v=1$; but since $g$ has the form (7.2),
$g(v)=\sigma(g(1))=g(\pi)={{\pi}^{2}}$, and so $v=\pi$, a contradiction.

Therefore, for the $T$-morphism $\sigma$, the answer to Question 7.1.9 is
negative.

Recall that near the end of Subsection 2.5, we claimed that for a $T$-space
$S$ and $H\subseteq\operatorname{End}_{T}(S)$, it is possible that ${{S}^{H}}$
is not a $T$-space. Now we can show this claim by this example.

Let $D={{\left\langle 1\right\rangle}_{T}}\bigcup\\{1\\}$. Then let us
restrict the domain of $T$ to $D$. We can tell that the “new” $T$, which we
still denote by $T$ for convenience, is an operator semigroup on $D$ and all
the above argument in this example still holds. In particular,
$\sigma\in\operatorname{End}_{T}(S)$ and $2\in{{S}^{\sigma}}$ because 2 has
the form $($7.7$)$. Suppose that ${{S}^{\sigma}}$ is a $T$-space. Then
$\exists U\subseteq D$ such that ${{\left\langle
U\right\rangle}_{T}}={{S}^{\sigma}}$. Hence $\exists u\in D=S\bigcup\\{1\\}$
such that $2\in{{\left\langle u\right\rangle}_{T}}$. It follows that $u$ must
be 1, and so ${{S}^{\sigma}}\supseteq{{\left\langle 1\right\rangle}_{T}}=S$.
However, $\sigma(\pi)=\sigma(g(1))=g(\pi)={{\pi}^{2}}\neq\pi$, and so $\pi\in
S\backslash{{S}^{\sigma}}$, a contradiction.

###  7.2. A construction of $T$-morphisms mapping from ${{\left\langle
U\right\rangle}_{T}}(U\subseteq D)$

$T$-morphisms discussed in Subsection 7.1 are mapping from $T$-spaces
generated by a single element of $D$. To study $T$-morphisms mapping from
other $T$-spaces, we generalize Notation 7.1.1 as follows.

######  Notation 7.2.1.

Let $D$ be a set, let $M$ be a set of maps from $D$ to $D$ and let
$\alpha:U(\subseteq D)\to V(\subseteq D)$ be a map. Then by
$U\xrightarrow{M,\alpha}V$ we mean that

$\forall f,g\in M$ and $u,w\in U$, if $f(u)=g(w)$, then
$f(\alpha(u))=g(\alpha(w))$.

Moreover, by $U\overset{M,\alpha}{\longleftrightarrow}V$ we mean that

$\alpha$ is bijective, $U\xrightarrow{M,\alpha}V$ and
$V\xrightarrow{M,{{\alpha}^{-1}}}U$,

where $\alpha^{-1}:V\to U$ denotes the inverse of $\alpha$;

or equivalently, by $U\overset{M,\alpha}{\longleftrightarrow}V$ we mean that

$\alpha$ is bijective, and $\forall f,g\in M$ and $u,w\in U$,
$f(u)=g(w)\Leftrightarrow f(\alpha(u))=g(\alpha(w))$.

Now Proposition 7.1.3 can be generalized to

######  Proposition 7.2.2.

Let $\sigma$ be a $T$-morphism from a $T$-space $S_{1}$ to a $T$-space
$S_{2}$. Then $\forall U\subseteq{{S}_{1}}$, $U\xrightarrow{T,\alpha}{S}_{2}$,
where $\alpha:=\sigma{{|}_{U}}$. In particular,
${{S}_{1}}\xrightarrow{T,\sigma}{S}_{2}$.

###### Proof.

Let $f,g\in T$ and let $u,w\in U\subseteq{{S}_{1}}$. If $f(u)=g(w)$, then

$f(\alpha(u))=f(\sigma(u))=\sigma(f(u))=\sigma(g(w))=g(\sigma(w))=g(\alpha(w)).$

Hence by Notation 7.2.1, $U\xrightarrow{T,\alpha}{S}_{2}$. ∎

However, the converse of the above proposition is not true, as shown below.

######  Example 7.2.3.

Let $R$, $T$ and $\sigma$ be defined as in Example 7.1.5. Let $U\subseteq R$
and let $\alpha=\sigma{{|}_{U}}$. Then $\forall u,w\in U$ and $f,g\in T$,
$f(u)=g(w)$ implies that $f(\alpha(u))=g(\alpha(w))$. Hence
$U\xrightarrow{T,\alpha}R$. But obviously $\exists f\in T$ and $u\in R$ such
that $f(\sigma(u))\neq\sigma(f(u))$. So $\sigma$ is not a $T$-morphism.

To construct $T$-morphisms, we generalize Theorem 7.1.7 as follows.

######  Theorem 7.2.4.

Let $\alpha:U(\subseteq D)\to V(\subseteq D)$ be a map.

  1. (a)

$U\xrightarrow{T,\alpha}V$ if and only if there exists a map
$\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle
V\right\rangle}_{T}}$ given by $f(u)\mapsto f(\alpha(u))$, $\forall u\in U$
and $f\in T$.

  2. (b)

The above map $\sigma$ is a $T$-morphism.

###### Proof.

For (a):

$U\xrightarrow{T,\alpha}V$;

$\Leftrightarrow$ $f(u)=g(w)$ implies that $f(\alpha(u))=g(\alpha(w))$,
$\forall f,g\in T$ and $u,w\in U$;

$\Leftrightarrow$ $\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle
V\right\rangle}_{T}}$ given by $f(u)\mapsto f(\alpha(u))$, $\forall u\in U$
and $f\in T$, is a well-defined map.

For (b): Let $a\in{{\left\langle U\right\rangle}_{T}}$. Then $\exists u\in U$
and $g\in T$ such that $a=g(u)$. Let $f\in T$. Then
$\sigma(f(a))=\sigma((f\circ g)(u))$. By Definition 2.1.1, $f\circ g\in T$,
and so by the definition of $\sigma$, $\sigma((f\circ g)(u))=(f\circ
g)(\alpha(u))$. Besides, $g(\alpha(u))=\sigma(g(u))$ also by the definition of
$\sigma$. Combining the above equations, we have

$\sigma(f(a))=f(g(\alpha(u)))=f(\sigma(g(u)))=f(\sigma(a)).$

Thus $\sigma$ is a $T$-morphism. ∎

######  Theorem 7.2.5.

Let $\alpha:U(\subseteq D)\to V(\subseteq D)$ be a map.

  1. (a)

The following are equivalent:

    1. (i)

$U\overset{T,\alpha}{\longleftrightarrow}V$.

    2. (ii)

$\alpha$ is bijective and there exists a bijective map $\sigma:{{\left\langle
U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}}$ given by
$f(u)\mapsto f(\alpha(u))$, $\forall u\in U$ and $f\in T$.

  2. (b)

The above map $\sigma$ is a $T$-isomorphism.

###### Proof.

For (a):

$\Leftrightarrow$ $U\overset{T,\alpha}{\longleftrightarrow}V$;

$\Leftrightarrow$ $\alpha$ is bijective, and $\forall f,g\in T$ and $u,w\in
U$, $f(u)=g(w)$ if and only if $f(\alpha(u))=g(\alpha(w))$;

$\Leftrightarrow$ $\alpha$ is bijective and $\sigma:{{\left\langle
U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}}$ given by
$f(u)\mapsto f(\alpha(u))$, $\forall u\in U$ and $f\in T$, is a well-defined
bijective map.

For (b): By (b) in Theorem 7.2.4, $\sigma$ is a $T$-morphism. Since $\sigma$
is bijective, by Definition 2.4.1, it is a $T$-isomorphism. ∎

The converse of statement (b) in Theorem 7.2.4 raises a question as follows,
which is a generalization of Question 7.1.9.

######  Question 7.2.6.

For $U\subseteq D$ and a $T$-morphism $\sigma$ from ${{\left\langle
U\right\rangle}_{T}}$ to a $T$-space, is there any map $\alpha:U\to D$ such
that $\sigma(f(u))=f(\alpha(u))$, $\forall u\in U$ and $f\in T$?

The significance of this question is that, in the cases where the answer is
yes, we can construct $\sigma$ from $\alpha$ and $T$. Example 7.1.10 shows a
case where the answer to this question is negative, but we will see that there
are cases where the answer is positive. Thus this property deserves a name.

######  Definition 7.2.7.

Let $T$ be an operator semigroup on $D$, let $U\subseteq D$, and let $\sigma$
be a $T$-morphism from ${{\left\langle U\right\rangle}_{T}}$ to a $T$-space.
If there exists a map $\alpha:U\to D$ such that $\sigma(f(u))=f(\alpha(u))$,
$\forall u\in U$ and $f\in T$, then we say that $\sigma$ is _constructible by_
$\alpha$, or just say that $\sigma$ is _constructible_ for brevity.

Recall that it is possible that ${{\left\langle U\right\rangle}_{T}}$ does not
contain $U$ (Proposition 2.2.6). But if the identity function lies in $T$,
which is true in the cases such as Examples 2.1.2, 2.1.4 and 2.1.5, then
$U\subseteq{{\left\langle U\right\rangle}_{T}}$, where we have a positive
answer to Question 7.2.6.

######  Theorem 7.2.8.

Let $U\subseteq D$. If $U\subseteq{{\left\langle U\right\rangle}_{T}}$, then
any $T$-morphism $\sigma$ from ${{\left\langle U\right\rangle}_{T}}$ to a
$T$-space is constructible by $\alpha:=\sigma{{|}_{U}}$.

###### Proof.

Since $U\subseteq{{\left\langle U\right\rangle}_{T}}$, by Definition 2.3.1,
$\sigma(f(u))=f(\sigma(u))=f(\alpha(u))$, $\forall u\in U$ and $f\in T$. Thus
$\sigma$ is constructible by $\alpha$. ∎

Moreover, for Question 7.2.6, if the operator semigroup can be generated by a
single element, then we also have a positive answer as follows.

######  Claim 7.2.9.

Let $\sigma$ be a $T$-morphism from ${{\left\langle U\right\rangle}_{T}}$ to
${{\left\langle V\right\rangle}_{T}}$. If $T=\left\langle g\right\rangle$
(Definition 2.1.6), where $g$ is a function from $D$ to $D$, then there exists
a map $\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{T}}$ such that
$\sigma$ is constructible by $\alpha$.

We can give a proof of the claim, which would be a little long. We omit the
proof for brevity because the claim is a straightforward corollary of Theorem
7.3.9.

###  7.3. A construction of $\theta$-morphisms

For $\theta$-morphisms, we generalize Notation 7.2.1 to

######  Notation 7.3.1.

Let $D_{1}$ (resp. $D_{2}$) be a set, let $M_{1}$ (resp. $M_{2}$) be a set of
maps from $D_{1}$ to $D_{1}$ (resp. a set of maps from $D_{2}$ to $D_{2}$),
let $\theta\subseteq{{M}_{1}}\times{{M}_{2}}$, and let
$\alpha:U(\subseteq{D_{1}})\to V(\subseteq{{D}_{2}})$ be a map. Then by
$U\xrightarrow{\theta,\alpha}V$ we mean that $\forall
f,g\in\operatorname{Dom}\theta$ and $u,w\in U$,

$f(u)=g(w)$ implies that $\theta(f)(\alpha(u))=\theta(g)(\alpha(w))$.

Moreover, by $U\overset{\theta,\alpha}{\longleftrightarrow}V$ we mean that

$\alpha$ is bijective, $U\xrightarrow{\theta,\alpha}V$ and
$V\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U$,

where ${{\theta}^{-1}}:=\\{(h,f)\,|\,(f,h)\in\theta\\}$.

######  Remark.

  1. (1)

$\forall f\in\operatorname{Dom}\theta$ and $u\in U$, $f(u)=f(u)$, and so
$U\xrightarrow{\theta,\alpha}V$ implies that $\forall
f\in\operatorname{Dom}\theta$ and $u\in U$,
$\theta(f)(\alpha(u))=\theta(f)(\alpha(u))$. Hence by Convention 6.1.7,
$U\xrightarrow{\theta,\alpha}V$ implies that $\forall
f\in\operatorname{Dom}\theta$ and $u\in U$, $\theta(f)(\alpha(u))$ is well-
defined. Thus by Proposition 6.1.6, $U\xrightarrow{\theta,\alpha}V$ implies
that $\theta{{|}_{\operatorname{Im}\alpha}}$ is a map.

  2. (2)

If ${{M}_{1}}={{M}_{2}}=:M$ and $\theta$ is the identity map on $M$, then
$U\xrightarrow{\theta,\alpha}V$ is equivalent to $U\xrightarrow{M,\alpha}V$.

The following proposition generalizes Proposition 7.2.2.

######  Proposition 7.3.2.

Let $T_{1}$ and $T_{2}$ be operator semigroups and let
$\theta\subseteq{{T}_{1}}\times{{T}_{2}}$. Let $\phi$ be a $\theta$-morphism
from a $T_{1}$-space $S_{1}$ to a $T_{2}$-space $S_{2}$. Then $\forall
U\subseteq S_{1}$, $U\xrightarrow{\theta,\alpha}S_{2}$, where
$\alpha:=\phi{{|}_{U}}$. In particular, $S_{1}\xrightarrow{\theta,\phi}S_{2}$.

###### Proof.

By Proposition 6.1.9, $\theta{{|}_{\operatorname{Im}\phi}}$ is a map, and so
is $\theta{{|}_{\operatorname{Im}\alpha}}$ (by Proposition 6.1.6).

Let $f,g\in\operatorname{Dom}\theta$ and let $u,w\in U\subseteq S_{1}$. If
$f(u)=g(w)$, then by Definition 6.1.8,

$\theta(f)(\alpha(u))=\theta(f)(\phi(u))=\phi(f(u))=\phi(g(w))=\theta(g)(\phi(w))=\theta(g)(\alpha(w))$.

Hence by Notation 7.3.1, $U\xrightarrow{\theta,\alpha}S_{2}$. ∎

For some reasons which will be clear soon, we are more interested when
$\theta$ (of $\theta$-morphisms) has the following property.

######  Definition 7.3.3.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$,
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$, and let
$A\subseteq{{D}_{2}}$. If $\theta(f\circ g)(a)=(\theta(f)\circ\theta(g))(a)$,
$\forall f,g\in\operatorname{Dom}\theta$ and $a\in A$, then we say that
$\theta$ is _distributive over_ $A$.

######  Remark.

By Convention 6.1.7, the condition

“$\theta(f\circ g)(a)=(\theta(f)\circ\theta(g))(a)$, $\forall
f,g\in\operatorname{Dom}\theta$ and $a\in A$”

implies that $\forall f,g\in\operatorname{Dom}\theta$ and $a\in A$, both
$\theta(f\circ g)(a)$ and $(\theta(f)\circ\theta(g))(a)$ are well-defined, and
so the condition implies that $\operatorname{Dom}\theta$ is an operator
semigroup on $D_{1}$ and $\theta{{|}_{A}}$ is a map.

Then to construct $\theta$-morphisms, we generalize Theorem 7.2.4 as follows.

######  Theorem 7.3.4.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$ with
$\operatorname{Dom}\theta={{T}_{1}}$, and let $\alpha:U(\subseteq{D_{1}})\to
V(\subseteq{{D}_{2}})$ be a map.

  1. (a)

$U\xrightarrow{\theta,\alpha}V$ if and only if there exists a map
$\phi:{{\left\langle U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle
V\right\rangle}_{{{T}_{2}}}}$ given by $f(u)\mapsto\theta(f)(\alpha(u))$,
$\forall u\in U$ and $f\in{{T}_{1}}$.

  2. (b)

The above map $\phi$ is a $\theta$-morphism if $\theta$ is distributive over
$V$.

###### Proof.

For (a):

$U\xrightarrow{\theta,\alpha}V$;

$\Leftrightarrow$ $\forall f,g\in\operatorname{Dom}\theta(={{T}_{1}})$ and
$a,b\in U$, $f(a)=g(b)$ implies that
$\theta(f)(\alpha(a))=\theta(g)(\alpha(b))$;

$\Leftrightarrow$ $\phi:{{\left\langle
U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}}$
given by $f(u)\mapsto\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$, is a well-defined map.

For (b): Suppose that $\theta$ is distributive over $V$.

Let $a\in{{\left\langle U\right\rangle}_{{{T}_{1}}}}$. Then $\exists u\in U$
and $g\in{{T}_{1}}$ such that $a=g(u)$. Let $f\in{{T}_{1}}$. Then $f\circ
g\in{{T}_{1}}$ by Definition 2.1.1. So $\phi(f(a))=\phi((f\circ
g)(u))=\theta(f\circ g)(\alpha(u))$ by the definition of $\phi$. Then because
$\theta$ is distributive over $V$, by Definition 7.3.3, $\theta(f\circ
g)(\alpha(u))=\theta(f)(\theta(g)(\alpha(u)))$. And
$\theta(g)(\alpha(u))=\phi(g(u))$ by the definition of $\phi$. Combining the
above equations, we have

$\phi(f(a))=\theta(f)(\phi(g(u)))=\theta(f)(\phi(a)).$

Thus by Definition 6.1.8, $\phi$ is a $\theta$-morphism from ${{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ to ${{\left\langle
V\right\rangle}_{{{T}_{2}}}}$. ∎

Then Theorem 7.2.5 is generalized as follows.

