Research Article
Pre A*-Homomorphism
Department of Mathematics, Andhra Loyola Institute of Engineering and Technology, Vijayawada-8, A.P. India
In a draft paper (Manes, 1989), the equational theory of disjoint alternatives, around 1989, Manes introduced the concept of Ada (Algebra of disjoint alternatives) (A, ∧, V, (-)I, (-)π, 0, 1, 2) (Where ∧,V are binary operations on A, (-)I, (-)π are unary operations and 0, 1, 2 are distinguished elements on A) which is however differ from the definition of the Ada of his later paper (Manes, 1993) Adas and the equational theory of if-then-else in 1993. While the Ada of the earlier draft seems to be based on extending the If-Then-Else concept more on the basis of Boolean algebras and the later concept is based on C-algebras (A, ∧,V, (-) ~ )) (where ∧,V are binary operations on A, (-) ~ is a unary operation) introduced by Guzman and Squier (1990). Rao (1994) first introduced the concept of A*-algebra (A, ∧, V, *, (-) ~, (-)π, 0, 1, 2)) (where ∧,V,* are binary operations on A, (-) ~ ,(-)π are unary opePrarooparations and 0, 1, 2 are distinguished elements on A) not only studied the equivalence with Ada, C-algebra, Adas connection with 3-ring, stone type representation but also introduced the concept of A*-clone, the If-Then-Else structure over A*-algebra and Ideal of A*-algebra. Rao (2000) introduced the concept pre A*-algebra (A, ∨, ∧, (-)~) (where ∧, V are binary operations on A, (-)~ is a unary operation on A analogous to C-algebra as a reduct of A*-algebra, studied their subdirect representations, obtained the results that 2 = {0, 1} and 3 = {0, 1, 2} are the subdirectly irreducible pre-A*-algebras and every pre-A*-algebra can be imbedded in 3X( where 3X is the set of all mappings from a nonempty set X into 3 = {0, 1, 2}). Praroopa (2012) introduced the specific concepts on pre A*-algebra and of the papers Praroopa and Rao (2011a), Praroopa and Rao (2011b) studied pre A*-algebra as a semilattice, lattice in pre A*-algebra.
PRELIMINARIES
Definition: Pre A*-Algebra (Rao, 2000): An algebra (A, ∨, ∧, (-)~) satisfying:
• | (x~)~ = x, ∀x∈A |
• | x∧x = x, ∀x∈A |
• | x∧y = y∧x, ∀ x, y∈A |
• | (x∧y)~ = x~∨y~, ∀x, y∈A |
• | x∧(y∧z) = (x∧y)∧z , ∀ x, y, z∈A |
• | x∧(y∨z) = (x∧y)∨(x∧z), ∀x, y, z∈A |
• | x∧y = x∧(x~∨y), ∀x, y∈A |
is called a pre A*-algebra.
Definition: (Praroopa, 2012) pre A*- Homomorphism: Let (A1, ∧, ∨, (-)~) and (A2, ∧, ∨, (-)~) be two pre A*-algebras. A mapping f: A1→A2 is called an pre A*homomorphism, if:
• | The homomorphism f: A1→A2 is onto, then f is called epimorphism |
• | The homorphism f: A1→A2 is one-one, then f is called monomorphism |
• | The homomorphism f : A1→A2 is one-one and onto then f is called an isomorphism and A1, A2 are isomorphic, denoted by A1 ≅ A2 |
Definition: (Praroopa, 2012) kernel of pre A*-homomorphism: By the definition of pre A*-homomorphism, define Kernel of pre A*-homomorphism Let A1, A2 be two pre A*-algebras and f: A1→A2be a pre A*-homomorphism then the set {x∈A/f(x) = 0} is called the kernel of f and it is denoted by kerf.
Example: Let A be a pre A*-algebra with 1, 0. Suppose that for every x∈A-{0, 1}, x∨x~≠1. Define f:A→{0, 1, 2} by f(1) = 1, f(0) = 0 and f(x) = 2 if x≠0, 1. Then f is a pre A*-homomorphism.
