Page 81 - Textos de Matemática Vol. 45

6.5. Schreier accessibility 73 = l(x1) + tg(b1) + l(x2) + tg(b2) = f(x1, b1) + f(x2, b2),
and, moreover
f((x1,b1)·(x2,b2)=f(x1 ·x2 +q′(tg(b1)·l(x2))+q′(l(x1)·tg(b2)),b1 ·b2)=
= l(x1 · x2) + lq′(tg(b1) · l(x2)) + lq′(l(x1) · tg(b2)) + tg(b1 · b2) = = l(x1) · l(x2) + tg(b1) · l(x2) + l(x1) · tg(b2) + tg(b1) · tg(b2) = = (l(x1) + tg(b1)) · (l(x2) + tg(b2)) = f(x1, b1) · f(x2, b2),
where we are using the fact that, by Proposition 6.0.11 (d), lq′(tg(b1)·l(x2)) = tg(b1) · l(x2) and lq′(l(x1) · tg(b2)) = l(x1) · tg(b2), since tg(b1) · l(x2) and l(x1) · tg(b2) belong to l(X). Hence f is a morphism of semirings.
Given another element d′ ∈ D, we can repeat the same construction as above, obtaining a Schreier split extension
0 //Xooq¯//A′oop′ //B′ //0, k′ s′
and a morphism of Schreier split extensions
0 //Xooq¯//A′oop′//B′ //0 k′ s′
1X f′ g′
// oo q′ r //
0 X //Coo D 0,
//
whereB′ =B,k=k′,p=p′,s=s′ andq¯=q,andmoreoverA′ =Aasa
monoid.
Suppose now that d and d′ are such that t(d) · l(x) = t(d′) · l(x) and l(x) · t(d) = l(x) · t(d′) for every x ∈ X. If we prove that, under this hypothesis, the multiplications defined on A and A′ are equal, then, by the faithfulness of the split extension
// oo q′ r // //
0 X //Coo D 0,
lt
we can conclude that g = g′, and hence d = d′. In order to do this, it suffices toprovethat,foreveryx∈X andeveryb∈B
tg(b) · l(x) = tg′(b) · l(x)
lt