######  Theorem 7.3.5.

Let $T_{1}$, $T_{2}$, $U$ and $\alpha$ be defined as in Theorem 7.3.4. Let
$\theta\subseteq{{T}_{1}}\times{{T}_{2}}$ such that
$\operatorname{Dom}\theta={{T}_{1}}$ and $\operatorname{Im}\theta={{T}_{2}}$.
Then

  1. (a)

The following are equivalent:

    1. (i)

$U\overset{\theta,\alpha}{\longleftrightarrow}V$.

    2. (ii)

$\alpha$ is bijective, there exists a bijective map $\phi:{{\left\langle
U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}}$
given by $f(u)\mapsto\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$, and its inverse $\phi^{-1}:{{\left\langle
V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}}$
can be given by $g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v))$, $\forall v\in V$
and $g\in{{T}_{2}}$.

  2. (b)

Suppose that $\theta$ is distributive over $V$. Then the bijective map $\phi$
given above is a $\theta$-isomorphism.

###### Proof.

For (a):

$U\overset{\theta,\alpha}{\longleftrightarrow}V$;

$\Leftrightarrow$ $\alpha$ is bijective, $U\xrightarrow{\theta,\alpha}V$ and
$V\xrightarrow{{{\theta}^{-1}},{{\alpha}^{-1}}}U$;

$\Leftrightarrow$ $\alpha$ is bijective, $\phi:{{\left\langle
U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}}$
given by $f(u)\mapsto\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$, is a well-defined map, and $\phi^{\prime}:{{\left\langle
V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle U\right\rangle}_{{{T}_{1}}}}$
given by $g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v))$, $\forall v\in V$ and
$g\in{{T}_{2}}$ is also a well-defined map;

$\Leftrightarrow$ $\alpha$ is bijective, $\phi:{{\left\langle
U\right\rangle}_{{{T}_{1}}}}\to{{\left\langle V\right\rangle}_{{{T}_{2}}}}$
given by $f(u)\mapsto\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$, is a well-defined bijective map, and its inverse
$\phi^{-1}:{{\left\langle V\right\rangle}_{{{T}_{2}}}}\to{{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ can be given by
$g(v)\mapsto\theta^{-1}(g)(\alpha^{-1}(v))$, $\forall v\in V$ and
$g\in{{T}_{2}}$.

The second equivalence relation follows from Theorem 7.3.4.

The third equivalence relation is explained as follows. The sufficiency
($\Leftarrow$) is obvious, so we only show the necessity ($\Rightarrow$).

Let $(f,g)\in\theta$ and $u\in U$. Then

$\phi^{\prime}(\phi(f(u)))$

$=\phi^{\prime}(\theta(f)(\alpha(u)))$ (by the definition of $\phi$)

$=\phi^{\prime}(g(\alpha(u)))$ (by Definition 6.1.5)

$=\theta^{-1}(g)(u)$ (by the definition of $\phi^{\prime}$)

$=f(u)$ (by (the $\theta^{-1}$ version of) Definition 6.1.5)

Then $\phi^{\prime}\circ\phi$ is the identity map on ${{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ (because $\operatorname{Dom}\theta={{T}_{1}}$).

Analogously, we can show that $\phi\circ\phi^{\prime}$ is the identity map on
${{\left\langle V\right\rangle}_{{{T}_{2}}}}$. Therefore, both $\phi$ and
$\phi^{\prime}$ are bijective and $\phi^{\prime}$ is the inverse of $\phi$.

For (b):

By (b) in Theorem 7.3.4, the map $\phi$ given in (ii) is a $\theta$-morphism.
Since $\phi$ is bijective, by Definition 6.2.1, $\phi$ is a
$\theta$-isomorphism. ∎

Theorem 7.3.4 raises a question as follows, which is a generalization of
Question 7.2.6.

######  Question 7.3.6.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$,
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$, let $U\subseteq
D_{1}$ and let $\phi$ be a $\theta$-morphism from ${{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ to a $T_{2}$-space. If
$\operatorname{Dom}\theta={{T}_{1}}$, is there any map $\alpha:U\to{{D}_{2}}$
such that $\phi(f(u))=\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$?

The significance of this question is that, in the cases where the answer is
yes, we can construct $\phi$ from $\alpha$ and $\theta$. This property
deserves a name as follows, which is a generalization of Definition 7.2.7.

######  Definition 7.3.7.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$,
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$, let $U\subseteq
D_{1}$ and let $\phi$ be a $\theta$-morphism from ${{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ to a $T_{2}$-space. If
$\operatorname{Dom}\theta={{T}_{1}}$ and there exists a map
$\alpha:U\to{{D}_{2}}$ such that $\phi(f(u))=\theta(f)(\alpha(u))$, $\forall
u\in U$ and $f\in{{T}_{1}}$, then we say that $\phi$ is _constructible by_
$\alpha$, or just say that $\phi$ is _constructible_ for brevity.

######  Remark.

By Convention 6.1.7, the condition “$\phi(f(u))=\theta(f)(\alpha(u))$,
$\forall u\in U$ and $f\in{{T}_{1}}$” implies that
$\theta{{|}_{\operatorname{Im}\alpha}}$ is a map.

Since $T$-morphisms are special cases of $\theta$-morphisms, Example 7.1.10
shows that it is possible that a $\theta$-morphism $\phi$ from ${{\left\langle
U\right\rangle}_{{{T}_{1}}}}$ to a $T_{2}$-space is not constructible by any
map $\alpha:U\to{{D}_{2}}$ even if $\theta$ is distributive over ${{D}_{2}}$.
Now let’s show some cases where $\theta$-morphisms are constructible. The
following generalizes Theorem 7.2.8.

######  Theorem 7.3.8.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$,
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$ and let
$U\subseteq{D_{1}}$. If $\operatorname{Dom}\theta={{T}_{1}}$ and
$U\subseteq{{\left\langle U\right\rangle}_{{{T}_{1}}}}$, then any
$\theta$-morphism $\phi$ from ${{\left\langle U\right\rangle}_{{{T}_{1}}}}$ to
a $T_{2}$-space is constructible by $\alpha:=\phi{{|}_{U}}$.

###### Proof.

Since $\operatorname{Dom}\theta={{T}_{1}}$ and $U\subseteq{{\left\langle
U\right\rangle}_{{{T}_{1}}}}$, by Definition 6.1.8, $\forall u\in U$ and
$f\in{{T}_{1}}$, $\phi(f(u))=\theta(f)(\phi(u))=\theta(f)(\alpha(u))$. ∎

As was asserted by Claim 7.2.9, a $T$-morphism is constructible if $T$ can be
generated by a single element. Now we generalize this claim for
$\theta$-morphisms.

######  Theorem 7.3.9.

Let $T_{1}$ and $T_{2}$ be operator semigroups on $D_{1}$ and $D_{2}$,
respectively, let $\theta\subseteq{{T}_{1}}\times{{T}_{2}}$, and let $\phi$ be
a $\theta$-morphism from ${{\left\langle U\right\rangle}_{{{T}_{1}}}}$ to
${{\left\langle V\right\rangle}_{{{T}_{2}}}}$. If
$\operatorname{Dom}\theta={{T}_{1}}=\left\langle g\right\rangle$,
$\operatorname{Im}\theta={{T}_{2}}$, and $\theta$ is distributive over
$V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}$, then there exists a map
$\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}$ such that
$\phi$ is constructible by $\alpha$.

######  Remark.

If $T:={{T}_{1}}={{T}_{2}}$ and $\theta$ is the identity map on $T$, then we
can tell that Theorem 7.3.9 degenerates into Claim 7.2.9.

###### Proof.

If $\phi$ is the empty map $\emptyset\to\emptyset$, then by Definition 7.3.7,
$\phi$ is constructible by the empty map trivially. Hence we only need to
prove the case where $U(\subseteq{D_{1}})$ and $V(\subseteq{{D}_{2}})$ are
nonempty.

By (the remark under) Definition 7.3.3, $\theta{{|}_{A}}$ is a map, where
$A:=V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}$, and so is
$\theta{{|}_{\operatorname{Im}\phi}}$ because
$\operatorname{Im}\phi\subseteq{{\left\langle
V\right\rangle}_{{{T}_{2}}}}\subseteq A$.

First, $\forall x\in{{D}_{2}}$ and $f\in{{T}_{1}}$, set

(7.9) ${{(\theta(f))}_{A}^{-1}}(x)=\\{a\in A\,|\,\theta(f)(a)=x\\}.$

It is well-defined because $\theta{{|}_{A}}$ is a map.

Let $u\in U$. Let
${{A}_{u}}=\bigcap\nolimits_{f\in{{T}_{1}}}{{{(\theta(f))}_{A}^{-1}}(\phi(f(u)))}$.
Suppose that ${{A}_{u}}\neq\emptyset$. Then by Equation $($7.9$)$,
$\theta(f)(a)=\phi(f(u))$, $\forall a\in{{A}_{u}}$ and $f\in{{T}_{1}}$. Let
$\alpha(u)$ be any $a\in{{A}_{u}}$. Then $\forall f\in{{T}_{1}}$,
$\phi(f(u))=\theta(f)(a)=\theta(f)(\alpha(u))$ and $\alpha(u)\in
A=V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}$, as desired for $\phi$
to be constructible by $\alpha:U\to V\bigcup{{\left\langle
V\right\rangle}_{{{T}_{2}}}}$.

Therefore, it suffices to show that ${{A}_{u}}\neq\emptyset$.

To show that ${{A}_{u}}\neq\emptyset$, we shall first show that
${{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\subseteq A_{u}$. Then we shall prove
that ${{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset$.

Since ${{T}_{1}}=\left\langle g\right\rangle$,
${{A}_{u}}=\bigcap\nolimits_{f\in{{T}_{1}}}{{{(\theta(f))}_{A}^{-1}}(\phi(f(u)))}=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u))})$,
where ${{g}^{k}}$ denotes the composite of $k(\in{{\mathbb{Z}}^{+}})$ copies
of $g$.

Let $k(\in{{\mathbb{Z}}^{+}})>1$. Because $g(u)\in{{\left\langle
U\right\rangle}_{{{T}_{1}}}}$, by Definition 6.1.8,

$\phi({{g}^{k}}(u))=\phi({{g}^{k-1}}(g(u)))=\theta({{g}^{k-1}})(\phi(g(u)))={{(\theta(g))}^{k-1}}(\phi(g(u)))$

because $\theta$ is assumed to be distributive over $V\bigcup{{\left\langle
V\right\rangle}_{{{T}_{2}}}}(\supseteq\operatorname{Im}\phi)$, and so

$\phi({{g}^{k}}(u))={{(\theta(g))}^{k-1}}(\phi(g(u))).$

Applying ${{(\theta({{g}^{k}}))}_{A}^{-1}}$ to both sides of the above
equation, we have

(7.10)
${{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u)))={{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u)))).$

Suppose that $\exists a\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))$. Then by
$($7.9$)$, $\theta(g)(a)=\phi(g(u))$. Hence
${{(\theta(g))}^{k-1}}(\theta(g)(a))={{(\theta(g))}^{k-1}}(\phi(g(u)))$. Since
$\theta$ is distributive over $A$,

${{(\theta(g))}^{k-1}}(\theta(g)(a))={{(\theta(g))}^{k}}(a)=\theta({{g}^{k}})(a).$

Combining the above three equations, we have
$\theta({{g}^{k}})(a)={{(\theta(g))}^{k-1}}(\phi(g(u)))$. Then by $($7.9$)$,
$a\in{{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u))))$. So

(7.11)
$(a\in){{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\subseteq{{(\theta({{g}^{k}}))}_{A}^{-1}}({{(\theta(g))}^{k-1}}(\phi(g(u)))).$

Combining $($7.10$)$ with $($7.11$)$, we have

${{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u)))\supseteq{{(\theta(g))}_{A}^{-1}}(\phi(g(u))).$

Hence
${{A}_{u}}(=\bigcap\nolimits_{k\in{{\mathbb{Z}}^{+}}}{{{(\theta({{g}^{k}}))}_{A}^{-1}}(\phi({{g}^{k}}(u))}))\supseteq(\theta(g))_{A}^{-1}(\phi(g(u))$.

Therefore, to show that ${{A}_{u}}\neq\emptyset$, it suffices to prove that
${{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset$.

Since $\phi(g(u))\in{{\left\langle V\right\rangle}_{{{T}_{2}}}}$ and
$\operatorname{Im}\theta={{T}_{2}}$, $\exists v\in V$ and
$h\in{{\mathbb{Z}}^{+}}$ such that $\phi(g(u))=\theta({{g}^{h}})(v)$. If
$h=1$, then by $($7.9$)$, $v\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))$. And if
$h>1$, then because $\theta$ is distributive over $V\bigcup{{\left\langle
V\right\rangle}_{{{T}_{2}}}}$,
$\phi(g(u))=\theta({{g}^{h}})(v)=\theta(g)(\theta({{g}^{h-1}})(v))$, and so by
$($7.9$)$ and the fact that $\theta({{g}^{h-1}})(v)\in{{\left\langle
V\right\rangle}_{{{T}_{2}}}}\subseteq A$,
$\theta({{g}^{h-1}})(v)\in{{(\theta(g))}_{A}^{-1}}(\phi(g(u)))$. In both cases
of $h$, ${{(\theta(g))}_{A}^{-1}}(\phi(g(u)))\neq\emptyset$, as desired.
Therefore, ${{A}_{u}}\neq\emptyset$.

In summary, for each $u\in U$, let $\alpha(u)$ be any element of ${A}_{u}$.
Then the map $\alpha:U\to V\bigcup{{\left\langle V\right\rangle}_{{{T}_{2}}}}$
satisfies $\phi(f(u))=\theta(f)(\alpha(u))$, $\forall u\in U$ and
$f\in{{T}_{1}}$.

Hence by Definition 7.3.7, $\phi$ is constructible by $\alpha$. ∎

###  7.4. Another construction of $T$-morphisms

Let $\sigma$ be a $T$-morphism mapping from a $T$-space $S$. If $\sigma(a)=b$,
where $a\in S$, then $\forall f\in T$, $\sigma(f(a))=f(\sigma(a))=f(b)$. In
other words, if an ordered pair $(a,b)\in\sigma$, where $\sigma$ is regarded
as a binary relation, then $\forall f\in T$, $(f(a),f(b))\in\sigma$. This
observation implies that there exists some sort of basic parts of
$T$-morphisms.

To facilitate our descriptions, we give the following notations only for this
subsection.

######  Notation 7.4.1.

Let $T^{*}=T\bigcup\\{$Id on $D$} and let
${{R}_{T}}(x,y)=\\{(f(x),f(y))\,|\,f\in T^{*}\\}$, where $(x,y)\in D\times D$.
Besides, both ${{S}_{a}}$ and ${{S}_{b}}$ denote $T$-spaces.

######  Remark.

Clearly, ${{R}_{T}}(x,y)\subseteq D\times D$ is a relation in $D$.

######  Proposition 7.4.2.

Let $R\subseteq{{S}_{a}}\times{{S}_{b}}$ such that $\operatorname{Dom}R$ is a
$T$-subspace of $S_{a}$. Then the following are equivalent:

  1. (i)

$R$ is a $T$-morphism from $\operatorname{Dom}R$ to ${{S}_{b}}$.

  2. (ii)

There exists a relation
$\bigcup\nolimits_{i}{\\{({{a}_{i}},{{b}_{i}})\\}}\subseteq{{S}_{a}}\times{{S}_{b}}$
such that $R=\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})}$ and $R$ is
a function.

###### Proof.

(i)$\Rightarrow$(ii): Assume that
$R=\bigcup\nolimits_{i}{\\{({{a}_{i}},{{b}_{i}})}\\}$. By Definition 2.3.1,
$\forall f\in T^{*}$, $({{a}_{i}},{{b}_{i}})\in R$ implies that
$R(f({{a}_{i}}))=f(R({{a}_{i}}))=f({{b}_{i}})$, and so
$(f({{a}_{i}}),f({{b}_{i}}))\in R$.

Therefore, $R=\bigcup\nolimits_{i}{\\{(f({{a}_{i}}),f({{b}_{i}}))\,|\,f\in
T^{*}\\}}=\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})}$. Hence (ii) is
true.

(ii)$\Rightarrow$(i): Let $g\in T^{*}$. Suppose that $(u,v)\in R$. Since
$R=\bigcup\nolimits_{i}{{{R}_{T}}({{a}_{i}},{{b}_{i}})}=\bigcup\nolimits_{i}{\\{(f({{a}_{i}}),f({{b}_{i}}))}\,|\,f\in
T^{*}\\}$ and $T$ is a semigroup, $(g(u),g(v))\in R$. Because $R$ is a
function, $R(g(u))=g(v)=g(R(u))$. Hence by Definition 2.3.1, $R$ is a
$T$-morphism from $\operatorname{Dom}R$ to ${{S}_{b}}$. ∎

The above proposition implies that ${{R}_{T}}(x,y)$, where
$(x,y)\in{{S}_{a}}\times{{S}_{b}}$, may be regarded as some sort of basic part
of a $T$-morphism from a $T$-subspace of ${{S}_{a}}$ to ${{S}_{b}}$. This
observation inspires us to define a topology on ${{S}_{a}}\times{{S}_{b}}$ as
follows.

######  Definition 7.4.3.

Let
${B_{T}}({S_{a}}\times{S_{b}})=\\{{R_{T}}(x,y)\,|\,(x,y)\in{S_{a}}\times{S_{b}}\\}$.
Let ${{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}})$ be the topology on
${{S}_{a}}\times{{S}_{b}}$ generated by ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$
as a subbasis; that is, let ${{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}})$ be
the collection of all unions of finite intersections of elements of
${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$.

