Page 83 - Textos de Matemática Vol. 45

6.5. Schreier accessibility 75
s(b1 · b2) · k(x) = s(b1) · (s(b2) · k(x)) = s(b1) · (s(b′2) · k(x)) = = s(b′1) · (s(b′2) · k(x)) = s(b′1 · b′2) · k(x),
and similarly for the other two equalities.
Then we can define D as the quotient B of B w.r.t. the congruence R,
R
and a morphism g: B → D, which is the canonical projection. Moreover, we
can construct a semiring C in the following way: as a monoid, it is the direct
product of X and B , and the multiplication is defined by R
(x1, [b1]) · (x2, [b2]) = (x1 · x2 + q(s(b1) · k(x2)) + q(k(x1) · s(b2)), [b1 · b2]).
It is easy to see that this definition does not depend on the choice of the representatives of the classes [b1] and [b2], and that in this way we obtain a semiring and a Schreier split extension
// oo q′ r // //
0 X //Coo D 0,
lt
where r = πD, q′ = πX, l = ⟨1,0⟩ and t = ⟨0,1⟩. It follows from Proposition 6.5.3 that it is a faithful Schreier split extension. It remains to build a morphism f : A → C making the following diagram commutative:
0 // X oo q // A oo p // B
s
1X f g
// oo q′ r //
0 X //Coo D 0
lt
in order to complete the index of our given Schreier split extension. We define f by putting f(a) = (q(a),[p(a)]). It clearly makes the diagram commutative, so we only need to prove that it is a morphism of semirings:
f(a1 +a2)=(q(a1 +a2),[p(a1 +a2)])=(q(a1)+q(a2),[p(a1)]+[p(a2)])= = (q(a1), [p(a1)]) + (q(a2), [p(a2)]) = f(a1) + f(a2),
indeed, since the upper Schreier split extension is a Schreier split epimorphism in the category of commutative monoids, the map q is a morphism of monoids. Moreover, A is isomorphic, as a monoid, to the direct product X × B (thanks to Proposition 5.5.1). As shown in [26], the multiplication in X × B is defined by
(x1, b1) · (x2, b2) = (x1 · x2 + q(s(b1) · k(x2)) + q(k(x1) · s(b2)), b1 · b2).
This immediately implies that f preserves the multiplication, and hence it is a morphism of semirings. This concludes the proof.
k
// 0 //