f = f$_{1}$ + · · · + f$_{m}$ and ∂f$_{i}$ = ∂f$_{j}$ if i = j . Fix i ≤ m . Since f$_{i}$ is nonzero there exist x, y ∈ X such that xρy and f$_{i}$ ( x, y ) = 0. By equation 3 we have f$_{j}$ ( x, y ) e$_{xy}$ = e$_{xx}$f$_{j}$e$_{yy}$ for j = 1 , . . . , m . If f$_{j}$ ( x, y ) = 0 then ∂e$_{xy}$ = ∂f$_{j}$ since ∂e$_{xx}$ = ∂e$_{yy}$ = 1. When j = i we conclude Φ ( x, y ) = ∂f$_{i}$ since Φ ( x, y ) = ∂e$_{xy}$ by construction and f$_{i}$ ( x, y ) = 0 by our choice of x, y ∈ X . But if j = i and f$_{j}$ ( x, y ) = 0 then ∂e$_{xy}$ = ∂f$_{j}$ so ∂f$_{i}$ = ∂f$_{j}$ , which is a contradiction. We are left with f$_{j}$ ( x, y ) = 0 for all j = i . Therefore f ( x, y ) = f$_{i}$ ( x, y ) = 0 and Φ ( x, y ) = a by the assumption on f . We already proved Φ ( x, y ) = ∂f$_{i}$ so ∂f$_{i}$ = a . Since i was arbitrarily chosen we have ∂f$_{i}$ = a for all i = 1 , . . . , m . Since ∂f$_{1}$, . . . , ∂f$_{m}$ are distinct, we must have m = 1. Thus f = f$_{1}$ ∈ S$_{a}$ , as desired. Now suppose g ∈ S$_{a}$ and g ( w, z ) = 0 for some w, z ∈ X such that wρz . Equation 3 becomes g ( w, z ) e$_{wz}$ = e$_{ww}$ge$_{zz}$ and we have ∂e$_{wz}$ = ∂g since ∂e$_{ww}$ = ∂e$_{zz}$ = 1 by assumption. Moreover a = ∂g and ∂e$_{wz}$ = Φ ( w, z ) so Φ ( w, z ) = a . Remark 3.5 Equation 4 gives ( e$_{xx}$ ) 2 = e$_{xx}$ and so Φ ( x, x ) 2 = Φ ( x, x ) for all x ∈ X . The condition e$_{xx}$ ∈ S$_{1}$ for all x ∈ X stated in part 2 of Theorem 3.4 may not hold for all monoids. But if G is a cancellative monoid then ( e$_{xx}$ ) 2 = e$_{xx}$ implies Φ ( x, x ) = 1 for all x ∈ X and the condition in part 2 is satisfied automatically. Definition 3.6 Let G be a semigroup and let ρ be a relation on X . 1. A subset β of ρ is a G -extendible set for ρ if for every function φ : β → G there exists a homomorphism Φ : ρ → G such that Φ |$_{β}$ = φ . If we may choose Φ so that Im Φ is finite then we say φ is grading admissible . 9