Page 74 - Textos de Matemática Vol. 45

66 Chapter 6. Semirings
Proposition 6.3.7. Given a split epimorphism (X, Y, f, s), its kernel equivalence relation R[f] is a Schreier one if and only if the split epimorphism is itself a Schreier one and the kernel K[f] is a ring.
Proposition 6.3.8. A Schreier reflexive graph in SRng:
d0 // X1oos0 //X0
d1
with q: X1 → K[d0] its associated Schreier retraction, has at most one structure of internal category. When it is the case, the composition of a composable pair (τ,σ) is given by: d1(τ,σ) = q(τ) + σ. This map is a morphism of semirings (and hence it is the composition of an internal category) if and only if, for any α ∈ K[d0] and any σ ∈ X1 the following conditions hold:
(1) q(α·s0d1(σ))=α·σ,
(2) q(s0d1(σ)·α)=σ·α.
In particular, these conditions imply that K[d0] and K[d1] commute in X1.
Proof. The fact that the unique possible composition is given by d1(τ,σ) = q(τ) + σ comes from Proposition 3.2.3, through Lemma 6.0.8. It is easy to check that such composition becomes a homomorphism of semirings if and only if conditions (1) and (2) are satisfied. Moreover, when σ is in K[d1], we get
α · σ = q(α · s0d1(σ)) = q(α · 0) = 0 σ · α = q(s0d1(σ) · α) = q(0 · α) = 0,
and
and hence K[d0] and K[d1] commute in X1.
Proposition 6.3.9. Given a Schreier internal category X1, the following condi- tions are equivalent:
(a) X1 is a Schreier groupoid;
(b) R[d0] is a Schreier equivalence relation;
(c) K[d0] is a ring.
d0 //
Proposition 6.3.10. Let X1 oo s0 // X0 be a Schreier reflexive graph such that
d1
K[d0] is a ring. The following conditions are equivalent:
(a) the kernels of d0 and d1 cooperate in SRng; (b) the reflexive graph is an internal category;