States on Subtraction Algebras 1 Introduction

1160 Sang Moon Lee and Kyung Ho Kim
let x, y ∈ X. Then
t(x)+t(y − x) =(s ◦ f)(x)+(s ◦ f)(y − x)
= s(f(x)) + s(f(y) − f(x))
= s(f(y)) + s(f(x) − f(y))
= s(f(y)) + s(f(x − y))
=(s ◦ f)(y)+(s ◦ f)(x − y)
= t(y)+t(x − y).
Hence t is a Bosbach state. Finally, we prove that t is unique. Suppose that
there exists an another Bosbach state r : X → [0, 1] such that r = s ◦ f. Then
r(x) =(s ◦ f)(x) =t(x) for all x ∈ X. Hence we have r = t.
Theorem 3.9. Let f : X → Y be a bijection homomorphism and s : X →
[0, 1] be a Bosbach state on X. Then there exists an unique Bosbach state
t : Y → [0, 1] such that s = t ◦ f.
X f Y
s
∃t
[0, 1]
Proof. Let y ∈ Y. Since f is onto, there exists x ∈ X such that f(x) =y. Put
t(y) =s(x). Then t(y) =t(f(x)) = (t◦f)(x) =s(x), and so t◦f = s. Since f is
one to one, then Ker(f) =0, and so t(0) = t(f(0)) = (t◦f)(0) = s(0) = 0. Let
y 1 ,y 2 ∈ Y. Then there exists x 1 ,x 2 ∈ X such that f(x 1 )=y 1 and f(x 2 )=y 2 .
Then we have
t(y 1 )+t(y 2 − y 1 )=t(f(x 1 )+t(f(x 2 ) − f(x 1 ))
= t(f(x 1 )+t(f(x 2 − x 1 ))
=(t ◦ f)(x 1 )+(t ◦ f)(x 2 − x 1 )
= s(x 1 )+s(x 2 − x 1 )
= s(x 2 )+s(x 1 − x 2 )
=(t ◦ f)(x 2 )+(t ◦ f)(x 1 − x 2 )
=(t(f(x 2 )+t(f(x 1 − x 2 ))
= t(f(x 2 )) + t(f(x 1 ) − f(x 2 ))
= t(y 2 )+t(y 1 − y 2 ).
Hence t is a Bosbach state. Suppose that there exists an another Bosbach
state r : X → [0, 1] such that s = r ◦ f. Let y ∈ Y. Then there exists x ∈ X