######  Remark.

Clearly, ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$ qualifies as a subbasis for a
topology on ${{S}_{a}}\times{{S}_{b}}$.

######  Proposition 7.4.4.

Let $\bigcap\nolimits_{i}{B_{i}}$ be an arbitrary intersection of elements of
${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$. Then $\bigcap\nolimits_{i}{B_{i}}$ is
also a union of elements of ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$.

###### Proof.

If $\bigcap\nolimits_{i}{B_{i}}=\emptyset$, then by convention,
$\bigcap\nolimits_{i}{B_{i}}$ may be regarded as the union of an empty family
of elements of ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$. So we only need to
consider the case where $\bigcap\nolimits_{i}{B_{i}}\neq\emptyset$.

Any element of ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$ is a set
${R_{T}}(x,y)=\\{(f(x),f(y))\,|\,f\in T^{*}\text{ }\\!\\!\\}\\!\\!\text{ }$,
where $(x,y)\in{{S}_{a}}\times{{S}_{b}}$. So for any $B_{i}$, $(u,v)\in B_{i}$
and $g\in T^{*}$, since $T$ is a semigroup, we can tell that $(g(u),g(v))\in
B_{i}$. Then $\forall g\in T^{*}$ and $(u,v)\in\bigcap\nolimits_{i}{B_{i}}$,
$(g(u),g(v))\in\bigcap\nolimits_{i}{B_{i}}$

Therefore,

$\bigcap\nolimits_{i}{B_{i}}$

$=\bigcup\nolimits_{(u,v)\in\bigcap\nolimits_{i}{B_{i}}}{\\{(g(u),g(v))\,|\,g\in
T^{*}\text{ }\\!\\!\\}\\!\\!\text{ }}$

$=\bigcup\nolimits_{(u,v)\in\bigcap\nolimits_{i}{B_{i}}}{{{R}_{T}}(u,v)}.$ ∎

Thus immediately we have

######  Corollary 7.4.5.

${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$ is a basis of
${{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}})$. So
${{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}})$ is the collection of all
unions of elements of ${{B}_{T}}({{S}_{a}}\times{{S}_{b}})$.

The following theorem is obvious when Proposition 7.4.2 is combined with the
above corollary.

######  Theorem 7.4.6.

Let $R\subseteq{{S}_{a}}\times{{S}_{b}}$ such that $\operatorname{Dom}R$ is a
$T$-subspace of $S_{a}$. Then $R$ is a $T$-morphism from
$\operatorname{Dom}(R)$ to ${{S}_{b}}$ if and only if
$R\in{{\mathcal{T}}_{T}}({{S}_{a}}\times{{S}_{b}})$ and $R$ is a function.

##  8\. Applications to topological spaces

In this section we shall study the Galois correspondences for topological
spaces. After introducing some notions and properties in Subsection 8.1, in
Subsection 8.2, we employ Theorems 3.2.5 and 3.2.6 and obtain the “general”
Galois correspondences for topological spaces (Corollaries 8.2.2 and 8.2.4).
In Subsection 8.3, we derive two results which “restrict” the Galois
correspondences to homeomorphisms from a topological space to itself (Theorems
8.3.11 and 8.3.20).

###  8.1. Basic notions and properties

To simplify our descriptions, we define some notations for Section 8 as
follows.

######  Notation 8.1.1.

Unless otherwise specified, $X$ and $Y$ always denote topological spaces. Let
$\mathcal{P}(X)$ and $\mathcal{P}(Y)$ denote the power sets of $X$ and $Y$,
respectively. Moreover,

$T:={{T}_{X}}:=\\{\text{I}{{\text{d}}_{X}}$,
$\text{C}{{\text{l}}_{X}}:\mathcal{P}(X)\to\mathcal{P}(X)$ given by
${{A}_{X}}\mapsto\overline{{{A}_{X}}}\\}$

and

${{T}_{Y}}:=\\{\text{I}{{\text{d}}_{Y}}$,
$\text{C}{{\text{l}}_{Y}}:\mathcal{P}(Y)\to\mathcal{P}(Y)$ given by
${{A}_{Y}}\mapsto\overline{{{A}_{Y}}}\\}$,

where $\text{I}{{\text{d}}_{X}}$ and $\text{I}{{\text{d}}_{Y}}$ denote the
identity functions on $\mathcal{P}(X)$ and $\mathcal{P}(Y)$, respectively.

######  Remark.

By Definition 2.1.1, it is easy to see that $T(={{T}_{X}})$ and ${{T}_{Y}}$
are operator semigroups on $\mathcal{P}(X)$ and $\mathcal{P}(Y)$,
respectively.

Then we have

######  Proposition 8.1.2.

$\mathcal{P}(X)$ is a $T$-space and

$\operatorname{End}_{T}(\mathcal{P}(X))=\\{\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall
A\in\mathcal{P}(X)\\}$.

###### Proof.

By Definition 2.2.1, $\mathcal{P}(X)$ is a $T$-space because
${{\left\langle\mathcal{P}(X)\right\rangle}_{T}}=P(X)$.

Since $T=\\{\text{I}{{\text{d}}_{X}}$,
$\text{C}{{\text{l}}_{X}}:\mathcal{P}(X)\to\mathcal{P}(X)$ given by
${{A}_{X}}\mapsto\overline{{{A}_{X}}}\\}$, by Definition 2.3.1, a map
$\sigma:\mathcal{P}(X)\to\mathcal{P}(X)$ is a $T$-morphism if and only if

$(\sigma(\text{C}{{\text{l}}_{X}}(A))=)\sigma(\overline{A})=\overline{\sigma(A)}(=\text{C}{{\text{l}}_{X}}(\sigma(A))),\forall
A\in\mathcal{P}(X)$.

So by Definition 2.3.7,

$\operatorname{End}_{T}(\mathcal{P}(X))=\\{\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall
A\in\mathcal{P}(X)\\}.$

∎

For brevity, we take out a part from Theorem 6.1.3 as follows.

######  Definition 8.1.3.

Let $f:X\to Y$ be a map. Then we define its _induced map_
$f^{*}:\mathcal{P}(X)\to\mathcal{P}(Y)$ by

  1. (1)

$\forall A\in\mathcal{P}(X)$ that is closed in $X$, let
$f^{*}(A)=\overline{f(A)}$, and

  2. (2)

$\forall A\in\mathcal{P}(X)$ that is not closed in $X$, let $f^{*}(A)=f(A)$,

where $f(A)$ denotes the set $\\{f(x)\,|\,x\in A\\}$ for convenience.

######  Remark.

In Section 8, unless otherwise specified, $f^{*}$ is defined by Definition
8.1.3 whenever $f$ is a map between topological spaces.

By Definition 8.1.3, we can restate Theorem 2.3.5 as follows.

######  Proposition 8.1.4.

A map $f:X\to X$ is continuous if and only if its induced map
$f^{*}:\mathcal{P}(X)\to\mathcal{P}(X)$ is a $T$-morphism.

To facilitate our following discussion, we give

######  Notation 8.1.5.

Let $F(X,Y)$ denote the set of all functions from $X$ to $Y$ and let
$F^{*}(X,Y)$ denote the set $\\{f^{*}\,|\,f\in F(X,Y)\\}$. Let $F(X)=F(X,X)$
and let $F^{*}(X)=F^{*}(X,X)$. Moreover, let $C(X)$ denote the set of all
continuous functions from $X$ to $X$, and let $C^{*}(X)$ denote the set
$\\{f^{*}\,|\,f\in C(X)\\}$.

Then by Proposition 8.1.4, the following is obvious.

######  Corollary 8.1.6.

$F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))=C^{*}(X)$.

It would be desirable if
$F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))=\operatorname{End}_{T}(\mathcal{P}(X))$.
However, it is possible that
$F^{*}(X)\bigcap\operatorname{End}_{T}(\mathcal{P}(X))\subsetneq\operatorname{End}_{T}(\mathcal{P}(X))$
(and so $C^{*}(X)\subsetneq\operatorname{End}_{T}(\mathcal{P}(X))$). That is,
there may exist a $T$-morphism $\sigma$ from $\mathcal{P}(X)$ to
$\mathcal{P}(X)$ such that $\sigma$ is not the induced map of any map from $X$
to $X$, as shown below.

######  Example 8.1.7.

Let $X$ be any topological space with the discrete topology. Then it follows
from Proposition 8.1.2 that any map from $\mathcal{P}(X)$ to $\mathcal{P}(X)$
is a $T$-morphism. Let $\sigma:\mathcal{P}(X)\to\mathcal{P}(X)$ be given by
$A\mapsto\emptyset,\forall A\in\mathcal{P}(X)$. If $X$ is nonempty, then by
Definition 8.1.3, it is not hard to see that $\sigma\notin F^{*}(X)$.

Before going further, we define

######  Notation 8.1.8.

We denote by ${{I}_{X,Y}}$ the map from $F(X,Y)$ to $F^{*}(X,Y)$ given by
$f\mapsto f^{*}$. Besides, let ${{I}_{X}}={{I}_{X,X}}$.

Obviously ${{I}_{X,Y}}$ is surjective. It would be desirable if ${{I}_{X,Y}}$
were always injective. However, there are cases where ${{I}_{X,Y}}$ is not
injective, as shown below.

######  Example 8.1.9.

Let $X=\\{1\\}$ and $Y=\\{2,3\\}$. Let ${{\mathcal{T}}_{X}}=\\{\emptyset,X\\}$
and ${{\mathcal{T}}_{Y}}=\\{\emptyset,Y\\}$ be the (trivial) topologies on $X$
and $Y$, respectively. Let two maps $f,g:X\to Y$ be given by $f(1)=2$ and
$g(1)=3$, respectively. Then $F(X,Y)=\\{f,g\\}$. And by Definition 8.1.3,
$f^{*}(\emptyset)=\emptyset=g^{*}(\emptyset)$ and
$f^{*}(\\{1\\})=\overline{\\{f(1)\\}}=\overline{\\{2\\}}=\\{2,3\\}=\overline{\\{3\\}}=\overline{\\{g(1)\\}}=g^{*}(\\{1\\})$.
So $f^{*}=g^{*}$. Hence the map ${{I}_{X,Y}}$ from $F(X,Y)$ to
$F^{*}(X,Y)=\\{f^{*}(=g^{*})\\}$ is not injective.

Surprisingly, there are two important cases where ${{I}_{X,Y}}$ is injective,
as shown by the following two propositions.

######  Proposition 8.1.10.

Let $X$ and $Y$ be topological spaces. Suppose that every finite point set in
$Y$ is closed, i.e., $Y$ satisfies the ${{T}_{1}}$ axiom. Then ${{I}_{X,Y}}$
is injective.

###### Proof.

Assume that ${{I}_{X,Y}}$ is not injective. Then there are two different maps
$f,g:X\to Y$ such that $f^{*}=g^{*}$. So $\exists x\in X$ such that $f(x)\neq
g(x)$.

By Definition 8.1.3, we can tell that $\\{x\\}$ is closed in $X$ (otherwise
$f^{*}(\\{x\\})=\\{f(x)\\}\neq\\{g(x)\\}=g^{*}(\\{x\\})$, which is contrary to
the assumption $f^{*}=g^{*}$). So again by Definition 8.1.3,
$\overline{\\{f(x)\\}}=f^{*}(\\{x\\})=g^{*}(\\{x\\})=\overline{\\{g(x)\\}}$.
However, since every finite point set in $Y$ is closed,
$\overline{\\{f(x)\\}}=\\{f(x)\\}$ and $\overline{\\{g(x)\\}}=\\{g(x)\\}$.
Thus $f(x)=g(x)$, which is contrary to $f(x)\neq g(x)$. ∎

######  Proposition 8.1.11.

For any topological space $X$, ${{I}_{X}}$ is injective.

###### Proof.

Assume that ${{I}_{X}}$ is not injective. Then there are two different maps
$f,g:X\to X$ such that $f^{*}=g^{*}$. So $\exists x\in X$ such that $f(x)\neq
g(x)$.

By Definition 8.1.3, we can tell that $\\{x\\}$ is closed in $X$ (otherwise
$f^{*}(\\{x\\})=\\{f(x)\\}\neq\\{g(x)\\}=g^{*}(\\{x\\})$, which is contrary to
the assumption $f^{*}=g^{*}$). So again by Definition 8.1.3,
$\overline{\\{f(x)\\}}=f^{*}(\\{x\\})=g^{*}(\\{x\\})=\overline{\\{g(x)\\}}$.
It follows that
$\overline{\\{f(x)\\}}=\overline{\\{g(x)\\}}\supseteq\\{f(x),g(x)\\}$. Hence
$\overline{\\{f(x)\\}}\neq\\{f(x)\\}$ and
$\overline{\\{g(x)\\}}\neq\\{g(x)\\}$ (because $f(x)\neq g(x)$), and so
$f(x)\neq x$ (otherwise
$\overline{\\{f(x)\\}}=\overline{\\{x\\}}=\\{x\\}=\\{f(x)\\}$, a
contradiction). Analogously, $g(x)\neq x$.

Then $\overline{\\{x,f(x)\\}}\supseteq\\{x,f(x),g(x)\\}\supsetneq\\{x,f(x)\\}$
because $\overline{\\{f(x)\\}}\supseteq\\{f(x),g(x)\\}$ and $f(x)\neq g(x)\neq
x$. So $\overline{\\{x,f(x)\\}}\neq\\{x,f(x)\\}$; that is, $\\{x,f(x)\\}$ is
not closed in $X$. Hence by Definition 8.1.3,

$f^{*}(\\{x,f(x)\\})=\\{f(x),f(f(x))\\}$ and
$g^{*}(\\{x,f(x)\\})=\\{g(x),g(f(x))\\}$.

Because it is assumed that $f^{*}=g^{*}$,
$\\{f(x),f(f(x))\\}=\\{g(x),g(f(x))\\}$.

Since $f(x)\neq g(x)$, we can tell from the above equation that
$f(f(x))=g(x)\neq f(x)=g(f(x))$. Because it is shown above that
$\overline{\\{f(x)\\}}\neq\\{f(x)\\}$, by Definition 8.1.3,
$f^{*}(\\{f(x)\\})=\\{f(f(x))\\}\neq\\{g(f(x))\\}=g^{*}(\\{f(x)\\})$, which is
contrary to the assumption that $f^{*}=g^{*}$. ∎

Moreover, there is an important case where ${{I}_{X}}$ commutes with
composition of functions, as shown below.

######  Lemma 8.1.12.

Let $f$ and $g$ be continuous maps from $X$ to $X$. Suppose that $g$ is
injective. Then $(f\circ g)^{*}=f^{*}\circ g^{*}$.

###### Proof.

It suffices to show that $(f\circ g)^{*}(A)=f^{*}(g^{*}(A)),\forall
A\in\mathcal{P}(X)$. There are two cases as follows.

(1) $A=\overline{A}$:

By Definition 8.1.3,

$(f\circ g)^{*}(A)=\overline{(f\circ g)(A)}=\overline{f(g(A))}$

and

$f^{*}(g^{*}(A))=f^{*}(\overline{g(A)})=\overline{f(\overline{g(A)})}.$

Moreover, since $f$ is continuous, by e.g. [10],
$f(\overline{g(A)})\subseteq\overline{f(g(A))}$. So
$\overline{f(\overline{g(A)})}\subseteq\overline{f(g(A))}$. On the other hand,
$f(\overline{g(A)})\supseteq f(g(A))$, and so
$\overline{f(\overline{g(A)})}\supseteq\overline{f(g(A))}$. Therefore,
$\overline{f(\overline{g(A)})}=\overline{f(g(A))}$.

Combining the above equations, we have $(f\circ g)^{*}(A)=f^{*}(g^{*}(A))$, as
desired.

(2) $A\neq\overline{A}$:

Assume that $g(A)$ is closed. Since $g$ is continuous, ${{g}^{-1}}(g(A))$ is
also closed. However, since $g$ is injective, ${{g}^{-1}}(g(A))=A$. Hence
${{g}^{-1}}(g(A))$ is not closed (because $A\neq\overline{A}$), a
contradiction. So $g(A)$ is not closed. Then by Definition 8.1.3,

$f^{*}(g^{*}(A))=f^{*}(g(A))=f(g(A))=(f\circ g)(A)=(f\circ g)^{*}(A),$

as desired. ∎

Now we can derive some results which shall be employed in Subsection 8.3.
Firstly, we define

######  Notation 8.1.13.

We denote by $\text{Hom}(X)$ the set of all homeomorphisms from $X$ to $X$.
Let $\text{Hom}^{*}(X)=\\{f^{*}\,|\,f\in\text{Hom}(X)\\}$.

Then we immediately have

######  Corollary 8.1.14.

$\emph{Hom}(X)$ constitutes a group, which we still denote by $\emph{Hom}(X)$,
with composition of functions as the binary operation.

$\text{Hom}^{*}(X)$ also constitutes a group, as shown below.

######  Proposition 8.1.15.

With composition of functions as the binary operation, $\emph{Hom}^{*}(X)$
constitutes a subgroup, which we still denote by $\emph{Hom}^{*}(X)$, of
$\operatorname{Aut}_{T}(\mathcal{P}(X))$ $($defined in Proposition 2.4.3$)$.

######  Remark.

Recall that $T$ is defined in Notation 8.1.1.

