Proposition 2

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

A f = { y ∈ E | ∀ x ∈ E , f ( x ) ≠ y }

Then,

∀ n ∈ ℕ , E = [ ∪ i = 0 i = n f i ( A f ) ] ∪ I f n + 1

where, f i ( A f ) = f ∘ f ∘ ⋯ ∘ f ( A f ) ︸ i   times

Proof

By induction:

For n = 0 , we have, by definition E = A f ∪ I f , and for n = 1 the proposition 1 states that, if E = F and f = g then, E = A f ∪ f ( A f ) ∪ I f 2 Suppose that the said property is true for n. Then, by composing by ƒ, we get:

I f = f ( E ) = [ ∪ i = 0 i = n f i + 1 ( A f ) ] ∪ I f n + 2 = [ ∪ i = 1 i = n + 1 f i ( A f ) ] ∪ I f n + 2

Then,

E = A f ∪ [ ∪ i = 1 i = n + 1 f i ( A f ) ] ∪ I f n + 2 = [ ∪ i = 0 i = n + 1 f i ( A f ) ] ∪ I f n + 2

Therefore,

∀ n ∈ ℕ , E = [ ∪ i = 0 i = n f i ( A f ) ] ∪ I f n + 1

Note 2

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

∀ n ∈ ℕ , A f n + 1 = ∪ i = 0 i = n f i ( A f )

Definition 1

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

· x ∈ E is a fixed point of ƒ if, f ( x ) = x .

· A ⊂ E is a fixed point set of ƒ if, f ( A ) = A .

Proposition 3

For all non-surjective injective maps f from an infinite set E to itself, ∀ n ∈ ℕ , f n ( A f ) contains no fixed points of ƒ.

Proof

By induction:

For n = 0 , let x ∈ A f , and by definition ∀ y ∈ E , f ( y ) ≠ x , particularly for y = x so f ( x ) ≠ x then A f contains no fixed points of f. Suppose that the aforementioned property is true for n, let x ∈ f n + 1 ( A f ) = f ( f n ( A f ) ) , then ∃ y ∈ f n ( A f ) such that x = f ( y ) , we have f ( y ) ≠ y (by inductive hypothesis), since ƒ is injective then f ( f ( y ) ) ≠ f ( y ) so f ( x ) ≠ x , then ∀ x ∈ f n + 1 ( A f ) , f ( x ) ≠ x . Therefore, ∀ n ∈ ℕ , f n ( A f ) 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:

F i x f = { x ∈ E | f ( x ) = x } and S t f ( E ) = { A ⊂ E | f ( A ) = A }

· ∀ n ∈ ℕ , F i x f ∩ f n ( A f ) = ∅

· ∀ n ∈ ℕ * , F i x f ⊆ F i x f n

· ∀ n ∈ ℕ * , F i x f n ⊆ F i x f 2 n

· ∀ p ∈ ℕ , ∀ n ∈ ℕ * , f p ( F i x f n ) = F i x f n

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

F i x f ∈ S t f ( E )

Proposition 4

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

∀ A ∈ S t f ( E ) , ∀ n ∈ ℕ , f n ( A f ) ∩ A = ∅

Proof

Let A ∈ S t f ( E ) , by induction, For n = 0 , we have, by definition A f ∩ I f = ∅ then A f ∩ A = A f ∩ f ( A ) = ∅ ˆ Suppose that the said property is true for n ∈ ℕ , let x ∈ f n + 1 ( A f ) ∩ A , then, ∃ y ∈ f n ( A f ) , x = f ( y ) and ∃ y 0 ∈ A , x = f ( y 0 ) . Since ƒ is injective so y = y 0 ∈ f n ( A f ) ∩ A , which contradicts the inductive hypothesis. Then,

f n + 1 ( A f ) ∩ A = ∅ .

