1. If G is a semigroup and Φ$_{1}$ : ρ$_{1}$ → G is a homomorphism then a homomorphism Φ$_{2}$ : ρ$_{2}$ → G is given by Φ$_{2}$ ( x, y ) = Φ$_{1}$ ( θ ( x ) , θ ( y )) for all x, y ∈ X with xρ$_{2}$y . 2. If S = I ( X$_{1}$, ρ$_{1}$, R ) and T = I ( X$_{2}$, ρ$_{2}$, R ) have gradings induced by Φ$_{1}$ and Φ$_{2}$ , respectively, then there is an injective homomorphism of G -graded rings h : S → T such that equation 7 holds for all f ∈ S and all x, y ∈ X$_{2}$ . ( h ( f )) ( x, y ) =      f ( θ ( x ) , θ ( y )) if xρ$_{2}$y in X$_{2}$ 0 otherwise (7) Proof. (1) Part 1 is trivial. (2) A routine check shows h is an R -module homomor- phism. In the definition of multiplication in equation 1, it is easy to see h is a ring homomorphism if θ ([ x, y ]) = [ θ ( x ) , θ ( y )] for all x, y ∈ X$_{2}$ such that xρ$_{2}$y . For any c ∈ [ θ ( x ) , θ ( y )] there is a transitive triple ( θ ( x ) , c, θ ( y )) ∈ Trans ( X$_{1}$ ) so by part 2 of definition 4.1 there exists u, v, w ∈ X$_{2}$ such that ( υ, v, w ) ∈ Trans ( X$_{2}$ ), θ ( u ) = θ ( x ), θ ( v ) = c , and θ ( w ) = θ ( y ). We find u = x and w = y since θ $^{∗}$( x, y ) = θ $^{∗}$( u, v ) and θ ∗ is bijective. Therefore v ∈ [ x, y ] and c ∈ θ ([ x, y ]). Since c was arbitrarily chosen we have θ ([ x, y ]) ⊆ θ ([ x, y ]). Moreover θ ([ x, y ]) ⊆ [ θ ( x ) , θ ( y )] since θ is order-preserving. Therefore θ ([ x, y ]) = [ θ ( x ) , θ ( y )] as desired. Given f ∈ S \ { 0 } there exist a$_{1}$, b$_{1}$ ∈ X$_{1}$ such that f ( a$_{1}$, b$_{1}$ ) = 0. If a$_{1}$ = b$_{1}$ then there exist x$_{1}$, y$_{1}$ ∈ X$_{2}$ such that x$_{1}$ρ ∗ $_{2}$y$_{1}$ and θ $^{∗}$( x$_{1}$, y$_{1}$ ) = ( a$_{1}$, b$_{1}$ ) ∈ ρ ∗ $_{2}$. If a$_{1}$ = b$_{1}$ then there exists x$_{1}$ ∈ X$_{2}$ such that a$_{1}$ = θ ( x$_{1}$ ) and we set y$_{1}$ = x$_{1}$ . In either case x$_{1}$ρ$_{2}$y$_{1}$ and h ( f ) ( x$_{1}$, y$_{1}$ ) = f ( a$_{1}$, b$_{1}$ ). Therefore h ( f ) ( x$_{1}$, y$_{1}$ ) = 0 and h ( f ) is nonzero. This proves h is injective. 12