###### Proof.

It follows from Lemma 8.1.12 that $\text{Hom}^{*}(X)$ is closed under
composition of functions as the binary operation; that is, $\forall
f^{*},g^{*}\in\text{Hom}^{*}(X)$, $f^{*}\circ g^{*}\in\text{Hom}^{*}(X)$.
Moreover, by Definition 8.1.3, the identity map on $X$ induces the identity
map on $\mathcal{P}(X)$, which we denote by $\operatorname{Id}^{*}$. $\forall
f\in\text{Hom}(X)$, let ${{(f^{*})}^{-1}}=({{f}^{-1}})^{*}$. Then by Lemma
8.1.12, ${{(f^{*})}^{-1}}\circ f^{*}=({{f}^{-1}})^{*}\circ
f^{*}=({{f}^{-1}}\circ f)^{*}=\operatorname{Id}^{*}$. Analogously,
$f^{*}\circ{{(f^{*})}^{-1}}=\operatorname{Id}^{*}$. Composition of functions
is associative. Hence $\text{Hom}^{*}(X)$ constitutes a group with composition
of functions as the binary operation.

${{\left\langle\mathcal{P}(X)\right\rangle}_{T}}=P(X)$, and so
$\mathcal{P}(X)$ is a $T$-space. From Proposition 8.1.4, we can tell that
$\text{Hom}^{*}(X)$ as the set $\\{f^{*}\,|\,f\in\text{Hom}(X)\\}$ is
contained in $\operatorname{End}_{T}(\mathcal{P}(X))$ as a set (defined in
Definition 2.3.7). Moreover, by Definition 8.1.3 and the definition of
homeomorphisms, it is not hard to see that
$\forall\sigma\in\text{Hom}^{*}(X)$, $\sigma$ is bijective. So
$\text{Hom}^{*}(X)\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X))$ (as a set
defined in Definition 2.4.1).

By Proposition 2.4.3, $\operatorname{Aut}_{T}(\mathcal{P}(X))$ constitutes a
group with composition of functions as the binary operation. So
$\text{Hom}^{*}(X)$ constitutes a subgroup of
$\operatorname{Aut}_{T}(\mathcal{P}(X))$. ∎

Moreover, $\text{Hom}^{*}(X)$ is isomorphic to $\text{Hom}(X)$, as shown
below.

######  Proposition 8.1.16.

${{I}_{X}}{{|}_{\emph{Hom}(X)}}$, which is the restriction of ${{I}_{X}}$ to
$\emph{Hom}(X)$, is a group isomorphism from $\emph{Hom}(X)$ to
$\emph{Hom}^{*}(X)$.

###### Proof.

By Proposition 8.1.11, ${{I}_{X}}{{|}_{\text{Hom}(X)}}$ is an injective map
from $\text{Hom}(X)$ to
$\text{Hom}^{*}(X)(=\\{f^{*}\,|\,f\in\text{Hom}(X)\\})$. It follows that
${{I}_{X}}{{|}_{\text{Hom}(X)}}$ is a bijective map between the two groups.
And by Lemma 8.1.12,

${{I}_{X}}(f\circ g)={{I}_{X}}(f)\circ{{I}_{X}}(g),\forall
f,g\in\text{Hom}(X).$

Therefore, ${{I}_{X}}{{|}_{\text{Hom}(X)}}$ is a group isomorphism. ∎

###  8.2. “General” Galois correspondences for topological spaces

Firstly, we need to recall some notations. By Definitions 2.5.1, 2.6.1, 3.2.1
and 3.2.3, we immediately have

######  Proposition 8.2.1.

Let $B\subseteq\mathcal{P}(X)$ and let
$H\subseteq\operatorname{End}_{T}(\mathcal{P}(X))$. Then

$\mathcal{P}{{(X)}^{H}}=\\{A\in\mathcal{P}(X)\,|\,\sigma(A)=A,\forall\sigma\in
H\\},$
$\operatorname{GMn}_{T}(\mathcal{P}(X)/B)=\\{\sigma\in\operatorname{End}_{T}(\mathcal{P}(X))\,|\,\sigma(A)=A,\forall
A\in B\\},$
$\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)=\\{\mathcal{P}{{(X)}^{L}}\,|\,B\subseteq\mathcal{P}{{(X)}^{L}},L\subseteq\operatorname{End}_{T}(\mathcal{P}(X))\\},$

and

$\operatorname{GSMn}_{T}(\mathcal{P}(X)/B)=\\{\operatorname{GMn}_{T}(\mathcal{P}(X)/K)\,|\,B\subseteq
K\subseteq\mathcal{P}(X)\\}.$

######  Remark.

$\operatorname{End}_{T}(\mathcal{P}(X))$ is given in Proposition 8.1.2.

Then applying Theorem 3.2.5 to a topological space $X$, we immediately get the
following, which, roughly speaking, is about the Galois correspondence between
the Galois $T$-monoids of $\mathcal{P}(X)$ and the fixed subsets of
$\mathcal{P}(X)$ under the actions of the $T$-endomorphisms of
$\mathcal{P}(X)$.

######  Corollary 8.2.2.

Let $X$ be a topological space, let $\mathcal{P}(X)$ be the power set of $X$,
and let $B\subseteq\mathcal{P}(X)$. Then the correspondences

$\gamma:H\mapsto\mathcal{P}{{(X)}^{H}}$ and
$\delta:K\mapsto\operatorname{GMn}_{T}(\mathcal{P}(X)/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSMn}_{T}(\mathcal{P}(X)/B)$ and
$\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)$.

Moreover, if we are not satisfied with the way in which
$\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)$ (resp.
$\operatorname{GSMn}_{T}(\mathcal{P}(X)/B)$) is given in Proposition 8.2.1,
then we may employ Theorem 5.1.2 (resp. Theorem 5.3.2) to investigate whether
we can endow $\mathcal{P}(X)$ with a topology such that

$\operatorname{Int}_{T}^{\operatorname{End}}(\mathcal{P}(X)/B)\backslash\\{\emptyset\\}=${${S}^{\prime}\,|\,B\subseteq{S}^{\prime}\leq_{q}\mathcal{P}(X)$,
${S}^{\prime}$ is closed and nonempty}

(resp. whether we can endow End${}_{T}(\mathcal{P}(X))$ with a topology such
that $\operatorname{GSMn}_{T}(\mathcal{P}(X)/B)$ is the set of all closed
submonoids of $\operatorname{GMn}_{T}(\mathcal{P}(X)/B)$).

However, we will not go further for this subject.

Analogously, by Proposition 8.1.2 and Definition 2.4.1, we have

Aut${}_{T}(\mathcal{P}(X))=\\{$bijective
$\sigma:\mathcal{P}(X)\to\mathcal{P}(X)\,|\,\sigma(\overline{A})=\overline{\sigma(A)},\forall
A\in\mathcal{P}(X)\\}$.

And by Definitions 2.6.1, 3.2.1 and 3.2.3, we immediately have

######  Proposition 8.2.3.

Let $B\subseteq\mathcal{P}(X)$. Then

$\operatorname{GGr}_{T}(\mathcal{P}(X)/B)=\\{\sigma\in\operatorname{Aut}_{T}(\mathcal{P}(X))\,|\,\sigma(A)=A,\forall
A\in B\\},$
$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)=\\{\mathcal{P}{{(X)}^{H}}\,|\,B\subseteq\mathcal{P}{{(X)}^{H}},H\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X))\\},$

and

$\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)=\\{\operatorname{GGr}_{T}(\mathcal{P}(X)/K)\,|\,B\subseteq
K\subseteq\mathcal{P}(X)\\}.$

Then applying Theorem 3.2.6 to a topological space $X$, we immediately have
the following, which is about the Galois correspondence between the Galois
$T$-groups of $\mathcal{P}(X)$ and the fixed subsets of $\mathcal{P}(X)$ under
the actions of the $T$-automorphisms of $\mathcal{P}(X)$

######  Corollary 8.2.4.

Let $X$ be a topological space, let $\mathcal{P}(X)$ be the power set of $X$,
and let $B\subseteq\mathcal{P}(X)$. Then the correspondences

$\gamma:H\mapsto\mathcal{P}{{(X)}^{H}}$ and
$\delta:K\mapsto\operatorname{GGr}_{T}(\mathcal{P}(X)/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)$ and
$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)$.

Moreover, if we are not satisfied with the way in which
$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)$ (resp.
$\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)$) is given in Proposition 8.2.3,
then we may employ Theorem 5.2.2 (resp. Theorem 5.4.2) to investigate whether
we can endow $\mathcal{P}(X)$ with a topology such that

$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)\backslash\\{\emptyset\\}=${${S}^{\prime}\,|\,B\subseteq{S}^{\prime}\leq_{q}\mathcal{P}(X)$,
${S}^{\prime}$ is closed and nonempty}

(resp. whether we can define a topology on Aut${}_{T}(\mathcal{P}(X))$ such
that

$\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)=\\{$all closed subgroups of
$\operatorname{GGr}_{T}(\mathcal{P}(X)/B)\\}$).

However, we shall not delve into this subject.

###  8.3. Galois correspondence for homeomorphisms

This subsection has two parts.

In the first part, we shall obtain a Galois correspondence where
$\operatorname{Aut}_{T}(\mathcal{P}(X)))$ is restricted to
$\operatorname{Hom*}(X)$ (which by Proposition 8.1.15 is a subgroup of
Aut${}_{T}(\mathcal{P}(X))$). In other words, we shall obtain a Galois
correspondence where all Galois groups are contained in
$\operatorname{Hom*}(X)$ (which is the set of the induced maps of the
homeomorphisms from $X$ to $X$).

In the second part, we shall restate the Galois correspondence obtained in the
first part in terms of $\operatorname{Hom}(X)$ (instead of
$\operatorname{Hom*}(X)$). In other words, we shall obtain a Galois
correspondence where all Galois groups are contained in
$\operatorname{Hom}(X)$ (which is the set of all homeomorphisms from $X$ to
$X$).

For the purpose of the first part, we need to go through a process which is
similar to parts of Subsections 2.5, 2.6 and 3.2.

Firstly, we define the following notation to replace
$\operatorname{GGr}_{T}(\mathcal{P}(X)/K)$ (in Corollary 8.2.4).

######  Definition 8.3.1.

$\forall K\subseteq\mathcal{P}(X)$, let

$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)=\\{\sigma\in\operatorname{Hom*}(X)\,|\,\sigma(A)=A,\forall
A\in K\\}.$

And informally, we call $\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ a
_Galois group contained in_ Hom*$(X)$.

######  Proposition 8.3.2.

$\forall K\subseteq\mathcal{P}(X)$,
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ constitutes a subgroup of
$\operatorname{Hom*}(X)$ with composition of functions as the binary
operation.

###### Proof.

Firstly, Proposition 8.1.15 tells us that $\operatorname{Hom*}(X)$ constitutes
a group with composition of functions as the binary operation. Secondly, we
can tell from Definition 8.3.1 that

$\text{I}{{\text{d}}_{\mathcal{P}(X)}}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)\subseteq\operatorname{Hom*}(X)$

and $\forall\sigma\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$, the inverse
${{\sigma}^{-1}}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$. Thus it
suffices to show that $\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ is closed
under composition of functions.

Let
${{\sigma}_{1}},{{\sigma}_{2}}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$.
By Definition 8.3.1, $\forall A\in K$,
${{\sigma}_{1}}(A)=A={{\sigma}_{2}}(A)$, and so
${{\sigma}_{1}}({{\sigma}_{2}}(A))=A$. Hence
${{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$
(since ${{\sigma}_{1}}\circ{{\sigma}_{2}}\in\operatorname{Hom*}(X)$ by
Proposition 8.1.15), as desired. ∎

######  Proposition 8.3.3.

Let $H\subseteq\operatorname{Hom*}(X)$. Then

$\mathcal{P}{{(X)}^{H}}=\\{A\in\mathcal{P}(X)\,|\,\sigma(A)=A,\forall\sigma\in
H\\}.$

###### Proof.

Since
$H\subseteq\operatorname{Hom*}(X)\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X))\subseteq\operatorname{End}_{T}(\mathcal{P}(X))$,
by Proposition 8.2.1, we get the equation. ∎

The following four lemmas are obvious.

######  Lemma 8.3.4.

$\forall B\subseteq K\subseteq\mathcal{P}(X)$,

$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/B)\supseteq\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K).$

######  Lemma 8.3.5.

$\forall H\subseteq G\subseteq\operatorname{Hom*}(X)$,
$\mathcal{P}{{(X)}^{H}}\supseteq\mathcal{P}{{(X)}^{G}}$.

######  Lemma 8.3.6.

$\forall K\subseteq\mathcal{P}(X),$
$K\subseteq\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}$.

######  Lemma 8.3.7.

$\forall H\subseteq\operatorname{Hom*}(X)$,
$H\subseteq\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})$.

Moreover, to replace
$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)$ and
$\operatorname{GSGr}_{T}(\mathcal{P}(X)/B)$ (in Corollary 8.2.4), we define

######  Definition 8.3.8.

$\forall B\subseteq\mathcal{P}(X)$, let

$\text{Int}{{}_{\text{Hom}^{*}}}(\mathcal{P}(X)/B)=\\{\mathcal{P}{{(X)}^{H}}\,|\,B\subseteq\mathcal{P}{{(X)}^{H}},H\subseteq\operatorname{Hom*}(X)\\}$

and

$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)=\\{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)\,|\,B\subseteq
K\subseteq\mathcal{P}(X)\\}$.

######  Remark.

$\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)\subseteq\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)$
as a result of
$\operatorname{Hom*}(X)\subseteq\operatorname{Aut}_{T}(\mathcal{P}(X))$, where
$\operatorname{Int}_{T}^{\operatorname{Aut}}(\mathcal{P}(X)/B)$ is given in
Proposition 8.2.3.

The following characterizes $\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$.

######  Lemma 8.3.9.

Let $B\subseteq K\subseteq\mathcal{P}(X)$. Then

$K\in\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)\Leftrightarrow
K=\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}$.

###### Proof.

1\. _Sufficiency_

If $K=\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}$, then
by Definition 8.3.8, $K\in\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$.

2\. _Necessity_

Suppose that $K\in\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$. Then $\exists
H\subseteq\operatorname{Hom*}(X)$ such that $K=\mathcal{P}{{(X)}^{H}}$, and so
by Lemma 8.3.7,

$H\subseteq\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})=\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K).$

Hence by Lemma 8.3.5,

$K=\mathcal{P}{{(X)}^{H}}\supseteq\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}.$

On the other hand,
$K\subseteq\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}$
by Lemma 8.3.6.

Therefore,
$K=\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}$. ∎

The following lemma characterizes
$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$.

######  Lemma 8.3.10.

Let $H\subseteq\operatorname{Hom*}(X)$ and let
$B\subseteq\mathcal{P}{{(X)}^{H}}\subseteq\mathcal{P}(X)$. Then

$H\in\operatorname{GSGr}{{}_{\emph{Hom}^{*}}}(\mathcal{P}(X)/B)\Leftrightarrow\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})=H.$

###### Proof.

1\. _Sufficiency_

If $\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})=H$, then
$H\in\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ by Definition 8.3.8.

2\. _Necessity_

Suppose that $H\in\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$. Then $\exists
K$ such that $B\subseteq K\subseteq\mathcal{P}(X)$ and
$H=\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$, and so by Lemma 8.3.6,

$K\subseteq\mathcal{P}{{(X)}^{\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)}}=\mathcal{P}{{(X)}^{H}}.$

Hence by Lemma 8.3.4,

$H=\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)\supseteq\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}}).$

On the other hand,
$H\subseteq\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})$
by Lemma 8.3.7.

Therefore,
$H=\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/\mathcal{P}{{(X)}^{H}})$. ∎

Then we have the following theorem, which, roughly speaking, is about the
Galois correspondence between the Galois groups contained in
$\operatorname{Hom*}(X)$ and the fixed subsets of $\mathcal{P}(X)$ under the
actions of the induced maps of the homeomorphisms from $X$ to $X$.

######  Theorem 8.3.11.

Let $X$ be a topological space, let $\mathcal{P}(X)$ be the power set of $X$,
and let $B\subseteq\mathcal{P}(X)$. Then the correspondences

$\gamma:H\mapsto\mathcal{P}{{(X)}^{H}}$ and
$\delta:K\mapsto\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ and
$\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$.

###### Proof.

$\gamma$ is inclusion-reversing by Lemma 8.3.5, and $\delta$ is inclusion-
reversing by Lemma 8.3.4.

Lemmas 8.3.9 and 8.3.10 imply that $\gamma\circ\delta$ and $\delta\circ\gamma$
are identity functions, respectively. Hence both $\delta$ and $\gamma$ are
bijective maps and ${{\delta}^{-1}}=\gamma$. ∎

Moreover, Proposition 8.1.16 implies that we may restate Theorem 8.3.11 in
terms of $\operatorname{Hom}(X)$ (instead of $\operatorname{Hom*}(X)$).

For this purpose, we need to define some notions for $\operatorname{Hom}(X)$
(to replace notions for $\operatorname{Hom*}(X)$).

Firstly, to replace $H(\subseteq\operatorname{Hom*}(X))$ and
$\mathcal{P}{{(X)}^{H}}$ in Theorem 8.3.11, we have

######  Definition 8.3.12.

$\forall F\subseteq\operatorname{Hom}(X)$, we denote by
$\mathcal{P}{{(X)}^{F}}$ the set

$\\{A\in\mathcal{P}(X)\,|\,\forall f\in F$, $\overline{f(A)}=A$ if
$A=\overline{A}$ and $f(A)=A$ if $A\neq\overline{A}$},

which equals $\\{A\in\mathcal{P}(X)\,|\,\forall f\in F$, $f(A)=A$} because
$F\subseteq\operatorname{Hom}(X)$.