Theorems on Pre A*-homomorphism (Praroopa, 2012): Theorem: Let f:A→B be a pre A*-homomorphism from a pre A*-algebra A into a pre A*-algebra B ad Kerf = {x∈A/f(x) = 0} is the kernel then kerf = {0} if and only if f is one- one.
Proof: - Suppose Kerf = {0}
To show that f is one-one:
For any x, y∈A, consider f(x) = f(y)
⇒ | f(x)-f(y) = 0 |
⇒f(x-y) = 0 | |
⇒x-y∈Kerf = {0} | |
⇒x-y = 0 | |
⇒f(x)-f(y) = 0 | |
⇒x = y, ∀x, y∈A |
Therefore f is one-one
Converse: Suppose that f is one-one
⇒f(x) = f(y) ⇒ x = y, ∀ x, y∈A
To show that Kert f = {0}
Let x∈ Kerf
⇒f(x) = 0
⇒ x = 0 (since f is one-one)
Therefore Kerf = {0}
Lemma (Rao, 2000): Let f: A1→A2 be pre A*-homomorphism where A1, A2 are pre A*-algebras with 11 and 12-Then:
(i) | If A1 has the element 2, then f(2) is the element of A2 |
(ii) | If a∈B(A1), then f(a)∈B(A2) |
Where:
B(A1) = {x/x∨ x ~ =1}
B(A2) = {x/x∨ x ~ =1}
Note: If f:A→B and g:B→C are pre A*-homomorphisms.
So their composition or product gof: A→C which is defined by gof (a) = g(f(a) is also pre A*-homomorphism.
Proposition (Praroopa, 2012): If f:A→B and g:B→C are pre A*-homomorphisms. Then:
(i) | If f and g are mono, so is gof |
(ii) | If f and g are epi, so is gof |
(iii) | If gof is mono, so is f |
(iv) | If gof is epi, so is g |
Proof: If f:A→B and g:B→C are pre A*-homomorphisms. Then gof: A→C is also a pre A*-homomorphisms.
(i) | Suppose f, g are one-one |
Now suppose (gof)(a1) = (gof)(a2)
⇒g(f(a1)) = g(f(a2))
⇒f(a1) = f(a2) ( Since g is oneone)
⇒a1 = a2 (Since f is one-one)
Therefore gof is mono |
(ii) | Suppose f, g are onto |
Let c∈C, since g is onto, there exists b∈B such that g(b) = c. | |
Since b∈B and f is onto there exists a∈A such that f(a) = b. | |
Therefore, g(b) = g(f(a)) = c, = (gof)(a) = c | |
Hence, for c∈C, there exists a∈A such that (gof)(a) = c | |
This is true for every c∈C | |
Therefore gof is onto | |
Therefore gof is epimorphism. |
(iii) | Suppose gof is mono i.e., gof is one-one |
We have to show f is one-one | |
Suppose f(a1) = f(a2) | |
⇒g(f(a1)) = g(f(a2)) | |
⇒(gof) (a1)) = (gof)(a2) | |
⇒a1 = a2 (Since gof is one-one ) | |
Therefore f is one-one. | |
Hence, f is mono. |
(iv) | Suppose gof is epi⇒ gof is onto |
We have to show g is onto | |
Since gof: A→C is onto, for any c∈C, there exist a∈A such that | |
(gof)(a) = c | |
⇒g(f(a)) = c where f(a)∈B | |
⇒f(a) = b, where b∈B | |
Therefore g(b) = c, for some b∈B | |
Therefore for c∈C, ∃b∈B∋g(b) = c , | |
this is true for all c∈C | |
Hence g is onto | |
Thus g is epimorphism. |
Corollary (Praroopa, 2012): The pre A*-homomorphisms f:A→B is an isomorphism, if and only if, there exists a pre A*-homomorphism g:B→C such that fog is an automorphism of B and gof is an automorphism of A.
