Partitioning of Any Infinite Set with the Aid of Non-Surjective Injective Maps and the Study of a Remarkable Semigroup

Charif Harrafa

Ecole Hassania des Travaux Publics, Casablanca, Morocco.

DOI: 10.4236/ojdm.2020.103008                

Abstract

In this article, we will present a particularly remarkable partitioning method of any infinite set with the aid of non-surjective injective maps. The non-surjective injective maps from an infinite set to itself constitute a semigroup for the law of composition bundled with certain properties allowing us to prove the existence of remarkable elements. Not to mention a compatible equivalence relation that allows transferring the said law to the quotient set, which can be provided with a lattice structure. Finally, we will present the concept of Co-injectivity and some of its properties.

Keywords

Partitioning, Non-Surjective, Injective, Infinite Set, Fixed Points, Lattice Structure

1. Introduction

The concept of map in mathematics has a primordial role in understanding the links that exist between the different mathematical fields and structures. A map is binary relation over two sets that associates to every element of the first set exactly one element of the second set, sometimes with a specific property. For instance, a “map” is a “linear transformation” in linear algebra, a “continuous function” in topology, operators in analysis and representations in group theory, etc. In this article we will show how non-surjective injective maps allow to partitioning an infinite set in several ways.

2. Part I

Proposition 1

Let E, and F be non-empty sets. If $f,g$ are two non-surjective injective maps from E to F and from F to E respectively, then:

· $\left(A,f\left(B\right),I_{f\circ g}\right)$ forms a partition of F.

· $\left(B,g\left(A\right),I_{g\circ f}\right)$ forms a partition of E.

where,

$A=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and $B=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$

and $I_f,I_g$ representing the image sets under $f,g$ respectively.

Proof

Let E and F be two non-empty sets, and let ƒ and g be two non-surjective injective maps from E to F, and from F to E respectively. Since ƒ and g are non-surjective, then the following sets:

$A=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and $B=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$ are non-empty.

Also, obviously $E=B\cup I_g$ such as $B\cap I_g=\varnothing$ as follows from the definition of $I_g$.

For any such map ƒ from E to F:

$I_f=f\left(E\right)=f\left(B\cup I_g\right)=f\left(B\right)\cup f\left(I_g\right)=f\left(B\right)\cup I_{f\circ g}$

So $F=A\cup f\left(B\right)\cup I_{f\circ g}$

Since ƒ is an injective map then:

$f\left(B\cap I_g\right)=f\left(B\right)\cap I_{f\circ g}=f(\varnothing )=\varnothing$

Therefore, $\left(A,f\left(B\right),I_{f\circ g}\right)$ forms a partition of F.

In analogy to the map g from F to E, $\left(B,g\left(A\right),I_{g\circ f}\right)$ forms a partition of E.

Note 1

This process could be applied repeatedly, and for each iteration, finer partitions of the sets E and F respectively will be obtained, e.g. after 2^nd iteration we have:

· $E=B\cup g\left(A\right)\cup \left(g\circ f\right)\left(B\right)\cup I_{g\circ f\circ g}$

· $F=A\cup f\left(B\right)\cup \left(f\circ g\right)\left(A\right)\cup I_{f\circ g\circ f}$

In case that ƒ and g are (two) (2) different non-surjective injective maps from an infinite set E to itself, we can compose either by ƒ or by g, or by both indefinitely. Thus several partition classes of E can obtained, for example after a second (2^nd) iteration

· $E=A_f\cup f\left(A_f\right)\cup f^2\left(A_f\right)\cup I_{f^3}$

· $E=A_g\cup g\left(A_g\right)\cup g^2\left(A_g\right)\cup I_{g^3}$

· $E=A_f\cup f\left(A_g\right)\cup \left(f\circ g\right)\left(A_f\right)\cup I_{f\circ g\circ f}$

· $E=A_g\cup g\left(A_f\right)\cup \left(g\circ f\right)\left(A_g\right)\cup I_{g\circ f\circ g}$

· $E=A_f\cup f\left(A_g\right)\cup \left(f\circ g\right)\left(A_g\right)\cup I_{f\circ g^2}$

· $E=A_g\cup g\left(A_f\right)\cup \left(g\circ f\right)\left(A_f\right)\cup I_{g\circ f^2}$

where,

$A_f=\left\{y\in F|\forall x\in E,f\left(x\right)\ne y\right\}$ and $A_g=\left\{x\in E|\forall y\in F,g\left(y\right)\ne x\right\}$

Example 1

If $E=F=\mathbb{N}$ i.e. the set of natural numbers, ƒ and g are two maps from $\mathbb{N}$ to itself (i.e. ƒ and g are non-surjective injective) and defined by: ˆ

· $\forall n\in \mathbb{N},f\left(n\right)=2n$

· $\forall n\in \mathbb{N},g\left(n\right)=2n+1$

Knowing that $A_f=2\mathbb{N}+1$ and $A_g=2\mathbb{N}$ then:

· $f\left(A_g\right)=4\mathbb{N}$

· $g\left(A_f\right)=4\mathbb{N}+3$

· $\left(f\circ g\right)\left(A_f\right)=8\mathbb{N}+6$

· $\left(g\circ f\right)\left(A_g\right)=8\mathbb{N}+1$

· $\left(f\circ g\circ f\right)\left(\mathbb{N}\right)=8\mathbb{N}+2$

· $\left(g\circ f\circ g\right)\left(\mathbb{N}\right)=8\mathbb{N}+5$

Therefore, we can partition the set $\mathbb{N}$ to the second (2^nd) order by ƒ and g such as:

$\mathbb{N}=\left(2\mathbb{N}+1\right)\cup \left(4\mathbb{N}\right)\cup \left(8\mathbb{N}+6\right)\cup \left(8\mathbb{N}+2\right)$

$\mathbb{N}=\left(2\mathbb{N}\right)\cup \left(4\mathbb{N}+3\right)\cup \left(8\mathbb{N}+1\right)\cup \left(8\mathbb{N}+5\right)$

2.1. Remarkable Partition

Proposition 2

Let E be an infinite set, $\mathbb{N}$ be the set of natural numbers, and ƒ be a non-surjective injective map from E to itself, such that:

$A_f=\left\{y\in E|\forall x\in E,f\left(x\right)\ne y\right\}$

Then,

$\forall n\in \mathbb{N},E=\left[{\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+1}}$

where, $f^i\left(A_f\right)=\underset{i\text{times}}{\underset{︸}{f\circ f\circ \cdots \circ f\left(A_f\right)}}$