Therefore, ∀ A ∈ S t f ( E ) , ∀ n ∈ ℕ , f n ( A f ) ∩ A = ∅

Lemma 1 (Generalization)

∀ p ∈ ℕ * , ∀ A ∈ S t f p ( E ) , ∀ n ∈ ℕ , f n ( A f ) ∩ A = ∅

Proof

By complete (strong) induction,

Let p ∈ ℕ * , and A ∈ S t f p ( E ) = { A ⊂ E | f p ( A ) = A } : For n = 0 , we have, by definition A f ∩ I f = ∅ , since ∃ p ∈ ℕ * , I f p ⊆ I f , then: A f ∩ A = A f ∩ f p ( A ) = ∅

Suppose that the said property is true for all i ∈ { 1 , ⋯ , n } , let x ∈ f n + 1 ( A f ) ∩ A , then, ∃ y ∈ A f , x = f n + 1 ( y ) and ∃ y 0 ∈ A , x = f p ( y 0 ) . Since ƒ is injective then,

· f n − p + 1 ( y ) = y 0 , if n + 1 > p , which contradicts the inductive hypothesis, because i = n − p + 1 ∈ { 1 , ⋯ , n }

· y = f p − n − 1 ( y 0 ) , if p > n + 1 , which contradicts A f ∩ I f p − n − 1 = ∅

· y = y 0 , if n + 1 = p , which contradicts A f ∩ A = ∅

Then,

f n + 1 ( A f ) ∩ A = ∅ ,

Therefore,

∀ p ∈ ℕ * , ∀ A ∈ S t f p ( E ) , ∀ n ∈ ℕ , f n ( A f ) ∩ A = ∅

QED

For all n ∈ ℕ , and for all ƒ non-surjective injective maps from an infinite set E to itself, we define the following:

· S f n ( E ) = { A ⊂ E | ∃ p ∈ { 1 , ⋯ , n + 1 } , f p ( A ) = A }

· S f ∞ ( E ) = { A ⊂ E | ∃ p ∈ ℕ * , f p ( A ) = A }

· S f n = { x ∈ A | A ∈ S f n ( E ) }

· S f ∞ = { x ∈ A | A ∈ S f ∞ ( E ) }

Theorem 1

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

E = [ ∪ n ∈ ℕ f n ( A f ) ] ∪ S f ∞

Proof

Note that for n = 0 , S f 0 ( E ) = S t f ( E ) , The sequence of subsets of E, ( S f n ) n ∈ ℕ is increasing by inclusion, so it is convergent, the limit is: ∪ n ∈ ℕ S f n = S f ∞

We have already established that,

∀ n ∈ ℕ , E = [ ∪ i = 0 i = n f i ( A f ) ] ∪ I f n + 1

Otherly, according to the Lemma 1, ∀ n ∈ ℕ , E = [ ∪ i = 0 i = n f i ( A f ) ] ∩ S f n = ∅

Let’s define the following, ∀ n ∈ ℕ , I f n + 1 ^ = I f n + 1 \ S f n the sequence of subsets of ( I f n + 1 ^ ) n ∈ ℕ is decreasing by inclusion then, it is convergent [1], the limit is:

∩ n ∈ ℕ I f n + 1 ^ = ∅

Because, let x ∈ ∩ n ∈ ℕ I f n + 1 ^ , then:

x ∈ ∩ n ∈ ℕ I f n + 1 \ S f n ⇔ ∀ n ∈ ℕ , x ∈ I f n + 1 \ S f n ⇔ ∀ n ∈ ℕ , x ∈ I f n + 1   and   x ∈ S f n ¯

⇔ ∀ n ∈ ℕ , x ∈ I f n + 1   and   ∀ n ∈ ℕ , x ∈ S f n ¯ ⇔ x ∈ ∩ n ∈ ℕ I f n + 1   and   x ∈ ∩ n ∈ ℕ S f n ¯

⇔ x ∈ ∩ n ∈ ℕ I f n + 1   and   x ∈ ∪ n ∈ ℕ S f n ¯ ⇔ x ∈ ∩ n ∈ ℕ I f n + 1   and   x ∈ S f ∞ ¯

⇔ x ∈ [ ∩ n ∈ ℕ I f n + 1 ] \ S f ∞

On the other hand, the sequence of subsets of E, ( I f n + 1 ) n ∈ ℕ is strictly decreasing by inclusion, then it is convergent and the limit is H = ∩ n ∈ ℕ I f n + 1 , and since ƒ is injective, so f ( H ) = H then ƒ is bijective from H to itself, and H ∈ S t f ( E ) ⊂ S f ∞ ( E ) .