Proof: Suppose that the pre A*-homomorphism f:A→B is an isomorphism.
i.e., f is a bijection | |
Then f-1:B→A is a bijection such that fof-1 = IB | |
and f-1of = IA. | |
Hence, f-1:B→A is a mapping such that fof-1 = IB which is an automoptism of B. | |
And f-1of = IA which is an automorphism of A | |
Now we show that f-1 is a Pre A*-homomorphism : | |
Since f is Pre A*-homomorphism, we have f(1) = 1 | |
⇒f-1(1) = 1 | |
Let b1, b2∈B. | |
Since f:A→B is isomorphism, we have f is onto | |
Then ∃ a1, a2∈A ∋ f(a1) = b1, f(a2) = b2. | |
Therefore f(a1va2) = f(a1) v f(a2) | |
= b1∨b2 (Since f is pre A*-homomorphism) | |
⇒a1∨a2 = f-1(b1∨b2) | |
⇒ f-1(b1∨b2) = f-1(b1)∨f-1(b1)∨ f-1(b2) | |
and f(a1∧a2) = f(a1)∧f(a2) | |
= b1∨b2 | |
⇒a1∧a2 = f-1(b1∧v2) | |
⇒f-1(b1∧b2) = f-1(b1) ∧f-1(b2) | |
Since f-1(b1~)f = (f-1(b1))~ | |
Therefore f-1 is a Pre A*-homomorphism |
By taking g = f-1 we have g:B→A is a Pre A*-homo, such that fog and gof are automorphisms of B and A, respectively.
Converse: Conversely, assume that g:B→A is a pre A*-homomorphism such that fog and gof are auto of B and A, respectively, where f:A→B is a pre A*-homomorphism
Since fog: B→B is an auto we have fog is an epimorphism.
⇒ f is epi (by proposition1.8)
⇒ f is onto
Since gof is auto, we have gof is mono
⇒ f is mono (Since by proposition1.8)
⇒f is oneone
Therefore f:A→B is an isomorphism
Theorem (Praroopa, 2012): Under any pre A*-homomorphism f of a pre A*-algebra A onto a pre A* algebra A1 with 0, the set kerf (kernel of f) is an ideal in A.
Proof: Let f : A→A1 be a pre A*-homomorphism
Then Kerf = {x∈A/f(x) = 0}
(i) | If f(a) = 0 ad f(b) = 0 |
Then f(a∨b) = f(a)∨f(b) = 0 V 0 = 0
⇒a∨b∈Kerf
i.e., if a∈Kerf, b∈Kerf⇒a∨b∈kerf
(ii) a∈Kerf ⇒f(a) = 0
for b∈B(A) f(a∧b)= f(a) ∧ f(b) = 0 ∧ f(b)
= 0 Since f(b)∈B(A1), . .
⇒a∧b∈Kerf
Therefore, from (i) and (ii), Kerf is an ideal in A
Proposition (Praroopa, 2012): If f is a pre A*-homomorphism of a pre A*-algebra A into another pre A*-algebra, then f(A)≅A/f-1(0). Where f(A) is called the image,
f-1(0) = {a∈A/f(a) = 0} the Kernel of f.
Proof : f: A→B is a pre A*-homomorphism.
Then by Proposition 3.6 there exists a congruence relation θx on A, an epimorphism αx: A→Ax and a monomorphism α: Ax→B such that αoαx = f
⇒αoαx(a) = f(a), ∀a∈A
⇒f(a) = αoαx(a)
= α(αx(a))
≅αx(a) (Since αis mono)
= Ax (Since αx is onto)
∴f(a)≅Ax, ∀a∈A
Hence, f(a)≅Ax→(a)
Since αx: A→Ax is onto then by fundamental theorem of homomorphism we have
Ker | αx={(s,t)∈AxA/αx (s) = αx (t)} |
={(s,t)∈AxA/x∧s = x∧t} | |
=θx | |
∴A/θx≅Ax | |
Since Kerαx = Ker f; | |
(Verification: Let s ∈Kerαx | |
⇒αx(s) = 0 | |
⇒x∧s = 0 | |
⇒α(x∧s) = 0 (Since α is one-one) | |
⇒f(s) = 0 (Since α: Ax→B by α(x∧s) = f(s), ∀s∈A) | |
⇒s∈Kerf) |
Hence A/ Kerf ≅Ax≅f(A) (by (1))
Therefore f(A)≅A/Kerf.
Established the concept of kernel of pre A*-homomorphism and proved some theorems on these pre A*-homomorphisms. Establish its useful theorems and related with these concepts of pre A*-homomorphisms.