Proof

By induction:

For $n=0$, we have, by definition $E=A_f\cup I_f$, and for $n=1$ the proposition 1 states that, if $E=F$ and $f=g$ then, $E=A_f\cup f\left(A_f\right)\cup I_{f^2}$ Suppose that the said property is true for n. Then, by composing by ƒ, we get:

$I_f=f\left(E\right)=\left[{\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^{i+1}\left(A_f\right)}\right]\cup I_{f^{n+2}}=\left[{\displaystyle \underset{i=1}{\overset{i=n+1}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+2}}$

Then,

$E=A_f\cup \left[{\displaystyle \underset{i=1}{\overset{i=n+1}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+2}}=\left[{\displaystyle \underset{i=0}{\overset{i=n+1}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+2}}$

Therefore,

$\forall n\in \mathbb{N},E=\left[{\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+1}}$

Note 2

For all non-surjective injective maps ƒ from an infinite set E to itself,

$\forall n\in \mathbb{N},A_{f^{n+1}}={\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^i(\; A\; f\; )}$

Definition 1

Let ƒ be a map from no-empty set E to itself,

· $x\in E$ is a fixed point of ƒ if, $f\left(x\right)=x$.

· $A\subset E$ is a fixed point set of ƒ if, $f\left(A\right)=A$.

Proposition 3

For all non-surjective injective maps f from an infinite set E to itself, $\forall n\in \mathbb{N}$, $f^n\left(A_f\right)$ contains no fixed points of ƒ.

Proof

By induction:

For $n=0$, let $x\in A_f$, and by definition $\forall y\in E$, $f\left(y\right)\ne x$, particularly for $y=x$ so $f\left(x\right)\ne x$ then $A_f$ contains no fixed points of f. Suppose that the aforementioned property is true for n, let $x\in f^{n+1}\left(A_f\right)=f\left(f^n\left(A_f\right)\right)$, then $\exists y\in f^n\left(A_f\right)$ such that $x=f\left(y\right)$, we have $f\left(y\right)\ne y$ (by inductive hypothesis), since ƒ is injective then $f\left(f\left(y\right)\right)\ne f\left(y\right)$ so $f\left(x\right)\ne x$, then $\forall x\in f^{n+1}\left(A_f\right)$, $f\left(x\right)\ne x$. Therefore, $\forall n\in \mathbb{N}$, $f^n\left(A_f\right)$ contains no fixed points of ƒ.

Note 3

Let E be an infinite set, and ƒ be a non-surjective injective map from E to itself, we define the following:

$Fi{x}_f=\left\{x\in E|f\left(x\right)=x\right\}$ and $S{t}_f\left(E\right)=\left\{A\subset E|f\left(A\right)=A\right\}$

· $\forall n\in \mathbb{N},Fi{x}_f\cap f^n(A_f)=\varnothing$

· $\forall n\in {\mathbb{N}}^{*},Fi{x}_f\subseteq Fi{x}_{f^n}$

· $\forall n\in {\mathbb{N}}^{*},Fi{x}_{f^n}\subseteq Fi{x}_{f^{2n}}$

· $\forall p\in \mathbb{N},\forall n\in {\mathbb{N}}^{*},f^p\left(Fi{x}_{f^n}\right)=Fi{x}_{f^n}$

· For all ƒ non-surjective injective maps from an infinite set E to itself,

$Fi{x}_f\in S{t}_f(\; E\; )$

Proposition 4

For all ƒ non-surjective injective maps from E to itself,

$\forall A\in S{t}_f\left(E\right),\forall n\in \mathbb{N},f^n\left(A_f\right)\cap A=\varnothing$

Proof

Let $A\in S{t}_f\left(E\right)$, by induction, For $n=0$, we have, by definition $A_f\cap I_f=\varnothing$ then $A_f\cap A=A_f\cap f\left(A\right)=\varnothing$ ˆ Suppose that the said property is true for $n\in \mathbb{N}$, let $x\in f^{n+1}\left(A_f\right)\cap A$, then, $\exists y\in f^n\left(A_f\right),x=f\left(y\right)$ and $\exists y_0\in A,x=f\left(y_0\right)$. Since ƒ is injective so $y=y_0\in f^n\left(A_f\right)\cap A$, which contradicts the inductive hypothesis. Then,

$f^{n+1}\left(A_f\right)\cap A=\varnothing$.

Therefore, $\forall A\in S{t}_f\left(E\right),\forall n\in \mathbb{N},f^n\left(A_f\right)\cap A=\varnothing$

Lemma 1 (Generalization)

$\forall p\in {\mathbb{N}}^{*},\forall A\in S{t}_{f^p}\left(E\right),\forall n\in \mathbb{N},f^n\left(A_f\right)\cap A=\varnothing$

Proof

By complete (strong) induction,

Let $p\in {\mathbb{N}}^{*}$, and $A\in S{t}_{f^p}\left(E\right)=\left\{A\subset E|f^p\left(A\right)=A\right\}$ : For $n=0$, we have, by definition $A_f\cap I_f=\varnothing$, since $\exists p\in {\mathbb{N}}^{*},I_{f^p}\subseteq I_f$, then: $A_f\cap A=A_f\cap f^p\left(A\right)=\varnothing$

Suppose that the said property is true for all $i\in \left\{1,\cdots ,n\right\}$, let $x\in f^{n+1}\left(A_f\right)\cap A$, then, $\exists y\in A_f,x=f^{n+1}\left(y\right)$ and $\exists y_0\in A,x=f^p\left(y_0\right)$. Since ƒ is injective then,

· $f^{n-p+1}\left(y\right)=y_0$, if $n+1>p$, which contradicts the inductive hypothesis, because $i=n-p+1\in \left\{1,\cdots ,n\right\}$

· $y=f^{p-n-1}\left(y_0\right)$, if $p>n+1$, which contradicts $A_f\cap I_{f^{p-n-1}}=\varnothing$

· $y=y_0$, if $n+1=p$, which contradicts $A_f\cap A=\varnothing$

Then,

$f^{n+1}\left(A_f\right)\cap A=\varnothing$,

Therefore,

$\forall p\in {\mathbb{N}}^{*},\forall A\in S{t}_{f^p}\left(E\right),\forall n\in \mathbb{N},f^n\left(A_f\right)\cap A=\varnothing$

QED

For all $n\in \mathbb{N}$, and for all ƒ non-surjective injective maps from an infinite set E to itself, we define the following:

· $S_{f^n}\left(E\right)=\left\{A\subset E|\exists p\in \left\{1,\cdots ,n+1\right\},f^p\left(A\right)=A\right\}$

· $S_{f^{\infty }}\left(E\right)=\left\{A\subset E|\exists p\in {\mathbb{N}}^{*},f^p\left(A\right)=A\right\}$

· $S_{f^n}=\left\{x\in A|A\in S_{f^n}\left(E\right)\right\}$

· $S_{f^{\infty }}=\left\{x\in A|A\in S_{f^{\infty }}\left(E\right)\right\}$

Theorem 1

Let E be an infinite set, and ƒ be a non-surjective injective map from E to itself, then:

$E=\left[{\displaystyle \underset{n\in \mathbb{N}}{\cup }{f}^n\left(A_f\right)}\right]\cup S_{f^{\infty }}$

Proof

Note that for $n=0$, $S_{f^0}\left(E\right)=S{t}_f\left(E\right)$, The sequence of subsets of E, ${\left(S_{f^n}\right)}_{n\in \mathbb{N}}$ is increasing by inclusion, so it is convergent, the limit is: ${\displaystyle {\cup }_{n\in \mathbb{N}}{S}_{f^n}}=S_{f^{\infty }}$

We have already established that,

$\forall n\in \mathbb{N},E=\left[{\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^i\left(A_f\right)}\right]\cup I_{f^{n+1}}$

Otherly, according to the Lemma 1, $\forall n\in \mathbb{N},E=\left[{\displaystyle {\cup }_{i=0}^{i=n}{f}^i\left(A_f\right)}\right]\cap S_{f^n}=\varnothing$

Let’s define the following, $\forall n\in \mathbb{N},\widehat{I_{f^{n+1}}}=I_{f^{n+1}}\backslash S_{f^n}$ the sequence of subsets of ${\left(\widehat{I_{f^{n+1}}}\right)}_{n\in \mathbb{N}}$ is decreasing by inclusion then, it is convergent [1], the limit is:

${\displaystyle \underset{n\in \mathbb{N}}{\cap }\widehat{I_{f^{n+1}}}}=\varnothing$

Because, let $x\in {\displaystyle {\cap }_{n\in \mathbb{N}}\widehat{I_{f^{n+1}}}}$, then:

$x\in {\displaystyle \underset{n\in \mathbb{N}}{\cap }{I}_{f^{n+1}}\backslash S_{f^n}}\iff \forall n\in \mathbb{N},x\in I_{f^{n+1}}\backslash S_{f^n}\iff \forall n\in \mathbb{N},x\in I_{f^{n+1}}\text{and}x\in \overline{S_{f^n}}$

$\iff \forall n\in \mathbb{N},x\in I_{f^{n+1}}\text{and}\forall n\in \mathbb{N},x\in \overline{S_{f^n}}\iff x\in {\displaystyle \underset{n\in \mathbb{N}}{\cap }{I}_{f^{n+1}}}\text{and}x\in {\displaystyle \underset{n\in \mathbb{N}}{\cap }\overline{S_{f^n}}}$

$\iff x\in {\displaystyle \underset{n\in \mathbb{N}}{\cap }{I}_{f^{n+1}}}\text{and}x\in \stackrel{¯}{{\displaystyle \underset{n\in \mathbb{N}}{\cup }{S}_{f^n}}}\iff x\in {\displaystyle \underset{n\in \mathbb{N}}{\cap }{I}_{f^{n+1}}}\text{and}x\in \overline{S_{f^{\infty }}}$

$\iff x\in \left[{\displaystyle \underset{n\in \mathbb{N}}{\cap }{I}_{f^{n+1}}}\right]\backslash S_{f^{\infty }}$

On the other hand, the sequence of subsets of E, ${\left(I_{f^{n+1}}\right)}_{n\in \mathbb{N}}$ is strictly decreasing by inclusion, then it is convergent and the limit is $H={\displaystyle {\cap }_{n\in \mathbb{N}}{I}_{f^{n+1}}}$, and since ƒ is injective, so $f\left(H\right)=H$ then ƒ is bijective from H to itself, and $H\in S{t}_f\left(E\right)\subset S_{f^{\infty }}\left(E\right)$.

$\forall n\in \mathbb{N},E=\left[{\displaystyle \underset{i=0}{\overset{i=n}{\cup }}{f}^i\left(A_f\right)}\right]\cup \widehat{I_{f^{n+1}}}\cup S_{f^n}$

Therefore,

$E=\left[{\displaystyle \underset{n\in \mathbb{N}}{\cup }{f}^n\left(A_f\right)}\right]\cup S_{f^{\infty }}$

N.B. If $S_{f^{\infty }}=\varnothing$, then we get:

$E={\displaystyle \underset{n\in \mathbb{N}}{\cup }{f}^n(\; A\; f\; )}$

QED

Example 2

$ℝ$ is the set of real numbers, and $P\left(ℝ\right)$ is the set of subsets of $ℝ$.

Let ƒ be a map from $ℝ$ to itself, defined by:

$f\left(x\right)=\{\begin{array}{l}x+1,\text{if}x\ge 0\\ x,\text{if}x<0\end{array}$

Therefore, ƒ is injective non-surjective from $ℝ$ to itself, because ƒ is injective and $A_f=\left[0,1\right[$.

where,

$\forall n\in {\mathbb{N}}^{*},f^n\left(x\right)=\{\begin{array}{l}x+n,\text{if}x\ge 0\\ x,\text{if}x<0\end{array}$

We have, $f\left(A_f\right)=\left[f\left(0\right),f\left(1\right)\right[=\left[1,2\right[\subset {ℝ}_{+}$ since ƒ is increasing Therefore,

$\forall n\in {\mathbb{N}}^{*},f^n\left(A_f\right)=\left[n,n+1\right[$

· $\forall n\in \mathbb{N},S_{f^n}=S_{f^{\infty }}={ℝ}_{-}^{*}$

· $\forall n\in \mathbb{N},S_{f^n}\left(E\right)=S_{f^{\infty }}\left(E\right)=P\left({ℝ}_{-}^{*}\right)$

According to the theorem 1, we can write:

$ℝ={ℝ}_{-}^{*}\cup \left[{\displaystyle \underset{n\in \mathbb{N}}{\cup }\left[n,n+1\right[}\right]$