∀ n ∈ ℕ , E = [ ∪ i = 0 i = n f i ( A f ) ] ∪ I f n + 1 ^ ∪ S f n

Therefore,

E = [ ∪ n ∈ ℕ f n ( A f ) ] ∪ S f ∞

N.B. If S f ∞ = ∅ , then we get:

E = ∪ n ∈ ℕ f n ( A f )

QED

Example 2

ℝ is the set of real numbers, and P ( ℝ ) is the set of subsets of ℝ .

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

f ( x ) = { x + 1 ,     if   x ≥ 0 x ,             if   x < 0

Therefore, ƒ is injective non-surjective from ℝ to itself, because ƒ is injective and A f = [ 0 , 1 [ .

where,

∀ n ∈ ℕ * , f n ( x ) = { x + n ,       if   x ≥ 0 x ,               if   x < 0

We have, f ( A f ) = [ f ( 0 ) , f ( 1 ) [ = [ 1 , 2 [ ⊂ ℝ + since ƒ is increasing Therefore,

∀ n ∈ ℕ * , f n ( A f ) = [ n , n + 1 [

· ∀ n ∈ ℕ , S f n = S f ∞ = ℝ − *

· ∀ n ∈ ℕ , S f n ( E ) = S f ∞ ( E ) = P ( ℝ − * )

According to the theorem 1, we can write:

ℝ = ℝ − * ∪ [ ∪ n ∈ ℕ [ n , n + 1 [ ]

Example 3

Remark

If h is an affine map from ℝ to ℝ , so that h ( x ) = a x + b / a , b ∈ ℝ and a ≠ 0 , then:

∀ n ∈ ℕ * , ∀ x ∈ ℝ , h n ( x ) = a n x + b ( 1 + a + ⋯ + a n − 1 )

If a ≠ 1 , then:

∀ n ∈ ℕ * , ∀ x ∈ ℝ , h n ( x ) = a n x + b ( 1 − a n 1 − a )

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

∀ x ∈ [ 0 , 4 ] , f ( x ) = { − x + 1 ,       if   x ∈ [ 0 , 1 ] x + 1 ,           if   x ∈ ] 1 , 2 ] 1 2 x + 2 ,     if   x ∈ [ 2 , 4 ]

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

· A f = ] 1 , 2 ]

· F i x f = { 4 }

· The set A = [ 0 , 1 ] ⊂ E , fulfills the condition f ( A ) = A

· We have A f = ] 1 , 2 ] , f ( A f ) = f ( ] 1 , 2 ] ) = ] 2 , 3 ] ⊂ [ 2 , 4 ] , f ( ] 2 , 3 ] ) = ] 3 , 7 2 ] , ⋯

· S f ∞ = { 4 } ∪ [ 0 , 1 ]

The restriction of ƒ to [ 2 , 4 ] is an affine function such as a = 1 2 , and b = 2 , then:

∀ n ∈ ℕ * , f n ( x ) = 1 2 n x + 2 ( 1 + 1 2 + ⋯ + 1 2 n − 1 )

We have: f n ( 2 ) = 4 − 1 2 n − 1 , and f n ( 3 ) = 4 − 1 2 n ,

According to Theorem 1:

[ 0 , 4 ] = { ] 1 , 2 ] ∪ n ∈ ℕ ] 4 − 1 2 n − 1 , 4 − 1 2 n ] } ∪ { 4 } ∪ [ 0 , 1 ]

Example 4

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

∀ x ∈ [ 0 , 3 ] , f ( x ) = { x + 2 ,             if   x ∈ [ 0 , 1 ] 1 2 x + 1 ,     if   x ∈ ] 1 , 2 [ x − 2 ,         if   x ∈ [ 2 , 3 ]

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

· A f = ] 1 , 3 2 ]

· F i x f = ∅

· The set A = [ 0 , 1 ] ⊂ E , fulfills the condition f ( A ) ≠ A , and f 2 ( A ) = A

· The set B = [ 2 , 3 ] ⊂ E , fulfills the condition f ( B ) ≠ B , and f 2 ( B ) = B

· S f ∞ = [ 0 , 1 ] ∪ [ 2 , 3 ]

We have,

· ∀ x ∈ ] 1 , 2 [ , ∀ n ∈ ℕ * , f n ( x ) = 1 2 n x + 2 ( 1 − 1 2 n )