######  Remark.

Since $\mathcal{P}{{(X)}^{F}}=\\{A\in\mathcal{P}(X)\,|\,\forall f\in F$,
$f(A)=A$}, $\mathcal{P}{{(X)}^{F}}$ is the set of $A(\subseteq X)$ such that
for any homeomorphism $f\in F$ from $X$ to $X$, $A$ is fixed (not necessarily
pointwisely) under the action of $f$.

######  Lemma 8.3.13.

Let $F\subseteq\operatorname{Hom}(X)$ and let $F^{*}=\\{f^{*}\,|\,f\in F\\}$.
Then $\mathcal{P}{{(X)}^{F}}=\mathcal{P}{{(X)}^{F^{*}}}$.

###### Proof.

Let $A\in\mathcal{P}(X)$. By Definition 8.1.3, $\forall f\in F$,

$f^{*}(A)=A\,\Leftrightarrow\,\overline{f(A)}=A$ if $A=\overline{A}$ and
$f(A)=A$ if $A\neq\overline{A}.$

By Proposition 8.2.1,
$\mathcal{P}{{(X)}^{F^{*}}}=\\{A\in\mathcal{P}(X)\,|\,f^{*}(A)=A,\,\forall
f^{*}\in F^{*}\\}$. Therefore, $A\in\mathcal{P}(X)^{F^{*}}$ if and only if
$A\in\mathcal{P}(X)^{F}$. ∎

Secondly, to replace $\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$ in Theorem
8.3.11, we have

######  Definition 8.3.14.

$\forall B\subseteq\mathcal{P}(X)$, let

$\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)=\\{\mathcal{P}{{(X)}^{F}}\,|\,B\subseteq\mathcal{P}{{(X)}^{F}},F\subseteq\operatorname{Hom}(X)\\}.$

Then by Definition 8.3.8 and Lemma 8.3.13, we immediately have

######  Lemma 8.3.15.

$\forall B\subseteq\mathcal{P}(X)$,
$\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)=\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$.

Thirdly, to replace $\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ in Theorem
8.3.11, we have

######  Definition 8.3.16.

$\forall K\subseteq\mathcal{P}(X)$, we denote by
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$ the set

$\\{f\in\operatorname{Hom}(X)\,|\,\forall A\in K$, $\overline{f(A)}=A$ if
$A=\overline{A}$ and $f(A)=A$ if $A\neq\overline{A}$},

which obviously is $\\{f\in\operatorname{Hom}(X)\,|\,\forall A\in K$,
$f(A)=A$}. And informally, we call
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$ a _Galois group contained in_
Hom$(X)$.

######  Remark.

Since
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)=\\{f\in\operatorname{Hom}(X)\,|\,\forall
A\in K$, $f(A)=A\\}$, $\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$ is the set
of homeomorphisms $f$ from $X$ to $X$ such that $\forall A\in K$, $A(\subseteq
X)$ is fixed (not necessarily pointwisely) under the action of $f$.

Recall that ${{I}_{X}}{{|}_{\operatorname{Hom}(X)}}$ is a group isomorphism
from $\operatorname{Hom}(X)$ to $\operatorname{Hom*}(X)$ (Proposition 8.1.16).
Now we have

######  Lemma 8.3.17.

Let $K\subseteq\mathcal{P}(X)$ and let
$G=\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$. Then $I_{X}|_{G}$, which is
the restriction of $I_{X}$ to $G$, is a group isomorphism from
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$ to
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$.

######  Remark.

So by Corollary 8.1.14, $\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$
constitutes a subgroup of $\operatorname{Hom}(X)$ with composition of
functions as the binary operation.

###### Proof.

Since $G\subseteq\operatorname{Hom}(X)$,
${{I}_{X}}{{|}_{\operatorname{Hom}(X)}}$ is a group isomorphism from
$\operatorname{Hom}(X)$ to $\operatorname{Hom*}(X)$ (Proposition 8.1.16), and
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ constitutes a subgroup of
$\operatorname{Hom*}(X)$ (Proposition 8.3.2), it suffices to show that
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ is the image of $I_{X}|_{G}$.

Let $f\in\operatorname{Hom}(X)$ and $A\in K$. Then by Definition 8.1.3,

$f^{*}(A)=A\,\Leftrightarrow\,\overline{f(A)}=A$ if $A=\overline{A}$ and
$f(A)=A$ if $A\neq\overline{A}.$

So by Definitions 8.3.1 and 8.3.16, we can tell that

$f^{*}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)\Leftrightarrow
f\in\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)(=G).$

It follows that $\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ is the image of
$I_{X}|_{G}$, as desired. ∎

Lastly, to replace $\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ in Theorem
8.3.11, we give

######  Definition 8.3.18.

$\forall B\subseteq\mathcal{P}(X)$, let

$\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)=\\{\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)\,|\,B\subseteq
K\subseteq\mathcal{P}(X)\\}.$

######  Lemma 8.3.19.

$\forall B\subseteq\mathcal{P}(X)$, the map

$\beta:\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)\to\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$

given by

$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)\mapsto\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$,
$\forall B\subseteq K\subseteq\mathcal{P}(X)$,

is bijective.

###### Proof.

Comparing the definition of $\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$ in
Definition 8.3.18 with that of $\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$
in Definition 8.3.8, we can tell that $\beta$ is a well-defined surjective
map. So we only need to show that $\beta$ is injective.

Let $G_{1},G_{2}\in\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$ with
$\beta(G_{1})=\beta(G_{2})=:G$.

As was shown in the proof of Lemma 8.3.17, $\forall K\subseteq\mathcal{P}(X)$
and $f\in\operatorname{Hom}(X)$,
$f^{*}\in\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)\Leftrightarrow
f\in\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$.

Hence $\forall f\in\operatorname{Hom}(X)$, $f\in G_{1}\Leftrightarrow f^{*}\in
G\Leftrightarrow f\in G_{2}$, and so $G_{1}=G_{2}$, as desired. ∎

Now we can obtain a main theorem of this section, which, roughly speaking, is
about the Galois correspondence between the Galois groups contained in
$\operatorname{Hom}(X)$ and the fixed subsets of $\mathcal{P}(X)$ under the
actions of the homeomorphisms from $X$ to $X$, where a fixed element of
$\mathcal{P}(X)$ is actually a fixed (not necessarily pointwisely) subset of
$X$.

######  Theorem 8.3.20.

Let $X$ be a topological space, let $\mathcal{P}(X)$ be the power set of $X$,
and let $B\subseteq\mathcal{P}(X)$. Then the correspondences

$\gamma:F\mapsto\mathcal{P}{{(X)}^{F}}$ and
$\delta:K\mapsto\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$ and
$\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)$.

###### Proof.

By Lemmas 8.3.13, 8.3.15, 8.3.17 and 8.3.19, respectively, each of
$\mathcal{P}{{(X)}^{F^{*}}}$, $\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$,
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ and
$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ has its equivalent counterpart
which is induced by the inverse of ${{I}_{X}}{{|}_{\operatorname{Hom}(X)}}$,
i.e., $\mathcal{P}{{(X)}^{F}}$, $\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)$,
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$, and
$\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$, respectively.

Roughly speaking, the theorem holds because it can be obtained by simply
replacing $H(\subseteq\operatorname{Hom}^{*}(X))$,
$\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$,
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$, and
$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ in Theorem 8.3.11 by their
equivalent counterparts $F(\subseteq\operatorname{Hom}(X))$,
$\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)$,
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$, and
$\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$, respectively. We explain in
detail as follows.

For the following arguments, consider the case where all the four replacements
are made simultaneously.

Firstly, Lemma 8.3.13 justifies the replacement of $H(=F^{*})$ and
$\mathcal{P}{{(X)}^{H}}(=\mathcal{P}{{(X)}^{F^{*}}})$ in Theorem 8.3.11 by $F$
and $\mathcal{P}{{(X)}^{F}}$, respectively.

With the replacement of $\mathcal{P}{{(X)}^{F^{*}}}$ by
$\mathcal{P}{{(X)}^{F}}$, $\operatorname{Int_{Hom*}}(\mathcal{P}(X)/B)$ is
replaced by $\operatorname{Int_{Hom}}(\mathcal{P}(X)/B)$. This replacement is
justified by Lemma 8.3.15.

One more replacement is the replacement of
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$, which is
$\\{f^{*}\,|\,f\in\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)\\}$, by
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$. By Lemma 8.3.17, the restriction
of the inverse of ${{I}_{X}}{{|}_{\operatorname{Hom}(X)}}$ to
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ is a group isomorphism from
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ to
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$, and so the isomorphism justifies
the replacement.

Finally, as a result of the replacement of
$\operatorname{GGr_{Hom*}}(\mathcal{P}(X)/K)$ by
$\operatorname{GGr_{Hom}}(\mathcal{P}(X)/K)$,
$\operatorname{GSGr_{Hom*}}(\mathcal{P}(X)/B)$ is replaced by
$\operatorname{GSGr_{Hom}}(\mathcal{P}(X)/B)$. This replacement is justified
by Lemma 8.3.19.

The inverse of ${{I}_{X}}{{|}_{\operatorname{Hom}(X)}}$ induces only the above
four replacements involved. Thus we conclude the proof. ∎

##  9\. Applications to dynamical systems

In this section, we first show that results in Sections 2 to 7 apply to
dynamical systems, then we obtain the Galois correspondences for dynamical
systems.

###  9.1. Basic notions and properties

In the literature, there are various definitions of dynamical systems. We take
the following as in [12].

######  Definition 9.1.1.

Let $M$ be a monoid with identity element $e$ acting on a set $S$. That is,
there is a map $\Phi:M\times S\to S$ given by $(g,x)\mapsto{{\Phi}_{g}}(x)$
such that ${{\Phi}_{g}}\circ{{\Phi}_{h}}={{\Phi}_{g\circ h}}$ and
${{\Phi}_{e}}=\operatorname{Id}$, where ${{\Phi}_{f}},\forall f\in M$, is a
map from $S$ to $S$. Then the triple $(M,S,\Phi)$ is called a _dynamical
system_.

######  Theorem 9.1.2.

Let $(M,S,\Phi)$ be a dynamical system and let $T=\\{{{\Phi}_{g}}\,|\,g\in
M\\}$. Then $T$ is an operator semigroup on $S$, $\operatorname{Id}\in T$ and
$S$ is a $T$-space.

###### Proof.

By Definition 9.1.1, $\forall{{\Phi}_{g}}\in T$, ${{\Phi}_{g}}$ is a map from
$S$ to $S$. And $\forall{{\Phi}_{g}},{{\Phi}_{h}}\in T$,
${{\Phi}_{g}}\circ{{\Phi}_{h}}={{\Phi}_{g\circ h}}\in T$. So by Definition
2.1.1, $T$ is an operator semigroup on $S$. Then ${{\left\langle
S\right\rangle}_{T}}\subseteq S$. Moreover, ${{\left\langle
S\right\rangle}_{T}}\supseteq S$ since ${{\Phi}_{e}}=\operatorname{Id}\in T$.
So ${{\left\langle S\right\rangle}_{T}}=S$ is a $T$-space. ∎

Theorem 9.1.2 implies that we can apply results in Sections 2 to 7 to
dynamical systems.

Conversely, for any operator semigroup $T$ with $\operatorname{Id}\in T$, each
$T$-space induces a dynamical system, as shown below.

######  Theorem 9.1.3.

Let $T$ be an operator semigroup on a set $D$ with $\operatorname{Id}\in T$,
let $S$ be a $T$-space, and let a map $\Phi:T\times S\to S$ be given by
$(f,x)\mapsto f(x)$. Then $\Phi$ is well-defined and $(T,S,\Phi)$ is a
dynamical system.

###### Proof.

By Proposition 2.2.7, ${{\left\langle S\right\rangle}_{T}}\subseteq S$, and so
$\Phi$ is well-defined. $\forall f\in T$, let ${{\Phi}_{f}}=f{{|}_{S}}$. Then
$\Phi$ can be equivalently given by $(f,x)\mapsto{{\Phi}_{f}}(x)$. Hence by
Definition 9.1.1, we can tell that $(T,S,\Phi)$ is a dynamical system. ∎

Theorem 9.1.3 implies that we can apply theories of dynamical systems to
$T$-spaces with $\operatorname{Id}\in T$. However, this subject exceeds the
scope of this paper.

###  9.2. Galois correspondences for dynamical systems

In this subsection, unless otherwise specified, we use the following notation.

######  Notation 9.2.1.

Let $M$, $S$, $\Phi$ and $(M,S,\Phi)$ be defined as in Definition 9.1.1. Let
$T=\\{{{\Phi}_{g}}\,|\,g\in M\\}$.

######  Remark.

By Theorem 9.1.2, $T$ is an operator semigroup on $S$, $\operatorname{Id}\in
T$ and $S$ is a $T$-space.

With Notation 9.2.1 and Theorem 9.1.2, we may apply Sections 2 to 7 to a
dynamical system $(M,S,\Phi)$ quite straightforwardly. We take Galois
correspondences as our example.

Firstly, we need to recall some notations.

By Definitions 2.3.1 and 2.3.7, we immediately have

######  Proposition 9.2.2.

$\operatorname{End}_{T}(S)=\\{\sigma:S\to
S\,|\,\sigma(f(x))=f(\sigma(x)),\forall x\in S\,and\,f\in T\\}$.

To get some more idea about $\operatorname{End}_{T}(S)$, we take the
prototypical example of a discrete dynamical system as our example:

######  Proposition 9.2.3.

Let $({{\mathbb{N}}_{0}},I,\Phi)$ be an iterated map $($as a dynamical
system$)$ on interval $I\subseteq\mathbb{R}$, where
${{\Phi}_{0}}=\operatorname{Id}$, ${{\Phi}_{n}}$ is the composite of $n$
copies of $h$, $\forall n\in{{\mathbb{Z}}^{+}}$, and $h$ is a map from $I$ to
itself. Let $T=\\{{{\Phi}_{n}}\,|\,n\in{{\mathbb{N}}_{0}}\\}$. Then
$T\subseteq\operatorname{End}_{T}(I)$.

###### Proof.

By Theorem 9.1.2, $T$ is an operator semigroup on $I$, $\operatorname{Id}\in
T$ and $I$ is a $T$-space. Since $T=\\{\operatorname{Id}\\}\bigcup\left\langle
h\right\rangle$, $\sigma(f(x))=f(\sigma(x))$, $\forall x\in I$ and
$\sigma,f\in T$. Hence by Proposition 9.2.2,
$T\subseteq\operatorname{End}_{T}(I)$. ∎

By Definitions 2.5.1, 2.6.1, 3.2.1 and 3.2.3, we immediately have

######  Proposition 9.2.4.

Let $B\subseteq S$ and let $H\subseteq\operatorname{End}_{T}(S)$. Then

${{S}^{H}}=\\{x\in S\,|\,\sigma(x)=x,\forall\sigma\in H\\},$
$\operatorname{GMn}_{T}(S/B)=\\{\sigma\in\operatorname{End}_{T}(S)\,|\,\sigma(x)=x,\forall
x\in B\\},$
$\operatorname{Int}_{T}^{\operatorname{End}}(S/B)=\\{{{S}^{L}}\,|\,B\subseteq{{S}^{L}},L\subseteq\operatorname{End}_{T}(S)\\},$

and

$\operatorname{GSMn}_{T}(S/B)=\\{\operatorname{GMn}_{T}(S/K)\,|\,B\subseteq
K\subseteq S\\}.$

Then the following theorem is a straightforward result by applying Theorem
3.2.5 to a dynamical system.

######  Theorem 9.2.5.

Let $(M,S,\Phi)$ be a dynamical system, let $T=\\{{{\Phi}_{g}}\,|\,g\in M\\}$,
and let $B\subseteq S$. Then the correspondences

$\gamma:H\mapsto{{S}^{H}}\,and\,\delta:K\mapsto\operatorname{GMn}_{T}(S/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSMn}_{T}(S/B)$ and
$\operatorname{Int}_{T}^{\operatorname{End}}(S/B)$.

If we are not satisfied with the way in which
$\operatorname{Int}_{T}^{\operatorname{End}}(S/B)$ (resp.
$\operatorname{GSMn}_{T}(S/B)$) is given in Proposition 9.2.4, then we may
apply Theorem 5.1.2 (resp. Theorem 5.3.2) to investigate whether we can endow
$S$ with a topology such that

$\operatorname{Int}_{T}^{\operatorname{End}}(S/B)\backslash\\{\emptyset\\}=\\{{S}^{\prime}\,|\,B\subseteq{S}^{\prime}\leq_{q}S$,
${S}^{\prime}$ is closed and nonempty}

(resp. whether we can endow $\operatorname{End}_{T}(S)$ with a topology such
that $\operatorname{GSMn}_{T}(S/B)$ is the set of all closed submonoids of
$\operatorname{GMn}_{T}(S/B)$).

Analogously, by Definitions 2.3.1 and 2.4.1, we have

######  Proposition 9.2.6.

$\operatorname{Aut}_{T}(S)=\\{$bijective $\sigma:S\to
S\,|\,\sigma(f(x))=f(\sigma(x)),\forall x\in S\,and\,f\in T\\}$.