Example 3

Remark

If h is an affine map from $ℝ$ to $ℝ$, so that $h\left(x\right)=ax+b/a,b\in ℝ$ and $a\ne 0$, then:

$\forall n\in {\mathbb{N}}^{*},\forall x\in ℝ,h^n\left(x\right)=a^{n}x+b\left(1+a+\cdots +a^{n-1}\right)$

If $a\ne 1$, then:

$\forall n\in {\mathbb{N}}^{*},\forall x\in ℝ,h^n\left(x\right)=a^{n}x+b\left(\frac{1-a^n}{1-a}\right)$

Let ƒ be a map defined from $E=\left[0,4\right]$ to itself by:

$\forall x\in \left[0,4\right],f\left(x\right)=\{\begin{array}{l}-x+1,\text{if}x\in \left[0,1\right]\\ x+1,\text{if}x\in \left]1,2\right]\\ \frac{1}{2}x+2,\text{if}x\in \left[2,4\right]\end{array}$

ƒ is non-surjective injective map from E to itself, so that:

· $A_f=\left]1,2\right]$

· $Fi{x}_f=\left\{4\right\}$

· The set $A=\left[0,1\right]\subset E$, fulfills the condition $f\left(A\right)=A$

· We have $A_f=\left]1,2\right]$, $f\left(A_f\right)=f\left(\left]1,2\right]\right)=\left]2,3\right]\subset \left[2,4\right]$, $f\left(\left]2,3\right]\right)=\left]3,\frac{7}{2}\right]$, $\cdots$

· $S_{f^{\infty }}=\left\{4\right\}\cup \left[0,1\right]$

The restriction of ƒ to $\left[2,4\right]$ is an affine function such as $a=\frac{1}{2}$, and $b=2$, then:

$\forall n\in {\mathbb{N}}^{*},f^n\left(x\right)=\frac{1}{2^n}x+2\left(1+\frac{1}{2}+\cdots +\frac{1}{2^{n-1}}\right)$

We have: $f^n\left(2\right)=4-\frac{1}{2^{n-1}}$, and $f^n\left(3\right)=4-\frac{1}{2^n}$,

According to Theorem 1:

$\left[0,4\right]=\left\{\left]1,2\right]{\displaystyle {\cup }_{n\in \mathbb{N}}\left]4-\frac{1}{2^{n-1}},4-\frac{1}{2^n}\right]}\right\}\cup \left\{4\right\}\cup \left[0,1\right]$

Example 4

Let ƒ be a map defined from $E=\left[0,3\right]$ to itself by:

$\forall x\in \left[0,3\right],f\left(x\right)=\{\begin{array}{l}x+2,\text{if}x\in \left[0,1\right]\\ \frac{1}{2}x+1,\text{if}x\in \left]1,2\right[\\ x-2,\text{if}x\in \left[2,3\right]\end{array}$

ƒ is non-surjective injective from E to itself, so that:

· $A_f=\left]1,\frac{3}{2}\right]$

· $Fi{x}_f=\varnothing$

· The set $A=\left[0,1\right]\subset E$, fulfills the condition $f\left(A\right)\ne A$, and $f^2\left(A\right)=A$

· The set $B=\left[2,3\right]\subset E$, fulfills the condition $f\left(B\right)\ne B$, and $f^2\left(B\right)=B$

· $S_{f^{\infty }}=\left[0,1\right]\cup \left[2,3\right]$

We have,

· $\forall x\in \left]1,2\right[,\forall n\in {\mathbb{N}}^{*},f^n\left(x\right)=\frac{1}{2^n}x+2\left(1-\frac{1}{2^n}\right)$

· ${\lim }_{x\to 1^{-}}{f}^n\left(x\right)=2-\frac{1}{2^n}$ and $f^n\left(\frac{3}{2}\right)=2-\frac{1}{2^{n+1}}$

Therefore, according to Theorem 1:

$\left[0,3\right]=\left\{{\displaystyle {\cup }_{n\in \mathbb{N}}\left]2-\frac{1}{2^n},2-\frac{1}{2^{n+1}}\right]}\right\}\cup \left[0,1\right]\cup \left[2,3\right]$

Corollary 1

Let $f,g$ be non-surjective injective maps from an infinite set E to itself such as $S_{f^{\infty }}=\overline{S_{g^{\infty }}}$, then:

$E={\displaystyle \underset{n\in \mathbb{N}}{\cup }{B}_n}$

where, $\forall n\in \mathbb{N},B_n=f^n\left(A_f\right)\cup g^n(\; A\; g\; )$

Therefore, ${\left(B_n\right)}_{n\in \mathbb{N}}$ forms a partition of E.

Definition 2

· Let E be an infinite set, we write: $Injns\left(E\right)$ as being the set of non-surjective injective maps from E to itself.

· $Injns\left(E,F\right)$ as the set of non-surjective injective maps from a set E to a set F, that being said, E and F are supposed to be non-empty and $\left|E\right|\le \left|F\right|$ ( $\left|E\right|$ is the cardinal of E).

Properties

1) $\left(Injns\left(E\right),\circ \right)$ is a semigroup, because the composite of 2 (two) injective maps ƒ and g is injective and $I_{f\circ g}\subset I_f$, so $\forall f,g\in Injns\left(E\right)$, $f\circ g\in Injns\left(E\right)$.

2) $\forall f,g\in Injns\left(E\right)$, $A_{f\circ g}=A_f\cup f\left(A_g\right)$, $\left(A_f,f\left(A_g\right),I_{f\circ g}\right)$ is, indeed, a partition of E and $E=A_{f\circ g}\cup I_{f\circ g}$, because $f\circ g\in Injns\left(E\right)$.

3) There is no idempotent element for the law of composition in $Injns\left(E\right)$ and if such an element exists, then $f^2=f$ and since ƒ is injective, then $f=I_d$ which is contradictory, because, $f\in Injns\left(E\right)$.