· lim x → 1 − f n ( x ) = 2 − 1 2 n and f n ( 3 2 ) = 2 − 1 2 n + 1

Therefore, according to Theorem 1:

[ 0 , 3 ] = { ∪ n ∈ ℕ ] 2 − 1 2 n , 2 − 1 2 n + 1 ] } ∪ [ 0 , 1 ] ∪ [ 2 , 3 ]

Corollary 1

Let f , g be non-surjective injective maps from an infinite set E to itself such as S f ∞ = S g ∞ ¯ , then:

E = ∪ n ∈ ℕ B n

where, ∀ n ∈ ℕ , B n = f n ( A f ) ∪ g n ( A g )

Therefore, ( B n ) n ∈ ℕ forms a partition of E.

Definition 2

· Let E be an infinite set, we write: I n j n s ( E ) as being the set of non-surjective injective maps from E to itself.

· I n j n s ( E , F ) 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 | E | ≤ | F | ( | E | is the cardinal of E).

Properties

1) ( I n j n s ( E ) , ∘ ) is a semigroup, because the composite of 2 (two) injective maps ƒ and g is injective and I f ∘ g ⊂ I f , so ∀ f , g ∈ I n j n s ( E ) , f ∘ g ∈ I n j n s ( E ) .

2) ∀ f , g ∈ I n j n s ( E ) , A f ∘ g = A f ∪ f ( A g ) , ( A f , f ( A g ) , I f ∘ g ) is, indeed, a partition of E and E = A f ∘ g ∪ I f ∘ g , because f ∘ g ∈ I n j n s ( E ) .

3) There is no idempotent element for the law of composition in I n j n s ( E ) and if such an element exists, then f 2 = f and since ƒ is injective, then f = I d which is contradictory, because, f ∈ I n j n s ( E ) .

4)Let f ∈ I n j n s ( E ) , and assuming that ƒ is a map from E to itself such as: f ˜ ( x ) = { x ,                 if   x ∈ A f f ( x ) ,     if   x ∈ I f

5) We have, ∀ f ∈ I n j n s ( E ) , f ˜ ∈ I n j n s ( E ) knowing that I f ˜ = A f ∪ I f 2 and A f ˜ = f ( A f )

6) Let f ∈ I n j n s ( E ) , we have, f ˜ ( A f ˜ ) = f ˜ ( f ( A f ) ) = f 2 ( A f ) , and

I f ˜ 2 = f ˜ ( I f ˜ ) = f ˜ ( A f ∪ I f 2 ) = f ˜ ( A f ) ∪ f ˜ ( I f 2 ) = A f ∪ I f 3

Then,

E = A f ˜ ∪ f ˜ ( A f ˜ ) ∪ I f ˜ 2 = A f ∪ f ( A f ) ∪ f 2 ( A f ) ∪ I f 3

Therefore, f ˜ allow us to reduce order of iteration.

7) ∀ f ∈ I n j n s ( E ) ,

· A f ˜ ∘ f = A f ˜ ∪ f ˜ ( A f ) = A f ∪ f ( A f ) = A f 2 , then I f ˜ ∘ f = I f 2

· A f ∘ f ˜ = A f ∪ f ( A f ˜ ) = A f ∪ f 2 ( A f ) ,then, I f ∘ f ˜ = f ( A f ) ∪ I f 3

· A f ˜ ˜ = f ˜ ( A f ˜ ) = f 2 ( A f ) , then I f ˜ ˜ = A f ∪ f ( A f ) ∪ I f 3

8) Let f ∈ I n j n s ( E , F ) and g ∈ I n j n s ( F , E ) , so:

f ∘ g ∈ I n j n s ( F ) and g ∘ f ∈ I n j n s ( E )

Example 5

Let f , g ∈ I n j n s ( E ) , we define the binary relation R on I n j n s ( E ) , by:

f R g ⇔ A f = A f ⇔ I f = I g

R is, indeed, an equivalence relation, because:

· R is reflexive, f R f , ∀ f ∈ I n j n s ( E )

· R is symmetric, f R g ⇔ A f = A g ⇔ g R f , ∀ f , g ∈ I n j n s ( E )

