Page 75 - Textos de Matemática Vol. 45

6.3. Schreier internal structures in semirings 67 (c) the reflexive graph is an internal groupoid.
Proof. We already proved (see Proposition 6.3.8) that, if a Schreier reflex- ive graph is an internal category, then the kernels of the domain and of the codomain cooperate. Hence (b) implies (a).
Conversely, suppose that we have a Schreier reflexive graph
d0 // K[d0] // X1 oo s0 // X0,
ooq k
d1
with the Schreier split epimorphism associated with the codomain d1 given by
oo t oo s0 K[d1] // X1
// X0,
such that k and l cooperate (we recall that (d1,s0) is a Schreier split epimor-
phism thanks to Proposition 3.1.13 and Lemma 6.0.8). This means that x·y=y·x=0 forall x∈K[d0], y∈K[d1].
We already know from Proposition 3.3.5 that the unique possible multiplication is the map defined by
m(a′, a) = q(a′) + a,
and that it is a morphism of monoids (since the additive monoids involved are all commutative). We only have to show that it is a morphism of semirings, i.e. that m(a′1 · a′2, a1, a2) = m(a′1, a1) · m(a′2, a2). We have that:
m(a′1 ·a′2,a1,a2)=q(a′1 ·a′2)+a1 ·a2 =
= q(a′1) · q(a′2) + q(s0d0(a′1) · q(a′2)) + q(q(a′1) · s0d0(a′2)) + a1 · a2,
while
m(a′1, a1) · m(a′2, a2) = (q(a′1) + a1) · (q(a′2) + a2) = =q(a′1)·q(a′2)+q(a′1)·a2 +a1 ·q(a′2)+a1 ·a2,
so it suffices to show that
q(s0d0(a′1) · q(a′2)) = a1 · q(a′2)
and
q(q(a′1) · s0d0(a′2)) = q(a′1) · a2.
a1 · q(a′2) = (t(a1) + s0d1(a1)) · q(a′2) = t(a1) · q(a′2) + s0d1(a1) · q(a′2) =
We have:
l d1