Then by Definitions 2.6.1, 3.2.1 and 3.2.3, we immediately have

######  Proposition 9.2.7.

Let $B\subseteq S$. Then

$\operatorname{GGr}_{T}(S/B)=\\{\sigma\in\operatorname{Aut}_{T}(S)\,|\,\sigma(x)=x,\forall
x\in B\\},$
$\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)=\\{{{S}^{H}}\,|\,B\subseteq{{S}^{H}},H\subseteq\operatorname{Aut}_{T}(S)\\},$

and

$\operatorname{GSGr}_{T}(S/B)=\\{\operatorname{GGr}_{T}(S/K)\,|\,B\subseteq
K\subseteq S\\}.$

Then applying Theorem 3.2.6 to a dynamical system, we immediately get

######  Theorem 9.2.8.

Let $(M,S,\Phi)$ be a dynamical system, let $T=\\{{{\Phi}_{g}}\,|\,g\in M\\}$,
and let $B\subseteq S$. Then the correspondences

$\gamma:H\mapsto{{S}^{H}}\,and\,\delta:K\mapsto\operatorname{GGr}_{T}(S/K)$

define inclusion-inverting mutually inverse bijective maps between
$\operatorname{GSGr}_{T}(S/B)$ and
$\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)$.

If we are not satisfied with the way in which
$\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)$ (resp.
$\operatorname{GSGr}_{T}(S/B)$) is given in Proposition 9.2.7, then we may
employ Theorem 5.2.2 (resp. Theorem 5.4.2) to investigate whether we can endow
$S$ with a topology such that

$\operatorname{Int}_{T}^{\operatorname{Aut}}(S/B)\backslash\\{\emptyset\\}=${${S}^{\prime}\,|\,B\subseteq{S}^{\prime}\leq_{q}S$,
${S}^{\prime}$ is closed and nonempty}

(resp. whether we can endow Aut${}_{T}(S)$ with a topology such that
$\operatorname{GSGr}_{T}(S/B)$ is the set of all closed subgroups of
$\operatorname{GGr}_{T}(S/B)$).

By Theorem 9.1.2, we may also apply other results in Sections 2 to 7 to
dynamical systems, but we will not go into it.

Part II. THEORY FOR THE CASE OF (PARTIAL) FUNCTIONS OF MORE THAN ONE VARIABLE

##  10\. Basic notions for the case of multivariable functions

Recall that, by Definition 2.1.1, each element of an operator semigroup has
only one variable. To make our theory more general, we shall generalize the
notion of operator semigroups to incorporate functions of more than one
variable. Then we shall find that some more familiar concepts in mathematics
such as ring homomorphisms, module homomorphisms, and group homomorphisms can
be characterized by $T$-morphisms or $\theta$-morphisms.

Subsection 10.1 generalizes the notion of operator semigroups. Accordingly,
Subsections 10.2, 10.3, 10.4, 10.5 and 10.6 generalize the notions of
$T$-spaces, $T$-morphisms, $T$-isomorphisms, $\theta$-morphisms and
$\theta$-isomorphisms, respectively.

###  10.1. Operator generalized-semigroup $T$

######  Definition 10.1.1.

Let $D$ be a set, let $n\in{{\mathbb{Z}}^{+}}$, and let $f$ be a function from
the cartesian product ${{D}^{n}}$ to $D$. $\forall i=1,\cdots,n$, let
$n_{i}\in{{\mathbb{Z}}^{+}}$ and let ${{g}_{i}}$ be a function from the
cartesian product ${{D}^{{{n}_{i}}}}$ to $D$. Let
$m=\sum\nolimits_{i}{{{n}_{i}}}$. Then the _composite_
$f\circ({{g}_{1}},\cdots,{{g}_{n}})$ is the function from ${{D}^{m}}$ to $D$
given by

$({{x}_{1}},\cdots,{{x}_{m}})\mapsto
f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}})),$

where
${{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{D}^{{{n}_{i}}}},\forall
i=1,\cdots,n$, ${{m}_{1}}:=0$ and
${{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n$.

######  Remark.

If $n=1$, we can tell that the $f\circ g_{1}$ defined above is the normal
composite of functions $f$ and $g_{1}$.

Now Definition 2.1.1 is generalized to

######  Definition 10.1.2.

Let $D$ be a set and let $T$ be a subset of

{all functions from the cartesian product ${{D}^{n}}$ to
$D\,|\,n\in{{\mathbb{Z}}^{+}}$}.

If $\forall n\in{{\mathbb{Z}}^{+}}$, $f\in T$ of $n$ variables, and
$({{g}_{1}},\cdots,{{g}_{n}})\in{{T}^{n}}$ (cartesian product), the composite
$f\circ({{g}_{1}},\cdots,{{g}_{n}})\in T$, then we call $T$ an _operator
generalized-semigroup_ on $D$ and $D$ is called the _domain_ of $T$.

######  Remark.

  1. (1)

The terminology “generalized-semigroup” is informal and just for convenience
because we have not defined it.

  2. (2)

Obviously, any operator semigroup is an operator generalized-semigroup.

The following example is a generalization of Example 2.1.2.

######  Example 10.1.3.

Let $B$ be a subfield of a field $F$. Let $T$ be the set

$\\{f^{*}:{{F}^{n}}\to F$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in
B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\\}$,

where $B[{{x}_{1}},\cdots,{{x}_{n}}]$ is the ring of polynomials over $B$ in
$n$ variables.

Then by Definition 10.1.1, $\forall
f^{*},{{g}_{1}}^{*},\cdots,{{g}_{n}}^{*}\in T$ such that $f\in
B[{{x}_{1}},\cdots,{{x}_{n}}]$ and ${{g}_{i}}\in
B[{{x}_{1}},\cdots,{{x}_{{{n}_{i}}}}]$, $\forall i=1,\cdots,n$, the composite
$f^{*}\circ({{g}_{1}}^{*},\cdots,{{g}_{n}}^{*})$ can be given by

$({{x}_{1}},\cdots,{{x}_{m}})\mapsto
f({{g}_{1}}({{\text{v}}_{1}}),\cdots,{{g}_{n}}({{\text{v}}_{n}}))$,

where $m:=\sum\nolimits_{i}{{{n}_{i}}}$,
$({{x}_{1}},\cdots,{{x}_{m}})\in{{F}^{m}}$,
${{\text{v}}_{i}}:=({{x}_{{{m}_{i}}+1}},\cdots,{{x}_{{{m}_{i}}+{{n}_{i}}}})\in{{F}^{{{n}_{i}}}},\forall
i=1,\cdots,n$, ${{m}_{1}}:=0$ and
${{m}_{i}}:=\sum\nolimits_{k=1}^{i-1}{{{n}_{k}}},\forall i=2,\cdots,n$.

For example, let $f={{g}_{1}}={{g}_{2}}={{x}_{1}}{{x}_{2}}$. Then
$f^{*}\circ({{g}_{1}}^{*},{{g}_{2}}^{*})$ can be induced by (with a slight
abuse of notation)
$f({{g}_{1}},{{g}_{2}})=({{x}_{1}}{{x}_{2}})({{x}_{3}}{{x}_{4}})={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}$.
That is, $f^{*}\circ({{g}_{1}}^{*},{{g}_{2}}^{*}):{{F}^{4}}\to F$ is given by
$({{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}})\mapsto{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}},\forall({{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}})\in{{F}^{4}}$.

Then by Definition 10.1.2, we can tell that $T$ is an operator generalized-
semigroup on $F$.

Note that the map which induces $T$, i.e.,

$\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]}\to
T$

given by $f\mapsto f^{*}$ , is surjective, but it is not necessarily
injective.

######  Example 10.1.4.

Let $R$ be a ring with identity and let
${{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle$ be the
ring of polynomials over the prime field of order 2 in $n$ noncommuting
variables. Let

$T=\\{f^{*}:{{R}^{n}}\to R$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\\}$.

Then we can tell that $T$ is an operator generalized-semigroup on $R$.

The map which induces $T$, i.e.,

$\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle}\to
T$

is given by $f\mapsto f^{*}$.

######  Example 10.1.5.

Let $M$ be a module over a commutative ring $R$ with identity. Let

$T=\\{f^{*}:{{M}^{n}}\to M$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\},n\in{{\mathbb{Z}}^{+}}\\}$,

(where $\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\in
R[{{x}_{1}},\cdots,{{x}_{n}}]$). Then $T$ is an operator generalized-semigroup
on $M$.

Let
$\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\}}\to T$ be the map inducing $T$ given by $f\mapsto f^{*}$.

######  Example 10.1.6.

Let $G$ be an abelian group. Let

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z},\forall
i=1,\cdots,n\\}$},

(where
$\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\in\operatorname{Frac}({{\mathbb{F}}_{2}}[{{x}_{1}},\cdots,{{x}_{n}}])$).
Then $T$ is an operator generalized-semigroup on $G$.

Let
$\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}}\to
T$ be the map inducing $T$ given by $f\mapsto f^{*}$.

For not necessarily abelian groups, we have

######  Example 10.1.7.

Let $G$ be a group. Recall that a word is a finite string of symbols, where
repetition is allowed. Let
$\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle$
denote the set of all words on
$\\{{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\\}$. Let

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\\}$.

Then we can tell that $T$ is an operator generalized-semigroup on $G$.

Let
$\tau:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle}\to
T$ be the map inducing $T$ given by $f\mapsto f^{*}$.

###  10.2. $T$-spaces

######  Definition 10.2.1.

Let $T$ be an operator generalized-semigroup on $D$ and let $S\subseteq D$. If
$\exists U\subseteq D$ such that

$S=\\{f(u_{1},\cdots,u_{n})\,|\,n\in{{\mathbb{Z}}^{+}}$, $f\in T$ has $n$
variables, and $u_{1},\cdots,u_{n}\in U\\}$,

then we call $S$ a _$T$ -space_ generated by $U$, and $S$ may be denoted by
${{\left\langle U\right\rangle}_{T}}$.

######  Example 10.2.2.

Let $B$ be a subfield of a field $F$. Let $T$ be the operator generalized-
semigroup on $F$ defined in Example 10.1.3. Then
$\forall(\emptyset\neq)U\subseteq F$, ${{\left\langle U\right\rangle}_{T}}$ is
the ring $B[U]$.

######  Example 10.2.3.

Let $R$ be a ring with identity. Let $T$ be the operator generalized-semigroup
defined in Example 10.1.4. Then $\forall(\emptyset\neq)U\subseteq R$,
${{\left\langle U\right\rangle}_{T}}$ is a subring of $R$. Specifically,
${{\left\langle R\right\rangle}_{T}}=R$ because $\operatorname{Id}\in T$.

######  Example 10.2.4.

Let $M$ be a module over a commutative ring $R$ with identity. Let $T$ be the
operator generalized-semigroup defined in Example 10.1.5. If $X$ is a nonempty
subset of $M$, then ${{\left\langle X\right\rangle}_{T}}$ is the submodule of
$M$ generated by $X$.

######  Example 10.2.5.

Let $G$ be an abelian group. Let $T$ be the operator generalized-semigroup
defined in Example 10.1.6. If $X$ is a nonempty subset of $G$, then
${{\left\langle X\right\rangle}_{T}}$ is the subgroup of $G$ generated by $X$.

######  Example 10.2.6.

Let $G$ be a group. Let $T$ be the operator generalized-semigroup defined in
Example 10.1.7. If $X$ is a nonempty subset of $G$, then ${{\left\langle
X\right\rangle}_{T}}$ is the subgroup of $G$ generated by $X$.

###  10.3. $T$-morphisms

######  Definition 10.3.1.

Let $T$ be an operator generalized-semigroup and let $\sigma$ be a map from a
$T$-space $S$ to a $T$-space. If $\forall n\in{{\mathbb{Z}}^{+}}$,
$(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and $f\in T$ of $n$ variables,

$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})),$

then we call $\sigma$ a _$T$ -morphism_.

######  Remark.

  1. (1)

By Definitions 10.1.2 and 10.2.1 (or by Proposition 12.1.1, which is a
generalization of Proposition 2.2.7), we can tell that
$f(a_{1},\cdots,a_{n})\in S$. So $\sigma(f(a_{1},\cdots,a_{n}))$ is well-
defined.

  2. (2)

Definition 2.3.7 still applies.

Note that in Examples 10.1.3 to 10.1.7, all operator generalized-semigroups
are induced from polynomials, rational functions, or words on a set. So we
introduce an (informal) notion as follows, which will facilitate our study on
$T$-morphisms.

######  Definition 10.3.2.

Let $f$ be a mathematical expression associated with a rule $R$. Let
$n\in\mathbb{Z}$ and let $C$ and $D$ be sets.

If each of $x_{1},\cdots,x_{n}$ arises as a symbol in $f$, and by the rule
$R$, $f$ yields a single element, which we denote by $f(a_{1},\cdots,a_{n})$,
in $C$ whenever every $x_{1},\cdots,x_{n}$ in $f$ is replaced by any
$a_{1},\cdots,a_{n}\in D$, respectively, then we call the couple $(f,R)$ a
(C-valued) _formal function_ of $n$ variables $x_{1},\cdots,x_{n}$ and $D$ is
called a _domain_ of $(f,R)$.

Moreover, we may call $f$ a _formal function_ if its associated rule $R$ is
clear from the context.

######  Remark.

  1. (1)

For example, $1+0{{x}_{1}}+0{{x}_{2}}\in\mathbb{Q}[{{x}_{1}},{{x}_{2}}]$ is a
formal function of two variables with the associated rule being the addition
and multiplication of field elements. In Examples 10.1.3 to 10.1.7, all
operator generalized-semigroups are induced from formal functions.

  2. (2)

By the definition, a formal function $(f,R)$ may have more than one domain.
For example, any field which contains $\mathbb{Q}$ is a domain of any formal
function $f\in\mathbb{Q}[{{x}_{1}},\cdots,{{x}_{n}}]$.

  3. (3)

Any function $f$ of $n$ variables $x_{1},\cdots,x_{n}$ may be regarded as the
formal function $f(x_{1},\cdots,x_{n})$, which is regarded as a mathematical
expression, with its associated rule being the mapping defined by the function
$f$. The converse is not true, however, because a formal function may have
more than one domain, and different domains of a formal function may induce
different functions. Therefore, formal functions may be regarded as a
generalization of functions.

When operator generalized-semigroups are induced from formal functions, the
following lemma will be useful.

######  Lemma 10.3.3.

Let $\mathcal{F}$ be a set of D-valued formal functions on a domain $D$ and
let $T$ be an operator generalized-semigroup on $D$ induced by $\mathcal{F}$;
that is, there is a surjective map $\tau:\mathcal{F}\to T$ being given by
$f\mapsto f^{*}$, where $f^{*}:{{D}^{n}}\to D$ is given by
$(a_{1},\cdots,a_{n})\mapsto f(a_{1},\cdots,a_{n})$ for $f$ of $n$ variables,
$\forall n\in{{\mathbb{Z}}^{+}}$. Let $\sigma$ be a map from a $T$-space $S$
to a $T$-space. Then the following are equivalent:

  1. (i)

$\sigma$ is a $T$-morphism.

  2. (ii)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and
$f\in\mathcal{F}$ of $n$ variables,

(10.1)
$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

###### Proof.

(i)$\Rightarrow$(ii): By Definition 10.3.1, $\forall n\in{{\mathbb{Z}}^{+}}$,
$(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and $f\in\mathcal{F}$ of $n$ variables,

$\sigma(\tau(f)(a_{1},\cdots,a_{n}))=\tau(f)(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

Then by the definition of $\tau$, (ii) must be true.

(ii)$\Rightarrow$(i): Since $\tau$ is surjective, $\forall f^{*}\in T$,
$\exists f\in\mathcal{F}$ such that $f^{*}=\tau(f)$. Then by statement (ii)
and the definition of $\tau$, we can tell that $\forall
n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and $f^{*}\in T$
of $n$ variables,
$\sigma(f^{*}(a_{1},\cdots,a_{n}))=f^{*}(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))$.
So by Definition 10.3.1, $\sigma$ must be a $T$-morphism. ∎

Lemma 10.3.3 characterizes a $T$-morphism by the formal functions which induce
$T$. When an operator generalized-semigroup is induced from formal functions,
the criterion for a $T$-morphism by Lemma 10.3.3 is normally more convenient
than that by Definition 10.3.1. Thus Lemma 10.3.3 simplifies our proofs of
some of our results as follows.

The following four theorems serve as examples of $T$-morphisms.

Firstly, Theorem 2.3.2 is generalized to

######  Theorem 10.3.4.

Let $B$ be a subfield of a field $F$, let $T$ be the operator generalized-
semigroup on $F$ defined in Example 10.1.3, and let nonempty $U,V\subseteq F$.
Then a map $\sigma:{{\left\langle U\right\rangle}_{T}}\to{{\left\langle
V\right\rangle}_{T}}$ is a ring homomorphism with $B$ fixed pointwisely if and
only if it is a $T$-morphism.

###### Proof.

We may give a proof which is analogous to that of Theorem 2.3.2. However, now
we employ Lemma 10.3.3 instead. By Example 10.1.3,

$T=\\{f^{*}:{{F}^{n}}\to F$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in
B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\\}$.

Thus ${{\left\langle U\right\rangle}_{T}}$ and ${{\left\langle
V\right\rangle}_{T}}$ are the rings $B[U]$ and $B[V]$, respectively. And Lemma
10.3.3 applies. Hence we only need to show that $\sigma:{{\left\langle
U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{T}}$ is a ring
homomorphism with $B$ fixed pointwisely if and only if statement (ii) in Lemma
10.3.3 is true in this case. Specifically, it suffices to show that the
following are equivalent:

  1. (1)

$\forall c\in B$, $\sigma(c)=c$, and $\forall a,b\in{{\left\langle
U\right\rangle}_{T}}(=B[U])$, $\sigma(a+b)=\sigma(a)+\sigma(b)$ and
$\sigma(ab)=\sigma(a)\sigma(b)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
U\right\rangle_{T}^{n}$ and $f\in B[{{x}_{1}},\cdots,{{x}_{n}}]$,

(10.2)
$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

(1)$\Rightarrow$(2): Since $f\in B[{{x}_{1}},\cdots,{{x}_{n}}]$ is a
polynomial, it is not hard to see that (2) is true.