· R is transitive, ∀ f , g and h ∈ I n j n s ( E ) , f R g and g R h ⇔ A f = A g and A g = A h ⇔ A f = A h ⇔ f R h

Example 6

Let f , g ∈ I n j n s ( E ) , we define the binary relation R on I n j n s ( E ) , by:

∀ f , g ∈ I n j n s ( E ) : f R g ⇔ ∀ n ∈ ℕ * , I f n = I g n

R is, indeed, an equivalence relation, because:

· R is reflexive, f R f , ∀ f ∈ I n j n s ( E )

· R is symmetric, f R g ⇔ ∀ n ∈ ℕ * , I f n = I g n ⇔ g R f , ∀ f , g ∈ I n j n s ( E )

· R is transitive, f , g and h ∈ I n j n s ( E ) , f R g and g R h ⇔ ∀ n ∈ ℕ * , I f n = I g n and ∀ n ∈ ℕ * , I g n = I h n ⇔ ∀ n ∈ ℕ * , I f n = I h n ⇔ f R h

Example 7

∀ f , g ∈ I n j n s ( E ) , f R g ⇔ f ( A f ) = g ( A g )

Example 8

∀ f , g ∈ I n j n s ( E ) , f R g ⇔ S f ∞ = S g ∞ ⇔ ∪ n ∈ ℕ f n ( A f ) = ∪ n ∈ ℕ g n ( A g )

∪ n ∈ ℕ f n ( A f ) : as being the ƒ-semicoverage of a set E.

Let E be an infinite set, and f , g ∈ I n j n s ( E ) . We define the binary relation R on I n j n s ( E ) , by:

f R g ⇔ 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 I n j n s ( E ) .

We define C l ( f ) = { g ∈ I n j n s ( E ) | | A g | = | A f | } the equivalence class of a map ƒ.

Note 6

· Let f ∈ I n j n s ( E ) , as A f ˜ = f ( A f ) so A f and A f ˜ have the same cardinal, because ƒ is injective then f ˜ ∈ C l ( f ) , therefore,

∀ f ∈ I n j n s ( E ) , f ˜ ∈ C l ( f )

· Let f , g ∈ I n j n s ( E ) , assuming that the cardinals of A f and A f are finite, and thus, | A f ∘ g | = | A f ∪ f ( A g ) | = | A f | + | A g | − | A f ∩ f ( A g ) | , and since A f ∩ f ( A g ) = ∅ , then: | A f ∘ g | = | A f ∪ f ( A g ) | = | A f | + | A g | .Since the composition of two (2) maps ƒ and g on I n j n s ( E ) yields a disjoint union, i.e. A f ∘ g = A f ∪ f ( A g ) , with A f ∩ f ( A g ) = ∅ ,then we can extend the sum of the cardinals even for infinite sets, such as:

| A f ∘ g | = | A f ∪ f ( A g ) | = | A f | + | A g | = sup { | A f | , | A g | }

· For all maps f , g , h and t ∈ I n j n s ( E ) so that f R g and h R t i.e. | A f | = | A g | and | A h | = | A t | . Since:

| A f ∘ h | = sup { | A f | , | A g | } = sup { | A g | , | A t | } = | A g ∘ t | , therefore,

f ∘ g R h ∘ t

· The map’s composition law is compatible under the equivalence relation R, then we can provide the quotient set ( I n j n s ( E ) / R , ∗ ) with a structure of a semigroup.

· ∀ C l ( f ) and C l ( g ) ∈ I n j n s ( E ) / R : C l ( f ) ∗ C l ( g ) = C l ( f ∘ g )

Let C l ( f ) , C l ( g ) ∈ I n j n s ( E ) / R , define a binary relation on I n j n s ( E ) / R by:

C l ( f ) ≤ C l ( g ) ⇔ | A f | ≤ | A g |

· ∀ C l ( f ) ∈ I n j n s ( E ) / R , C l ( f ) ≤ C l ( f )

So ≤ is reflexive.

· ∀ C l ( f ) , C l ( g ) ∈ I n j n s ( E ) \ R , C l ( f ) ≤ C l ( g ) and C l ( g ) ≤ C l ( f ) ⇔ | A f | ≤ | A g | and | A g | ≤ | A f | ⇔ | A f | = | A g | so C l ( f ) = C l ( g )

So ≤ is asymmetric.

