States on Subtraction Algebras 1 Introduction

<strong>States</strong> on subtraction algebras 1159
Proposition 3.5. Let s : X → [0, 1] be a Bosbach state on X. Then If x ≤ y,
then s(x) ≤ s(y) and s(x)+s(y − x) =s(y).
Proof. Let s be a Bosbach state on X and x ≤ y. Then x − y =0. Since
s(x)+s(y−x) =s(y)+s(x−y), we have 0 ≤ s(y−x) =s(x−y)+s(y)−s(x)=
0+s(y) − s(x) =s(y) − s(x), and so s(y) ≥ s(x).
Proposition 3.6. Let s : X → [0, 1] be a Bosbach state on X. Then Ker(s)
is a subalgebra of X.
Proof. Let x, y ∈ Ker(s). Then s(x) =s(y) =0. Since x − y ≤ x, we have
s(x − y) ≤ s(x) by Proposition 3.5, and so 0 ≤ s(x − y) ≤ s(x) =0. This
implies s(x − y) =0, i.e., x − y ∈ Ker(s).
Proposition 3.7.
is an ideal of X.
Let s : X → [0, 1] be a Bosbach state on X. Then Ker(s)
Proof. Since s(0) = 0, we have 0 ∈ Ker(s). Let x − y ∈ Ker(s) and y ∈
Ker(s). So, s(x − y) =s(y) =0. Since y − x ≤ y, we have s(y − x) ≤ s(y) by
Proposition 3.5, which implies 0 ≤ s(x − y) ≤ s(y) =0. Hence s(y − x) =0.
Now, since s(x)+s(y − x) =s(y)+s(x − y), we obtain s(x)+0=0+0. This
implies s(x) =0, i.e., x ∈ Ker(s).
Theorem 3.8. Let f : X → Y be a homomorphism and s : Y → [0, 1] be a
Bosbach state on Y. Then there exists an unique Bosbach state t : X → [0, 1]
such that the following diagram is commutative (i.e., t = s ◦ f).
X f
Y
s
∃t
[0, 1]
Proof. Let t := s ◦ f. It is clear that t is well-defined. Now we show that t is
a Bosbach state. We know that t(0) = (s ◦ f)(0) = s(f(0)) = s(0) = 0. Now,