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68 Chapter 6. Semirings = 0 + s0d1(a1) · q(a′2) = q(s0d1(a1) · q(a′2)),
where the last equality holds because s0d1(a1) · q(a′2) ∈ K[d0]. Similarly, we have:
q(a′1) · a2 = q(a′1) · (t(a2) + s0d1(a2)) = q(a′1) · t(a2) + q(a′1) · s0d1(a2) = = 0 + q(a′1) · s0d1(a2) = q(q(a′1) · s0d1(a2)).
The equivalence between (b) and (c) comes from Proposition 6.3.9. Proposition 6.3.11. Consider the following commutative diagram
d′0 //
′′ X1′oos0 X
// 0 d′1
uv d0 //
X, X1 oo s // 0
0
d1
where the two rows are Schreier reflexive graphs, u is a regular epimorphism, and the kernels of d0 and d′0 are rings. If the upper row is an internal groupoid, then so is the lower one.
Proof. Thanks to Proposition 6.3.10, we only have to show that k: X → A and l: Y → A cooperate, where k is the kernel of d0 and l is the kernel of
k′ : K[d′0] → X1′ and l′ : K[d′1] → X1′ cooperate, where k′ is the kernel of d′0 and l′ is the kernel of d′1. This means that
x′·y′ =y′·x′ =0 forall x′ ∈K[d′0], y′ ∈K[d′1].
Consider the following diagrams, where i and j are induced by u via the uni-
versal property of the kernels:
d′0 // ′′
d1. Since X1′ oo s0 // X0 is a Schreier internal groupoid, we already know that d′1
s′ ′ k′ //′oo0
′
s′
′ l′ //′oo0 ′
K[d0] X1
d′0
// X0
K[d1] X1
// X0
d′1 iuvjuv
// oo s0
K[d0] k X1
d0

// X0,
K[d1]
// oo s0
l
X1
// X0.
Since u is a regular epimorphism, i.e. it is surjective, also i is. Indeed, there exists a set-theoretical map r : X1 → X1′ such that ur = 1X1 . Then the map
d1