4)Let $f\in Injns\left(E\right)$, and assuming that ƒ is a map from E to itself such as: $\tilde{f}\left(x\right)=\{\begin{array}{l}x,\text{if}x\in A_f\\ f\left(x\right),\text{if}x\in I_f\end{array}$

5) We have, $\forall f\in Injns\left(E\right),\tilde{f}\in Injns\left(E\right)$ knowing that $I_{\tilde{f}}=A_f\cup I_{f^2}$ and $A_{\tilde{f}}=f(\; A\; f\; )$

6) Let $f\in Injns\left(E\right)$, we have, $\tilde{f}\left(A_{\tilde{f}}\right)=\tilde{f}\left(f\left(A_f\right)\right)=f^2\left(A_f\right)$, and

$I_{{\tilde{f}}^2}=\tilde{f}\left(I_{\tilde{f}}\right)=\tilde{f}\left(A_f\cup I_{f^2}\right)=\tilde{f}\left(A_f\right)\cup \tilde{f}\left(I_{f^2}\right)=A_f\cup I_{f^3}$

Then,

$E=A_{\tilde{f}}\cup \tilde{f}\left(A_{\tilde{f}}\right)\cup I_{{\tilde{f}}^2}=A_f\cup f\left(A_f\right)\cup f^2\left(A_f\right)\cup I_{f^3}$

Therefore, $\tilde{f}$ allow us to reduce order of iteration.

7) $\forall f\in Injns\left(E\right)$,

· $A_{\tilde{f}\circ f}=A_{\tilde{f}}\cup \tilde{f}\left(A_f\right)=A_f\cup f\left(A_f\right)=A_{f^2}$, then $I_{\tilde{f}\circ f}=I_{f^2}$

· $A_{f\circ \tilde{f}}=A_f\cup f\left(A_{\tilde{f}}\right)=A_f\cup f^2\left(A_f\right)$,then, $I_{f\circ \tilde{f}}=f\left(A_f\right)\cup I_{f^3}$

· $A_{\widetilde{\stackrel{˜}{f}}}=\tilde{f}\left(A_{\tilde{f}}\right)=f^2\left(A_f\right)$, then $I_{\widetilde{\stackrel{˜}{f}}}=A_f\cup f\left(A_f\right)\cup I_{f^3}$

8) Let $f\in Injns\left(E,F\right)$ and $g\in Injns\left(F,E\right)$, so:

$f\circ g\in Injns\left(F\right)$ and $g\circ f\in Injns(\; E\; )$

2.2. Equivalence Relations

Example 5

Let $f,g\in Injns\left(E\right)$, we define the binary relation R on $Injns\left(E\right)$, by:

$fRg\iff A_f=A_f\iff I_f=I_g$

R is, indeed, an equivalence relation, because:

· R is reflexive, $fRf,\forall f\in Injns(\; E\; )$

· R is symmetric, $fRg\iff A_f=A_g\iff gRf,\forall f,g\in Injns(\; E\; )$

· R is transitive, $\forall f,g$ and $h\in Injns\left(E\right)$, $fRg$ and $gRh\iff A_f=A_g$ and $A_g=A_h\iff A_f=A_h\iff fRh$

Example 6

Let $f,g\in Injns\left(E\right)$, we define the binary relation R on $Injns\left(E\right)$, by:

$\forall f,g\in Injns\left(E\right):fRg\iff \forall n\in {\mathbb{N}}^{*},I_{f^n}=I_{g^n}$

R is, indeed, an equivalence relation, because:

· R is reflexive, $fRf,\forall f\in Injns(\; E\; )$

· R is symmetric, $fRg\iff \forall n\in {\mathbb{N}}^{*},I_{f^n}=I_{g^n}\iff gRf,\forall f,g\in Injns(\; E\; )$

· R is transitive, $f,g$ and $h\in Injns\left(E\right)$, $fRg$ and $gRh\iff \forall n\in {\mathbb{N}}^{*},I_{f^n}=I_{g^n}$ and $\forall n\in {\mathbb{N}}^{*},I_{g^n}=I_{h^n}\iff \forall n\in {\mathbb{N}}^{*},I_{f^n}=I_{h^n}\iff fRh$

Example 7

$\forall f,g\in Injns\left(E\right),fRg\iff f\left(A_f\right)=g(\; A\; g\; )$

Example 8

$\forall f,g\in Injns\left(E\right),fRg\iff S_{f^{\infty }}=S_{g^{\infty }}\iff {\displaystyle \underset{n\in \mathbb{N}}{\cup }{f}^n\left(A_f\right)}={\displaystyle \underset{n\in \mathbb{N}}{\cup }{g}^n(\; A\; g\; )}$

${\displaystyle \underset{n\in \mathbb{N}}{\cup }{f}^n\left(A_f\right)}$ : as being the ƒ-semicoverage of a set E.

3. Part II

Let E be an infinite set, $f\in Injns\left(E\right)$, and $\bar{f}=\left\{g\in Injns\left(E\right)|I_g\cap I_f\ne \varnothing \right\}$

We get the following:

· $\forall f\in Injns\left(E\right),\forall n\in {\mathbb{N}}^{*},f^n\in \bar{f}$

· $\forall f,g\in Injns\left(E\right):I_f\subset I_g\Rightarrow \bar{f}\subset \bar{g}$

Let $h\in \bar{f}$ i.e. $I_f\cap I_h\ne \varnothing$ let $x\in I_h\cap I_f$, so $x\in I_h$ and $x\in I_f\subset I_g$, so $x\in I_h$ and $x\in I_g$, so $x\in I_h\cap I_g$ so $I_h\cap I_g\ne \varnothing$, then $h\in \bar{g}$ therefore $\bar{f}\subset \bar{g}$

Theorem 2

Let E be an infinite set, then there exists a non-surjective injective map $\psi$ from E to itself, so that for any such non-surjective injective map $\varphi$ from E to itself, we have:

$I_{\psi }\cap I_{\varphi }\ne \varnothing$

Proof

By contradiction, let’s suppose that for all $\psi$ non-surjective injective maps from E to itself, there exists a map ${\varphi }_{\psi }\in Injns\left(E\right)$ such as $I_{\psi }\cap I_{{\varphi }_{\psi }}=\varnothing$ so that $I_{{\varphi }_{\psi }}\subseteq A_{\psi }$. Additionally E is an infinite set, so E is equipotential [2] to $E\backslash \left\{a\right\},\forall a\in E$. Considering this bijection ƒ as a map from E to itself, then $f\in Injns\left(E\right)$, and $A_f=\left\{a\right\}$, according to all of the above, there exists a map $f_{\varphi }\in Injns\left(E\right)$ such as $I_{f_{\varphi }}\subseteq Af=\left\{a\right\}$, which contradicts the fact that $f_{\varphi }$ injective.

QED

Note 4

· $\exists \psi \in Injns\left(E\right)$, so that: $\forall \phi \in Injns\left(E\right)$, ${\psi }^{-1}\left(I_{\phi }\right)\ne \varnothing$

· If $\psi$ applies to the former theorem then: $\bar{\psi }=Injns\left(E\right)$.