· ∀ C l ( f ) , C l ( g ) and C l ( h ) ∈ I n j n s ( E ) / R : C l ( f ) ≤ C l ( g ) and C l ( g ) ≤ C l ( h ) ⇔ | A f | ≤ | A g | and | A g | ≤ | A h | ⇒ | A f | ≤ | A h | ⇒ C l ( f ) ≤ C l ( h )

So ≤ is transitive.

Therefore, the relation ≤ is a partial order on I n j n s ( E ) / R .

Note 7

· Since ∀ C l ( f ) , C l ( g ) ∈ I n j n s ( E ) / R , C l ( f ) ≤ C l ( g ) or C l ( g ) ≤ C l ( f ) then ≤ is a total partial order on I n j n s ( E ) / R .

· The partial order on the semigroup ( I n j n s ( E ) / R , ∗ ) is, indeed, compatible [3] with the equivalence class’s composition law of composition *, then:

∀ C l ( f ) , C l ( g ) and C l ( h ) ∈ I n j n s ( E ) / R ,

If C l ( f ) ≤ C l ( g ) then C l ( f ) ∗ C l ( h ) = C l ( f ∘ h ) ≤ C l ( g ∘ h ) = C l ( g ) ∗ C l ( h ) and C l ( h ) ∗ C l ( f ) = C l ( h ∘ f ) ≤ C l ( h ∘ g ) = C l ( h ) ∗ C l ( g )

· I n j n s ( E ) / R is well-ordered, because any non-empty subset has a smallest element.

· I n j n s ( E ) / R is a lattice, because it is ordered and every pair of elements has both upper bound and lower bound [4].

· I n j n s ( E ) / R provided with the order relation has a minimal element C l ( f ) , so that | A f | = 1 , and also maximal element C l ( g ) , so that A g has the same cardinality as E.

· If E is an infinite countable set, the map φ defined by:

φ : I n j n s ( E ) → ℕ * ∪ { ℵ 0 } = M , ∀ f ∈ I n j n s ( E ) , φ ( f ) = | A f | ,

where, ℵ 0 represents the cardinal of ℕ .

Considering the convention: ∀ n ∈ ℕ * , n + ℵ 0 = ℵ 0 , φ is a morphism of semigroups of ( I n j n s ( E ) , ∘ ) on ( M , + ) .

∀ f , g ∈ I n j n s ( E ) , φ ( f ∘ g ) = | A f ∘ g | = | A f | + | A g | = φ ( f ) + φ ( g )

Complement

Let f , g ∈ I n j n s ( E ) so that | A f | < | A g | , with the assumption that A f and A g are considered finite, I f and I g are equipotential because, the map ψ defined from I f to I g by:

∀ x ∈ I f , ψ ( x ) = ( g ∘ f − 1 ) ( x ) is bijective, where f − 1 is defined from I f to E, and for all x ∈ I f , we associated its inputs by ƒ, we have:

· ∀ x , y ∈ I f , ψ ( x ) = ψ ( y ) ⇒ ( g ∘ f − 1 ) ( x ) = ( g ∘ f − 1 ) ( y ) ⇒ g ( f − 1 ( x ) ) = g ( f − 1 ( y ) ) ⇒ f − 1 ( x ) = f − 1 ( y ) ⇒ x = y (because both g and f − 1 are injectives). Therefore ψ is injective.

· On the other hand ψ ( I f ) = ( g ∘ f − 1 ) ( I f ) = g ( f − 1 ( I f ) ) = g ( E ) = I g so ψ is surjective.

Let φ ∈ I n j n s ( A f , A g ) so that the map θ is defined by:

θ ( x ) = { φ ( x ) ,     if   x ∈ A f ψ ( x ) ,     if   x ∈ I f

Belong to I n j n s ( E ) and A θ = A φ .

Note 8

· If g = f 2 so ∀ x ∈ I f , ψ ( x ) = f ( x ) so if φ = f / A f then we have, θ = f , and if φ = I d / A f so θ = f ˜

· We considered, previously, that, A f and A g are finite. That said, we can build the φ map even in case that A f and A g are infinite and that | A f | = | A g | .