(2)$\Rightarrow$(1): Considering the case where $f\in
B[{{x}_{1}},\cdots,{{x}_{n}}]$ is a constant polynomial, we can tell from (2)
that $\forall c\in B$, $\sigma(c)=c$.

Let $a,b\in{{\left\langle U\right\rangle}_{T}}$. Let $g={{x}_{1}}+{{x}_{2}}$
and $h={{x}_{1}}{{x}_{2}}$. Then

$\sigma(a+b)$

$=\sigma(g(a,b))$

$=g(\sigma(a),\sigma(b))$ (by Equation $($10.2$)$ in (2))

$=\sigma(a)+\sigma(b)$.

And

$\sigma(ab)=\sigma(h(a,b))$

$=h(\sigma(a),\sigma(b))$ (by Equation $($10.2$)$ in (2))

$=\sigma(a)\sigma(b)$. ∎

However, the above ring homomorphism is between extensions over the same
field. We will deal with ring homomorphisms between extensions over different
fields in Subsection 10.5.

######  Theorem 10.3.5.

Let $M$ be a module over a commutative ring $R$ with identity. Let $T$ be the
operator generalized-semigroup defined in Example 10.1.5. If $X$ and $Y$ are
nonempty subsets of $M$, then a map $\sigma:{{\left\langle
X\right\rangle}_{T}}\to{{\left\langle Y\right\rangle}_{T}}$ is an
$R$-homomorphism of $R$-modules if and only if it is a $T$-morphism.

###### Proof.

By Example 10.1.5,

$T=\\{f^{*}:{{M}^{n}}\to M$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\},n\in{{\mathbb{Z}}^{+}}\\}$.

So ${{\left\langle X\right\rangle}_{T}}$ and ${{\left\langle
Y\right\rangle}_{T}}$ are the submodules of $M$ generated by $X$ and $Y$,
respectively. And Lemma 10.3.3 applies. Thus we only need to show that
$\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle
Y\right\rangle}_{T}}$ is an $R$-homomorphism of $R$-modules if and only if
statement (ii) in Lemma 10.3.3 is true in this case. Specifically, it suffices
to show that the following are equivalent:

  1. (1)

$\forall a,b\in{{\left\langle X\right\rangle}_{T}}$ and $r\in R$,
$\sigma(a+b)=\sigma(a)+\sigma(b)$ and $\sigma(ra)=r\sigma(a)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
X\right\rangle_{T}^{n}$ and
$f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\\}$,

(10.3)
$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

(1)$\Rightarrow$(2): Because
$f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\\}$ is
a linear polynomial whose constant term is zero, it is not hard to see that
(2) is true.

(2)$\Rightarrow$(1): Let $a,b\in{{\left\langle X\right\rangle}_{T}}$ and $r\in
R$. Let $g={{x}_{1}}+{{x}_{2}}$ and $h=r{{x}_{1}}$. Then

$\sigma(a+b)$

$=\sigma(g(a,b))$

$=g(\sigma(a),\sigma(b))$ (by Equation $($10.3$)$ in (2))

$=\sigma(a)+\sigma(b)$.

And

$\sigma(ra)=\sigma(h(a))$

$=h(\sigma(a))$ (by Equation $($10.3$)$ in (2))

$=r\sigma(a)$. ∎

The above $R$-homomorphism is between submodules of the same $R$-module. We
will deal with general $R$-homomorphisms of $R$-modules in Subsection 10.5.

######  Theorem 10.3.6.

Let $G$ be an abelian group and let $T$ be the operator generalized-semigroup
on $G$ defined in Example 10.1.6. If $X$ and $Y$ are nonempty subsets of $G$,
then a map $\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle
Y\right\rangle}_{T}}$ is a group homomorphism if and only if it is a
$T$-morphism.

###### Proof.

By Example 10.1.6,

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}$}.

Obviously, ${{\left\langle X\right\rangle}_{T}}$ and ${{\left\langle
Y\right\rangle}_{T}}$ are the subgroups of $G$ generated by $X$ and $Y$,
respectively.

We may apply Lemma 10.3.3 as well in this proof. However, our argument as
follows seems more straightforward.

Suppose that $\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle
Y\right\rangle}_{T}}$ is a $T$-morphism. Then by Definition 10.3.1, $\forall
n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
X\right\rangle_{T}^{n}$ and $f\in T$ of $n$ variables,

$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

In words, $\sigma$ commutes with the (multiplicative) operation equipped by
$G$. Hence $\sigma$ is a group homomorphism.

Apparently the converse of the above argument is also true. That is, if
$\sigma$ is a group homomorphism, then $\sigma$ commutes with the operation
equipped by $G$, and so it is a $T$-morphism. ∎

The above group homomorphism is between subgroups of the same abelian group.
We will deal with group homomorphisms between different abelian groups in
Subsection 10.5.

For not necessarily abelian groups, we have

######  Theorem 10.3.7.

Let $G$ be a group and let $T$ be the operator generalized-semigroup on $G$
defined in Example 10.1.7. If $X$ and $Y$ are nonempty subsets of $G$, then a
map $\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle
Y\right\rangle}_{T}}$ is a group homomorphism if and only if it is a
$T$-morphism.

###### Proof.

The proof is comparable with that of Theorem 10.3.6.

By Example 10.1.7,

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\\}$,

where
$\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle$
denotes the set of all words on $\\{{{x}_{1}},\cdots,$
${{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\\}$.

Obviously, ${{\left\langle X\right\rangle}_{T}}$ and ${{\left\langle
Y\right\rangle}_{T}}$ are the subgroups of $G$ generated by $X$ and $Y$,
respectively.

Suppose that $\sigma:{{\left\langle X\right\rangle}_{T}}\to{{\left\langle
Y\right\rangle}_{T}}$ is a $T$-morphism. Then by Definition 10.3.1, $\forall
n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
X\right\rangle_{T}^{n}$ and $f\in T$ of $n$ variables,

$\sigma(f(a_{1},\cdots,a_{n}))=f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}})).$

In words, $\sigma$ commutes with the multiplication equipped by $G$. Hence
$\sigma$ is a group homomorphism.

It is not hard to see the converse of the above argument is also true.
Specifically, if $\sigma$ is a group homomorphism, then $\sigma$ commutes with
the multiplication equipped by $G$, hence it is a $T$-morphism. ∎

The above group homomorphism is between subgroups of the same group. We will
deal with general group homomorphisms in Subsection 10.5.

###  10.4. $T$-isomorphisms

Definition 2.4.1 still applies: if a $T$-morphism is bijective, then we call
it a _$T$ -isomorphism_. To justify the definition, we need to show
Proposition 2.4.2 with a different proof as follows.

######  Proposition 10.4.1.

_(Proposition 2.4.2)_ Let ${{S}_{1}}$ and ${{S}_{2}}$ be $T$-spaces, let
$\sigma\in\operatorname{Iso}_{T}({{S}_{1}},{{S}_{2}})$ and let
${{\sigma}^{-1}}$ be the inverse map of $\sigma$. Then
${{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}})$.

###### Proof.

By Definition 10.3.1, $\forall n\in{{\mathbb{Z}}^{+}}$, $f\in T$ of $n$
variables and $(a_{1},\cdots,a_{n})\in{S_{1}^{n}}$,

$f(a_{1},\cdots,a_{n})={\sigma}^{-1}(f(\sigma({{a}_{1}}),\cdots,\sigma({{a}_{n}}))).$

So $\forall n\in{{\mathbb{Z}}^{+}}$, $f\in T$ of $n$ variables and
$(b_{1},\cdots,b_{n})\in{S_{2}^{n}}$,

$f({\sigma}^{-1}({b}_{1}),\cdots,{\sigma}^{-1}({b}_{n}))={\sigma}^{-1}(f(b_{1},\cdots,b_{n})).$

Then by Definition 10.3.1, ${\sigma}^{-1}$ is a $T$-morphism from $S_{2}$ to
$S_{1}$, and so
${{\sigma}^{-1}}\in\operatorname{Iso}_{T}({{S}_{2}},{{S}_{1}})$ by Definition
2.4.1. ∎

###  10.5. $\theta$-morphisms

To generalize Definition 6.1.8, we first generalize Notation 6.1.4 to

######  Notation 10.5.1.

Let $D_{1}$ (resp. $D_{2}$) be a set, let $M_{1}$ (resp. $M_{2}$) be a subset
of {all functions from the cartesian product $D_{1}^{n}$ to
${D_{1}}\,|\,n\in{{\mathbb{Z}}^{+}}$} (resp. a subset of {all functions from
$D_{2}^{n}$ to ${{D}_{2}}\,|\,n\in{{\mathbb{Z}}^{+}}$}), let
$\theta\subseteq{{M}_{1}}\times{{M}_{2}}$, and let $A\subseteq{{D}_{2}}$. We
denote by $\theta{{|}_{A}}$ the set $\\{(f,g{{|}_{A}})\,|\,(f,g)\in\theta\\}$,
where $g{{|}_{A}}$ denotes the function obtained by restricting every variable
of $g$ to $A$.

Definition 6.1.5 is generalized to

######  Definition 10.5.2.

Let $\theta$ and $A$ be defined as in Notation 10.5.1. Let
$f\in\operatorname{Dom}\theta$, $n\in{{\mathbb{Z}}^{+}}$ and
$(a_{1},\cdots,a_{n})\in A^{n}$. If $\forall(f,g_{1}),(f,g_{2})\in\theta$,
$g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n})$, then we say that
$\theta(f)(a_{1},\cdots,a_{n})$ is _well-defined_ and let
$\theta(f)(a_{1},\cdots,a_{n})=g_{1}(a_{1},\cdots,a_{n})$.

######  Remark.

By Convention 6.1.7, “$\forall(f,g_{1}),(f,g_{2})\in\theta$,
$g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n})$” implies that
$\forall(f,g)\in\theta$, $g$ has $n$ variables.

Hence Proposition 6.1.6 is generalized to

######  Proposition 10.5.3.

Let $\theta$ and $A$ be defined as in Notation 10.5.1. Then the following are
equivalent:

  1. (i)

$\theta{{|}_{A}}$ is a map.

  2. (ii)

$\forall f\in\operatorname{Dom}\theta$, $\exists n\in{{\mathbb{Z}}^{+}}$ such
that $\forall(a_{1},\cdots,a_{n})\in A^{n}$, $\theta(f)(a_{1},\cdots,a_{n})$
is well-defined.

###### Proof.

$\theta{{|}_{A}}$ is a map;

$\Leftrightarrow$ $\forall(f,g_{1}),(f,g_{2})\in\theta$,
$g_{1}{{|}_{A}}=g_{2}{{|}_{A}}$;

$\Leftrightarrow$ $\forall(f,g_{1}),(f,g_{2})\in\theta$, $g_{1}$ and $g_{2}$
have the same number of variables, assumed $n$, and
$\forall(a_{1},\cdots,a_{n})\in A^{n}$,
$g_{1}(a_{1},\cdots,a_{n})=g_{2}(a_{1},\cdots,a_{n})$;

$\Leftrightarrow$ $\forall f\in\operatorname{Dom}\theta$, $\exists
n\in{{\mathbb{Z}}^{+}}$ such that $\forall(a_{1},\cdots,a_{n})\in A^{n}$,
$\theta(f)(a_{1},\cdots,a_{n})$ is well-defined. ∎

Then Definition 6.1.8 is generalized to

######  Definition 10.5.4.

Let $T_{1}$ and $T_{2}$ be operator generalized-semigroups and let
$\theta\subseteq{{T}_{1}}\times{{T}_{2}}$. Let $\phi$ be a map from a
$T_{1}$-space $S$ to a $T_{2}$-space. If $\forall n\in{{\mathbb{Z}}^{+}}$,
$(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and $f\in\operatorname{Dom}\theta$ of $n$
variables,

$\phi(f(a_{1},\cdots,a_{n}))=\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})),$

then $\phi$ is called a _$\theta$ -morphism_.

######  Proposition 10.5.5.

Let $\theta$ and $\phi$ be given as in Definition 10.5.4. Suppose that $\phi$
is a $\theta$-morphism. Then $\forall(f,g)\in\theta$, $f$ and $g$ have the
same number of variables. Moreover, $\theta{{|}_{\operatorname{Im}\phi}}$ is a
map.

###### Proof.

Let $n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and
$n$-variable $f\in\operatorname{Dom}\theta$. Then by Definition 10.5.4 and
Convention 6.1.7, $\theta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ is well-
defined. Hence by Definition 10.5.2, $\forall(f,g)\in\theta$, $g$ has $n$
variables. Moreover, by Proposition 10.5.3,
$\theta{{|}_{\operatorname{Im}\phi}}$ is a map. ∎

When operator generalized-semigroups are induced from formal functions, the
following lemma, which generalizes Lemma 10.3.3, will be useful for
$\theta$-morphisms.

######  Lemma 10.5.6.

Let $\mathcal{F}$ $($resp. ${\mathcal{F}}^{\prime})$ be a set of D-valued
$($resp. ${D}^{\prime}$-valued$)$ formal functions on $D$ $($resp.
${D}^{\prime})$. Let $T$ $($resp. ${T}^{\prime})$ be an operator generalized-
semigroup on $D$ $($resp. ${D}^{\prime}$$)$ such that there is a map
$\tau:\mathcal{F}\to T$ being given by $f\mapsto f^{*}$, where
$f^{*}:{{D}^{n}}\to D$ is given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})$ for $f$ of $n$ variables, $\forall
n\in{{\mathbb{Z}}^{+}}$ $($resp. such that there is a map
${\tau}^{\prime}:{\mathcal{F}}^{\prime}\to{T}^{\prime}$ being given
analogously$)$. Let $\vartheta:\mathcal{F}\to{\mathcal{F}}^{\prime}$ be a map.
Let $\theta=\\{(\tau(f),{\tau}^{\prime}(\vartheta(f)))\,|\,f\in\mathcal{F}\\}$
and let $\phi$ be a map from a $T$-space $S$ to a ${T}^{\prime}$-space.

Then the following are equivalent:

  1. (i)

$\phi$ is a $\theta$-morphism.

  2. (ii)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and
$f\in\mathcal{F}$ of n variables,

(10.4)
$\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).$

###### Proof.

By the definitions of $\theta$, $\tau$ and $\tau^{\prime}$, $\theta\subseteq
T\times T^{\prime}$.

(i) $\Rightarrow$ (ii): Let $n\in{{\mathbb{Z}}^{+}}$,
$(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and $n$-variable $f\in\mathcal{F}$. Then

$\phi(f(a_{1},\cdots,a_{n}))$

$=\phi(\tau(f)(a_{1},\cdots,a_{n}))$

$=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ (because
$\tau(f)\in\operatorname{Dom}\theta$ and by Definition 10.5.4)

$={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$
(because $(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta$ and by Proposition
10.5.5, $\theta{{|}_{\operatorname{Im}\phi}}$ is a map)

$=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$, as desired.

(ii) $\Rightarrow$ (i): To show that $\phi$ is a $\theta$-morphism, we first
need to show that $\theta{{|}_{\operatorname{Im}\phi}}$ is a map.

Let
$(h,g{{|}_{\operatorname{Im}\phi}}),(h,{g}^{\prime}{{|}_{\operatorname{Im}\phi}})\in\theta{{|}_{\operatorname{Im}\phi}}$
with $(h,g),(h,{g}^{\prime})\in\theta$. By the definition of $\theta$,
$\exists l,m\in\mathcal{F}$ such that
$(h,g)=(\tau(l),{\tau}^{\prime}(\vartheta(l)))$ and
$(h,{g}^{\prime})=(\tau(m),{\tau}^{\prime}(\vartheta(m)))$, and so
$\tau(l)=\tau(m)$. Thus $l$ and $m$ have the same number of variables, which
we assume to be $n$. Then by Equation (10.4) and Convention 6.1.7, both
$\vartheta(l)$ and $\vartheta(m)$ have $n$ variables.

Assume that
$g{{|}_{\operatorname{Im}\phi}}\neq{g}^{\prime}{{|}_{\operatorname{Im}\phi}}$;
that is,
${\tau}^{\prime}(\vartheta(l)){{|}_{\operatorname{Im}\phi}}\neq{\tau}^{\prime}(\vartheta(m)){{|}_{\operatorname{Im}\phi}}$.
Then $\exists({{z}_{1}},\cdots,{{z}_{n}})\in{{S}^{n}}$ such that

$\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}}))\neq\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})).$

However,
$\phi(l({{z}_{1}},\cdots,{{z}_{n}}))=\phi(m({{z}_{1}},\cdots,{{z}_{n}}))$
because $\tau(l)=\tau(m)$, and so by Equation $($10.4$)$, we can tell that

$\vartheta(l)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}}))=\vartheta(m)(\phi({{z}_{1}}),\cdots,\phi({{z}_{n}})),$

a contradiction.