Proposition 5

$\forall f\in Injns\left(E\right)$, $\exists g\in Injns\left(E\right)\backslash \left\{f\right\}$, so that $A_f\cap A_g\ne \varnothing$

Proof

By contradiction, assuming that exists a map $f\in Injns\left(E\right)$, so that $\forall g\in Injns\left(E\right)\backslash \left\{f\right\}$, $A_f\cap A_g=\varnothing$, then $\forall g\in Injns\left(E\right)\backslash \left\{f\right\}$, $I_f\cup I_g=E$.

Let $g=f^2$, so $I_f\cup I_g=I_f⊊E$ because $f\in Injns\left(E\right)$ which is contradictory.

Proposition 6

$\forall f\in Injns\left(E\right)$, $\exists g\in Injns\left(E\right)\backslash \left\{f\right\}$, such as $A_f\cap I_g\ne \varnothing$

Proof

By contradiction, assuming that exists a map $f\in Injns\left(E\right)$, so that, $\forall g\in Injns\left(E\right)\backslash \left\{f\right\}$, $A_f\cap I_g=\varnothing$ then $\forall g\in Injns\left(E\right)\backslash \left\{f\right\},I_f\cup A_g=E$. On the other hand, if $g=\tilde{f},I_f\cup A_g=E$, additionally $A_g=f\left(A_f\right)\subset I_f$ so $I_f\cup A_g=I_f=E$ which is contradictory.

Note 5

We can define another composition law T for $Injns\left(E\right)$ so that,

$\forall f,g\in Injns\left(E\right),\left(fTg\right)\left(x\right)=\{\begin{array}{l}g\left(x\right),\text{if}x\in f^{-1}\left(I_g\right)\\ f\left(x\right),\text{if}x\in f^{-1}(\; A\; g\; )\end{array}$

· $fTf=f,\forall f\in Injns\left(E\right)$ then every element is idempotent under the law T.

· $\forall f,g\in Injns\left(E\right)$, if $I_f\cap I_g=\varnothing$, then $fTg=f$.

· $\forall f,g\in Injns\left(E\right)$, if $I_f\subseteq I_g$, then $fTg=g$.

· $\forall f\in Injns\left(E\right)$, $fT\tilde{f}=f$ and $\tilde{f}Tf=\tilde{f}$.

· Generally, $\forall f,g\in Injns\left(E\right)$, $fTg\ne gTf$.

4. Part III

4.1. Study of the Quotient Set

Let E be an infinite set, and $f,g\in Injns\left(E\right)$. We define the binary relation R on $Injns\left(E\right)$, by:

$fRg\iff A_f$ and $A_g$ have the same cardinal

The binary relation R is reflexive, symmetric and transitive, so R is an equivalence relation on $Injns\left(E\right)$.

We define $Cl\left(f\right)=\left\{g\in Injns\left(E\right)|\left|A_g\right|=\left|A_f\right|\right\}$ the equivalence class of a map ƒ.

Note 6

· Let $f\in Injns\left(E\right)$, as $A_{\tilde{f}}=f\left(A_f\right)$ so $A_f$ and $A_{\tilde{f}}$ have the same cardinal, because ƒ is injective then $\tilde{f}\in Cl\left(f\right)$, therefore,

$\forall f\in Injns\left(E\right),\tilde{f}\in Cl(\; f\; )$

· Let $f,g\in Injns\left(E\right)$, assuming that the cardinals of $A_f$ and $A_f$ are finite, and thus, $\left|A_{f\circ g}\right|=\left|A_f\cup f\left(A_g\right)\right|=\left|A_f\right|+\left|A_g\right|-\left|A_f\cap f\left(A_g\right)\right|$, and since $A_f\cap f\left(A_g\right)=\varnothing$, then: $\left|A_{f\circ g}\right|=\left|A_f\cup f\left(A_g\right)\right|=\left|A_f\right|+\left|A_g\right|$.Since the composition of two (2) maps ƒ and g on $Injns\left(E\right)$ yields a disjoint union, i.e. $A_{f\circ g}=A_f\cup f\left(A_g\right)$, with $A_f\cap f\left(A_g\right)=\varnothing$,then we can extend the sum of the cardinals even for infinite sets, such as:

$\left|A_{f\circ g}\right|=\left|A_f\cup f\left(A_g\right)\right|=\left|A_f\right|+\left|A_g\right|=\sup \left\{\left|A_f\right|,\left|A_g\right|\right\}$

· For all maps $f,g,h$ and $t\in Injns\left(E\right)$ so that $fRg$ and $hRt$ i.e. $\left|A_f\right|=\left|A_g\right|$ and $\left|A_h\right|=\left|A_t\right|$. Since:

$\left|A_{f\circ h}\right|=\sup \left\{\left|A_f\right|,\left|A_g\right|\right\}=\sup \left\{\left|A_g\right|,\left|A_t\right|\right\}=\left|A_{g\circ t}\right|$, therefore,

$f\circ gRh\circ t$

· The map’s composition law is compatible under the equivalence relation R, then we can provide the quotient set $\left(Injns\left(E\right)/R,\ast \right)$ with a structure of a semigroup.

· $\forall Cl\left(f\right)$ and $Cl\left(g\right)\in Injns\left(E\right)/R$ : $Cl\left(f\right)\ast Cl\left(g\right)=Cl\left(f\circ g\right)$

4.2. Partial Order

Let $Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)/R$, define a binary relation on $Injns\left(E\right)/R$ by:

$Cl\left(f\right)\le Cl\left(g\right)\iff \left|A_f\right|\le \left|A_g\right|$

· $\forall Cl\left(f\right)\in Injns\left(E\right)/R,Cl\left(f\right)\le Cl(\; f\; )$

So $\le$ is reflexive.

· $\forall Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)\backslash R$, $Cl\left(f\right)\le Cl\left(g\right)$ and $Cl\left(g\right)\le Cl\left(f\right)$ $\iff$ $\left|A_f\right|\le \left|A_g\right|$ and $\left|A_g\right|\le \left|A_f\right|$ $\iff$ $\left|A_f\right|=\left|A_g\right|$ so $Cl\left(f\right)=Cl(\; g\; )$

So $\le$ is asymmetric.

· $\forall Cl\left(f\right),Cl\left(g\right)$ and $Cl\left(h\right)\in Injns\left(E\right)/R$ : $Cl\left(f\right)\le Cl\left(g\right)$ and $Cl\left(g\right)\le Cl\left(h\right)$ $\iff$ $\left|A_f\right|\le \left|A_g\right|$ and $\left|A_g\right|\le \left|A_h\right|$ $\Rightarrow$ $\left|A_f\right|\le \left|A_h\right|$ $\Rightarrow$ $Cl\left(f\right)\le Cl(\; h\; )$

So $\le$ is transitive.

Therefore, the relation $\le$ is a partial order on $Injns\left(E\right)/R$.

Note 7

· Since $\forall Cl\left(f\right),Cl\left(g\right)\in Injns\left(E\right)/R$, $Cl\left(f\right)\le Cl\left(g\right)$ or $Cl\left(g\right)\le Cl\left(f\right)$ then $\le$ is a total partial order on $Injns\left(E\right)/R$.

· The partial order on the semigroup $\left(Injns\left(E\right)/R,\ast \right)$ is, indeed, compatible [3] with the equivalence class’s composition law of composition *, then:

$\forall Cl\left(f\right),Cl\left(g\right)$ and $Cl\left(h\right)\in Injns\left(E\right)/R$,

If $Cl\left(f\right)\le Cl\left(g\right)$ then $Cl\left(f\right)\ast Cl\left(h\right)=Cl\left(f\circ h\right)\le Cl\left(g\circ h\right)=Cl\left(g\right)\ast Cl\left(h\right)$ and $Cl\left(h\right)\ast Cl\left(f\right)=Cl\left(h\circ f\right)\le Cl\left(h\circ g\right)=Cl\left(h\right)\ast Cl(\; g\; )$

· $Injns\left(E\right)/R$ is well-ordered, because any non-empty subset has a smallest element.

· $Injns\left(E\right)/R$ is a lattice, because it is ordered and every pair of elements has both upper bound and lower bound [4].

· $Injns\left(E\right)/R$ provided with the order relation has a minimal element $Cl\left(f\right)$, so that $\left|A_f\right|=1$, and also maximal element $Cl\left(g\right)$, so that $A_g$ has the same cardinality as E.

· If E is an infinite countable set, the map $\phi$ defined by:

$\phi :Injns\left(E\right)\to {\mathbb{N}}^{*}\cup \left\{\aleph 0\right\}=M,\forall f\in Injns\left(E\right),\phi \left(f\right)=\left|A_f\right|$,

where, $\aleph 0$ represents the cardinal of $\mathbb{N}$.

Considering the convention: $\forall n\in {\mathbb{N}}^{*}$, $n+\aleph 0=\aleph 0$, $\phi$ is a morphism of semigroups of $\left(Injns\left(E\right),\circ \right)$ on $\left(M,+\right)$.

$\forall f,g\in Injns\left(E\right),\phi \left(f\circ g\right)=\left|A_{f\circ g}\right|=\left|A_f\right|+\left|A_g\right|=\phi \left(f\right)+\phi (\; g\; )$

Complement

Let $f,g\in Injns\left(E\right)$ so that $\left|A_f\right|<\left|A_g\right|$, with the assumption that $A_f$ and $A_g$ are considered finite, $I_f$ and $I_g$ are equipotential because, the map $\psi$ defined from $I_f$ to $I_g$ by:

$\forall x\in I_f$, $\psi \left(x\right)=\left(g\circ f^{-1}\right)\left(x\right)$ is bijective, where $f^{-1}$ is defined from $I_f$ to E, and for all $x\in I_f$, we associated its inputs by ƒ, we have:

· $\forall x,y\in I_f$, $\psi \left(x\right)=\psi \left(y\right)$ $\Rightarrow$ $\left(g\circ f^{-1}\right)\left(x\right)=\left(g\circ f^{-1}\right)\left(y\right)$ $\Rightarrow$ $g\left(f^{-1}\left(x\right)\right)=g\left(f^{-1}\left(y\right)\right)$ $\Rightarrow$ $f^{-1}\left(x\right)=f^{-1}\left(y\right)\Rightarrow x=y$ (because both g and $f^{-1}$ are injectives). Therefore $\psi$ is injective.

· On the other hand $\psi \left(I_f\right)=\left(g\circ f^{-1}\right)\left(I_f\right)=g\left(f^{-1}\left(I_f\right)\right)=g\left(E\right)=I_g$ so $\psi$ is surjective.

Let $\phi \in Injns\left(A_f,A_g\right)$ so that the map $\theta$ is defined by:

$\theta \left(x\right)=\{\begin{array}{l}\phi \left(x\right),\text{if}x\in A_f\\ \psi \left(x\right),\text{if}x\in I_f\end{array}$

Belong to $Injns\left(E\right)$ and $A_{\theta }=A_{\phi }$.

Note 8

· If $g=f^2$ so $\forall x\in I_f,\psi \left(x\right)=f\left(x\right)$ so if $\phi =f/A_f$ then we have, $\theta =f$, and if $\phi =I_d/A_f$ so $\theta =\tilde{f}$

· We considered, previously, that, $A_f$ and $A_g$ are finite. That said, we can build the $\phi$ map even in case that $A_f$ and $A_g$ are infinite and that $\left|A_f\right|=\left|A_g\right|$.

5. Part IV

E Permutations Group Action

Let $\sigma \left(E\right)$ be the permutations group of E, $f\in Injns\left(E\right)$ and $\sigma \in \sigma \left(E\right)$, $f\circ \sigma \in Injns\left(E\right)$, because, since ƒ and $\sigma$ are injective then $f\circ \sigma$ is injective and $\forall \sigma \in \sigma \left(E\right)$, $\left(f\circ \sigma \right)\left(E\right)=f\left(\sigma \left(E\right)\right)=f\left(E\right)=I_f⊊E$ then $f\circ \sigma$ is non-surjective.

where $\forall \sigma \in \sigma \left(E\right),I_{f\circ \sigma }=I_f$ then $A_{f\circ \sigma }=A_f$.

We have,

· ˆ $\forall {\sigma }_1,{\sigma }_2\in \sigma \left(E\right)$ and $f\in Injns\left(E\right)$ : $\left(f\circ {\sigma }_1\right)\circ {\sigma }_2=f\circ \left({\sigma }_1\circ {\sigma }_2\right)$

· $\forall f\in Injns\left(E\right)$, $f\circ I_d=f$

Let $\theta :\sigma \left(E\right)\times Injns\left(E\right)\to Injns(\; E\; )$

so that $\forall \sigma \in \sigma \left(E\right)$, $\forall f\in Injns\left(E\right)$, $\theta \left(\sigma ,f\right)=f\circ \sigma$

Therefore, the E permutations group operates on the rightmost on $Injns\left(E\right)$.