Hence $\theta{{|}_{\operatorname{Im}\phi}}$ must be a map. Then $\forall
n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{S}^{n}}$ and
$f\in\mathcal{F}$ of $n$ variables,

$\phi(\tau(f)(a_{1},\cdots,a_{n}))$

$=\phi(f(a_{1},\cdots,a_{n}))$ (by the definition of $\tau$)

$=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ (by Equation (10.4))

$={\tau}^{\prime}(\vartheta(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ (by
the definition of ${\tau}^{\prime}$)

$=\theta(\tau(f))(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ (because
$(\tau(f),{\tau}^{\prime}(\vartheta(f)))\in\theta$ and
$\theta{{|}_{\operatorname{Im}\phi}}$ is a map).

Since $\operatorname{Dom}\theta=\\{\tau(f)\,|\,f\in\mathcal{F}\\}$, by
Definition 10.5.4, $\phi$ is a $\theta$-morphism. ∎

Lemma 10.5.6 characterizes a $\theta$-morphism by formal functions. When
operator generalized-semigroups are induced from formal functions, the
criterion for a $\theta$-morphism by Lemma 10.5.6 is normally more convenient
than that by Definition 10.5.4. Thus Lemma 10.5.6 simplifies our proofs of the
following five theorems.

The following result is a generalization of both Theorem 6.1.10 and Theorem
10.3.4.

######  Theorem 10.5.7.

Let $F/B$ $($resp. ${F}^{\prime}/{B}^{\prime}$$)$ be a field extension, let an
operator generalized-semigroup $T$ $($resp. ${T}^{\prime}$$)$ be defined as in
Example 10.1.3 on $F$ $($resp. ${F}^{\prime}$$)$ over $B$ $($resp.
${B}^{\prime}$$)$, let $\varphi:B\to{B}^{\prime}$ be a field isomorphism, and
let
$\vartheta:\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]}\to\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}]}$
be the map given by $f\mapsto{f}^{\prime}$ where

$f({{x}_{1}},\cdots,{{x}_{n}})=\sum{{c_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots
x_{n}^{{{p}_{n}}}},$

and

${f}^{\prime}({{x}_{1}},\cdots,{{x}_{n}})=\sum{\varphi({c_{p_{1},\cdots,p_{n}}})x_{1}^{{{p}_{1}}}\cdots
x_{n}^{{{p}_{n}}}}.$

Let $\theta=\\{(\tau(f),{\tau}^{\prime}(\vartheta(f)))\,|\,f\in
B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\\}$, where $\tau$ and
${\tau}^{\prime}$ are the maps inducing $T$ and ${T}^{\prime}$, respectively,
defined as in Example 10.1.3.

Then $\forall(\emptyset\neq)U\subseteq F$ and
$(\emptyset\neq)V\subseteq{F}^{\prime}$, a map $\phi:{{\left\langle
U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}}$ is a
ring homomorphism extending $\varphi$ if and only if $\phi$ is a
$\theta$-morphism.

###### Proof.

We may prove this theorem in a way analogous to that for Theorem 6.1.10.
However, now we may employ Lemma 10.5.6 instead.

By Example 10.1.3,

$T=\\{f^{*}:{{F}^{n}}\to F$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in
B[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\\}$

and

${T}^{\prime}=\\{f^{*}:{{{F}^{\prime}}^{n}}\to{F}^{\prime}$ given by
$(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}],n\in{{\mathbb{Z}}^{+}}\\}$.

Thus ${{\left\langle U\right\rangle}_{T}}$ and ${{\left\langle
V\right\rangle}_{{{T}^{\prime}}}}$ are the rings $B[U]$ and ${B}^{\prime}[V]$,
respectively.

If let
$\mathcal{F}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{B[{{x}_{1}},\cdots,{{x}_{n}}]}$
and
${\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{B}^{\prime}[{{x}_{1}},\cdots,{{x}_{n}}]}$,
then Lemma 10.5.6 applies. So it suffices to show that $\phi:{{\left\langle
U\right\rangle}_{T}}\to{{\left\langle V\right\rangle}_{{{T}^{\prime}}}}$ is a
ring homomorphism extending $\varphi$ if and only if statement (ii) in Lemma
10.5.6 is true in this case. Specifically, we only need to show that the
following are equivalent:

  1. (1)

$\forall c\in B$, $\phi(c)=\varphi(c)$, and $\forall a,b\in{{\left\langle
U\right\rangle}_{T}}$, $\phi(a+b)=\phi(a)+\phi(b)$ and
$\phi(ab)=\phi(a)\phi(b)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
U\right\rangle_{T}^{n}$ and $f\in B[{{x}_{1}},\cdots,{{x}_{n}}]$,

(10.5)
$\phi(f(a_{1},\cdots,a_{n}))=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).$

(1) $\Rightarrow$ (2): $\forall n\in{{\mathbb{Z}}^{+}}$,
$f({{x}_{1}},\cdots,{{x}_{n}})=\sum{{c_{p_{1},\cdots,p_{n}}}x_{1}^{{{p}_{1}}}\cdots
x_{n}^{{{p}_{n}}}}\in B[{{x}_{1}},\cdots,{{x}_{n}}]$ and
$(a_{1},\cdots,a_{n})\in\left\langle U\right\rangle_{T}^{n}$,

$\phi(f(a_{1},\cdots,a_{n}))$

$=\phi(\sum{{c_{p_{1},\cdots,p_{n}}}a_{1}^{{{p}_{1}}}\cdots
a_{n}^{{{p}_{n}}}})$

$=\sum{\phi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}}$
(because (1) is true)

$=\sum{\varphi({c_{p_{1},\cdots,p_{n}}})(\phi(a_{1}))^{p_{1}}\cdots(\phi(a_{n}))^{p_{n}}}$
(because $\forall c\in B$, $\phi(c)=\varphi(c)$)

$=\vartheta(f)(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}}))$ (by the definition of
$\vartheta$).

(2) $\Rightarrow$ (1): Considering the case where $f\in
B[{{x}_{1}},\cdots,{{x}_{n}}]$ is a constant polynomial, we can tell from (2)
that $\forall c\in B$, $\phi(c)=\varphi(c)$.

Let $a,b\in{{\left\langle U\right\rangle}_{T}}$. Let $g={{x}_{1}}+{{x}_{2}}$
and let $h={{x}_{1}}{{x}_{2}}$. Then

$\phi(a+b)$

$=\phi(g(a,b))$

$=\vartheta(g)(\phi(a),\phi(b))$ (by Equation $($10.5$)$)

$=g(\phi(a),\phi(b))$ ($\vartheta(g)=g$ by the definition of $\vartheta$)

$=\phi(a)+\phi(b)$.

And

$\phi(ab)$

$=\phi(h(a,b))$

$=\vartheta(h)(\phi(a),\phi(b))$ (by Equation $($10.5$)$)

$=h(\phi(a),\phi(b))$ ($\vartheta(h)=h$ by the definition of $\vartheta$)

$=\phi(a)\phi(b)$. ∎

The following result is for ring homomorphisms between not necessarily
commutative rings.

######  Theorem 10.5.8.

Let $A$ and $R$ be rings with identity, let ${{T}_{A}}$ $($resp.
${{T}_{R}}$$)$ be the operator generalized-semigroup on $A$ $($resp. $R$$)$
defined as in Example 10.1.4, and let
$\theta=\\{({{\tau}_{A}}(f),{{\tau}_{R}}(f))\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\\}$,
where ${{\tau}_{A}}$ and ${{\tau}_{R}}$ are the maps inducing ${{T}_{A}}$ and
${{T}_{R}}$, respectively, defined as in Example 10.1.4. Then a map $\phi:A\to
R$ is a ring homomorphism if and only if $\phi$ is a $\theta$-morphism.

###### Proof.

By Example 10.1.4,

${{T}_{A}}=\\{f^{*}:{{A}^{n}}\to A$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\\}$,

and

${{T}_{R}}=\\{f^{*}:{{R}^{n}}\to R$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle,n\in{{\mathbb{Z}}^{+}}\\}$.

Obviously ${{\left\langle A\right\rangle}_{{{T}_{A}}}}=A$ and ${{\left\langle
R\right\rangle}_{{{T}_{R}}}}=R$. Hence $\phi:A\to R$ is $\phi:{{\left\langle
A\right\rangle}_{{{T}_{A}}}}\to{{\left\langle R\right\rangle}_{{{T}_{R}}}}$.

Let
$\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle}$
and let $\vartheta=\operatorname{Id}$. Then Lemma 10.5.6 applies. By Lemma
10.5.6, we only need to show that $\phi:A\to R$ is a ring homomorphism if and
only if statement (ii) in Lemma 10.5.6 is true in this case. Specifically, it
suffices to show that the following are equivalent:

  1. (1)

$\phi({{1}_{A}})={{1}_{R}}$ and $\forall a,b\in A$,
$\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in{{A}^{n}}$ and
$f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle$,

(10.6)
$\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).$

(1) $\Rightarrow$ (2): Since
$f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle$ is
a polynomial, it is not hard to see that (2) is true.

(2) $\Rightarrow$ (1): Considering the case where
$f\in{{\mathbb{F}}_{2}}\left\langle{{x}_{1}},\cdots,{{x}_{n}}\right\rangle$ is
the constant polynomial 1, we can tell from (2) that
$\phi({{1}_{A}})={{1}_{R}}$.

Let $a,b\in A$. Let $g={{x}_{1}}+{{x}_{2}}$ and $h={{x}_{1}}{{x}_{2}}$. Then

$\phi(a+b)$

$=\phi(g(a,b))$

$=g(\phi(a),\phi(b))$ (by Equation $($10.6$)$)

$=\phi(a)+\phi(b)$.

And

$\phi(ab)=\phi(h(a,b))$

$=h(\phi(a),\phi(b))$ (by Equation $($10.6$)$)

$=\phi(a)\phi(b)$. ∎

Theorem 10.3.5 is generalized to

######  Theorem 10.5.9.

Let $M$ and $N$ be modules over a commutative ring $R$ with identity, let
${{T}_{M}}$ $($resp. ${{T}_{N}}$$)$ be the operator generalized-semigroup on
$M$ $($resp. $N$$)$ defined as in Example 10.1.5, and let

$\theta=\\{({{\tau}_{M}}(f),{{\tau}_{N}}(f))\,|\,f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\},n\in{{\mathbb{Z}}^{+}}\\},$

where ${{\tau}_{M}}$ and ${{\tau}_{N}}$ are the maps inducing ${{T}_{M}}$ and
${{T}_{N}}$, respectively, defined as in Example 10.1.5.

If $X$ and $Y$ are nonempty subsets of $M$ and $N$, respectively, then a map
$\phi:{{\left\langle X\right\rangle}_{{{T}_{M}}}}\to{{\left\langle
Y\right\rangle}_{{{T}_{N}}}}$ is an R-homomorphism of R-modules if and only if
$\phi$ is a $\theta$-morphism.

###### Proof.

Our proof is comparable with that of Theorem 10.3.5.

By Example 10.1.5,

${{T}_{M}}=\\{f^{*}:{{M}^{n}}\to M$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\},n\in{{\mathbb{Z}}^{+}}\\}$

and

${{T}_{N}}=\\{f^{*}:{{N}^{n}}\to N$ given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\},n\in{{\mathbb{Z}}^{+}}\\}$.

Clearly, ${{\left\langle X\right\rangle}_{{{T}_{M}}}}$ and ${{\left\langle
Y\right\rangle}_{{{T}_{N}}}}$ are the submodules of $M$ and $N$ generated by
$X$ and $Y$, respectively.

Let
$\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in
R\\}}$ and let $\vartheta=\operatorname{Id}$. Then Lemma 10.5.6 applies. By
Lemma 10.5.6, it suffices to show that $\phi:{{\left\langle
X\right\rangle}_{{{T}_{M}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{N}}}}$ is
an $R$-homomorphism of $R$-modules if and only if statement (ii) in Lemma
10.5.6 is true in this case. Specifically, we only need to show that the
following are equivalent:

  1. (1)

$\forall a,b\in{{\left\langle X\right\rangle}_{{{T}_{M}}}}$ and $r\in R$,
$\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ra)=r\phi(a)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
X\right\rangle_{{{T}_{M}}}^{n}$ and
$f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\\}$,

(10.7)
$\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).$

(1) $\Rightarrow$ (2): Because
$f\in\\{\sum\nolimits_{i=1}^{n}{{{r}_{i}}{{x}_{i}}}\,|\,{{r}_{i}}\in R\\}$ is
a linear polynomial whose constant term is zero, it is not hard to see that
(2) is true.

(2) $\Rightarrow$ (1): Let $a,b\in{{\left\langle X\right\rangle}_{{{T}_{M}}}}$
and let $r\in R$. Let $g={{x}_{1}}+{{x}_{2}}$ and let $h=r{{x}_{1}}$. Then

$\phi(a+b)$

$=\phi(g(a,b))$

$=g(\phi(a),\phi(b))$ (by Equation $($10.7$)$)

$=\phi(a)+\phi(b)$.

And

$\phi(ra)$

$=\phi(h(a))$

$=h(\phi(a))$ (by Equation $($10.7$)$)

$=r\phi(a)$. ∎

Theorem 10.3.6 is generalized as follows.

######  Theorem 10.5.10.

Let $G$ and $H$ be abelian groups, let ${{T}_{G}}$ $($resp. ${{T}_{H}}$$)$ be
the operator generalized-semigroup on $G$ $($resp. $H$$)$ defined as in
Example 10.1.6, and let
$\theta=\\{({{\tau}_{G}}(f),{{\tau}_{H}}(f))\,|\,f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\},n\in{{\mathbb{Z}}^{+}}\\}$
, where ${{\tau}_{G}}$ and ${{\tau}_{H}}$ are the maps inducing ${{T}_{G}}$
and ${{T}_{H}}$, respectively, defined as in Example 10.1.6.

If $X$ and $Y$ are nonempty subsets of $G$ and $H$ respectively, then a map
$\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle
Y\right\rangle}_{{{T}_{H}}}}$ is a group homomorphism if and only if $\phi$ is
a $\theta$-morphism.

###### Proof.

By Example 10.1.6,

${{T}_{G}}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$
given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}$}

and

${{T}_{H}}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{H}^{n}}\to H}$
given by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}$}.

Besides, ${{\left\langle X\right\rangle}_{{{T}_{G}}}}$ and ${{\left\langle
Y\right\rangle}_{{{T}_{H}}}}$ are the subgroups of $G$ and $H$ generated by
$X$ and $Y$, respectively.

Let
$\mathcal{F}={\mathcal{F}}^{\prime}=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}}$
and let $\vartheta=\operatorname{Id}$. Then Lemma 10.5.6 applies. By Lemma
10.5.6, it suffices to show that $\phi:{{\left\langle
X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle Y\right\rangle}_{{{T}_{H}}}}$ is
a group homomorphism if and only if statement (ii) in Lemma 10.5.6 is true in
this case. Specifically, we only need to show that the following are
equivalent:

  1. (1)

$\forall a,b\in{{\left\langle X\right\rangle}_{{{T}_{G}}}}$,
$\phi(ab)=\phi(a)\phi(b)$.

  2. (2)

$\forall n\in{{\mathbb{Z}}^{+}}$, $(a_{1},\cdots,a_{n})\in\left\langle
X\right\rangle_{{{T}_{G}}}^{n}$ and
$f\in\\{\prod\nolimits_{i=1}^{n}{x_{i}^{{{p}_{i}}}}\,|\,{{p}_{i}}\in\mathbb{Z}\\}$,

$\phi(f(a_{1},\cdots,a_{n}))=f(\phi({{a}_{1}}),\cdots,\phi({{a}_{n}})).$

It is not hard to see that $(\text{1})\Leftrightarrow(\text{2})$. ∎

Theorem 10.3.7 is generalized to

######  Theorem 10.5.11.

Let $G$ and $H$ be groups, let ${{T}_{G}}$ $($resp. ${{T}_{H}}$$)$ be the
operator generalized-semigroup on $G$ $($resp. $H$$)$ defined as in Example
10.1.7, and let
$\theta=\\{({{\tau}_{G}}(f),{{\tau}_{H}}(f))\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle,n\in{{\mathbb{Z}}^{+}}\\}$,
where ${{\tau}_{G}}$ and ${{\tau}_{H}}$ are the maps inducing ${{T}_{G}}$ and
${{T}_{H}}$, respectively, defined as in Example 10.1.7.

If $X$ and $Y$ are nonempty subsets of $G$ and $H$, respectively, then a map
$\phi:{{\left\langle X\right\rangle}_{{{T}_{G}}}}\to{{\left\langle
Y\right\rangle}_{{{T}_{H}}}}$ is a group homomorphism if and only if $\phi$ is
a $\theta$-morphism.

###### Proof.

Our proof is comparable with that of Theorem 10.5.10.

By Example 10.1.7,

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{G}^{n}}\to G}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\\}$.

and

$T=\bigcup\nolimits_{n\in{{\mathbb{Z}}^{+}}}{\\{f^{*}:{{H}^{n}}\to H}$ given
by $(a_{1},\cdots,a_{n})\mapsto
f(a_{1},\cdots,a_{n})\,|\,f\in\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle\\}$.

where
$\left\langle{{x}_{1}},\cdots,{{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\right\rangle$
denotes the set of all words on $\\{{{x}_{1}},\cdots,$
${{x}_{n}},x_{1}^{-1},\cdots,x_{n}^{-1}\\}$.

Clearly, ${{\left\langle X\right\rangle}_{{{T}_{G}}}}$ and ${{\left\langle
Y\right\rangle}_{{{T}_{H}}}}$ are the subgroups of $G$ and $H$ generated by
$X$ and $Y$, respectively.

Let