Note 9

The relation R defined on $Injns\left(E\right)$ by: $fRg\iff \exists \sigma \in \sigma \left(E\right)$ such as $g=f\circ \sigma$ is an equivalence relation, that is called Intransitivity relation [5].

Proposition 7

Let be, $g\in Injns\left(E\right)$ : $\exists \sigma \in \sigma \left(E\right)$ so that $g=f\circ \sigma \iff I_f=I_g$.

Proof

$\Rightarrow$ ) If there exists $\sigma \in \sigma \left(E\right)$ so that $g=f\circ \sigma \Rightarrow I_g=I_{f\circ \sigma }=I_f$.

$\Leftarrow$ ) If $I_f=I_g$, then the map $\sigma :E\to E$, so that $x\mapsto \sigma \left(x\right)=f^{-1}\left(g\left(x\right)\right)$ is bijective, because:

· $\sigma \left(E\right)=f^{-1}\left(g\left(E\right)\right)=E$ so $\sigma$ is surjective.

· $\forall x,y\in E$ : $\sigma \left(x\right)=\sigma \left(y\right)$ $\Rightarrow$ $f^{-1}\left(g\left(x\right)\right)=f^{-1}\left(g\left(y\right)\right)$ $\Rightarrow$ $g\left(x\right)=g\left(y\right)$ $\Rightarrow$ $x=y$ so $\sigma$ is injective.

· $\forall x\in E$, $f\circ \sigma \left(x\right)=f\left(\sigma \left(x\right)\right)=g(\; x\; )$

Note 10

· The equivalence class (Intransitivity relation) of the element ƒ is called the orbit of ƒ, $Cl\left(f\right)=\left\{f\circ \sigma |\sigma \in \sigma \left(E\right)\right\}=\left\{g\in Injns\left(E\right)|I_g=I_f\right\}$.

· The stabilizer of ƒ is: $\Delta f=\left\{\sigma \in \sigma \left(E\right)|f\circ \sigma =f\right\}=\left\{I_d\right\}$, i.e. the morphism associated with the said action is injective.

6. Part V

Let $f,g\in Injns\left(E\right)$.

Definition 3

ƒ and g are said to be Co-injectives, if,

$I_f\cap I_g\ne \varnothing$ and $\forall x,y\in E,f\left(x\right)=g\left(y\right)\Rightarrow x=y$

Let $\hat{f}=\left\{g\in Injns\left(E\right)|g\text{is}\text{Co-injective}\text{with}f\right\}$

We have $\hat{f}\ne \varnothing$ because $\forall f\in Injns\left(E\right)$, $I_f\cap I_f\ne \varnothing$ and $\forall x,y\in E$, $f\left(x\right)=f\left(y\right)\Rightarrow x=y$ so ƒ is Co-injective with itself.

Therefore $\forall f\in Injns\left(E\right)$, $f\in \hat{f}$, then, $\forall f\in Injns\left(E\right)$ : $\hat{f}\ne \varphi$

Proposition 8

Let $h\in Injns\left(E\right)$, $\forall f,g\in Injns\left(E\right)$ :

$f,g$ are Co-injectives $\Rightarrow$ $h\circ f$ and $h\circ g$ are Co-injectives.

Proof:

Let $h\in Injns\left(E\right)$, and $f,g\in Injns\left(E\right)$ such as ƒ and g are Co-injective. $I_f\cap I_g\ne \varnothing$, additionally $h\left(I_f\cap I_g\right)=I_{h\circ f}\cap I_{h\circ g}\ne \varnothing$ (h is injective).

Let $x,y\in E$, $h\circ f\left(x\right)=h\circ g\left(y\right)\Rightarrow f\left(x\right)=g\left(y\right)\Rightarrow x=y$ (because h is injective and $f,g$ are Co-injectives).

Therefore $h\circ f$ and $h\circ g$ are Co-injectives.

Note 11

For all $f,g\in Injns\left(E\right)$, so that $f,g$ are Co-injectives, so: ˆ

· $f^2$ and $f\circ g$ are Co-injectives

· $\forall z\in I_f\cap I_g$, $\exists !x\in E$, so that, $z=f\left(x\right)=g(\; x\; )$

· $f^{-1}\left(I_g\right)=g^{-1}(\; I\; f\; )$

· If, $I_f=I_g$, then $f=g$ ˆ

· Let $h\in Injns\left(E\right)$, if $f\left(I_h\right)\cap g\left(I_h\right)\ne \varnothing$, then $f\circ h$ and $g\circ h$ are Co-injectives.

Proposition 9

$\forall f,g\in Injns\left(E\right)$, if $I_f⊊I_g$, then $f,g$ are not Co-injective.

Proof

$\forall f,g\in Injns\left(E\right)$, suppose that $I_f⊊I_g$, then we have $I_g=I_f\cup \left(I_g\backslash I_f\right)$. If $f,g$ are Co-injectives, then $f^{-1}\left(I_g\right)=g^{-1}\left(I_f\right)=E$, which is contradictory because:

$E=g^{-1}\left(I_f\right)\cup g^{-1}\left(I_g\backslash I_f\right)$, and $g^{-1}\left(I_g\backslash I_f\right)\ne \varnothing$, so $g^{-1}\left(I_f\right)⊊E$.

Proposition 10

Let $f,g\in Injns\left(E\right)$, so that ƒ and g are Co-injectives, $\forall h\in Injns\left(E\right)$, if g and h are Co-injectives and $I_f\cap I_h=I_f\cap I_g$ then ƒ and h are Co-injectives.

Proof

$I_f\cap I_h\ne \varnothing$, let $x,y\in E$, so that $f\left(x\right)=h\left(y\right)$. We have $f\left(x\right)=h\left(y\right)\in I_f\cap I_h=I_f\cap I_g$, then $f\left(x\right)=g\left(x\right)=h\left(y\right)$ (because $f,g$ are Co-injectives), and since g and h are Co-injectives then $x=y$. So $\forall x,y\in E$, $f\left(x\right)=h\left(y\right)\Rightarrow x=y$. Therefore $f,h$ are Co-injectives.

Acknowledgements

I would like to thank my dear friends, especially Chafik Bouazzaoui for us discussing the Cantor-Bernstein theorem and Mai Mohammed mainly who for his help with translating this document.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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