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46 Chapter 4. Mal’tsev aspects of Mon
Proof. Since R[f] is a Schreier equivalence relation, K[f] is a group. By the reflection of commutativity (Theorem 2.4.6), we have [R[f],R[f]] = 0 if and only if K[f] is commutative.
The following result is analogous to a known one in the context of Mal’tsev categories (see [17]):
Proposition 4.3.7. Consider a Schreier reflexive graph such that K[d0] is a group:
d0 // X1oos0 //X0
d1
The following conditions are equivalent:
1) this graph is underlying a Schreier category;
2) this graph is underlying a Schreier internal groupoid;
3) the kernel equivalence relations R[d0] and R[d1] centralize each other; 4) the kernels K[d0] and K[d1] commute in X1.
Proof. Since the split epimorphism (X1, X0, d0, s0) is a Schreier one and K[d0] is a group, the kernel equivalence relation R[d0] is a Schreier one, and the fact that any other equivalence relation on X1 centralizes with it makes sense.
1) ⇒ 2) Suppose the graph is underlying a Schreier internal category. Since K[d0] is a group, it is a Schreier internal groupoid.
2) ⇒ 3) Suppose the graph is underlying a Schreier internal groupoid. Let us first observe that the split epimorphism (X1, X0, d1, s0) is isomorphic to (X1, X0, d0, s0). The isomorphism is induced by the isomorphism γ : X1 → X1 which associates with any element of X1 (that is, with any arrow of the in- ternal groupoid) its inverse with respect to the composition of arrows in the internal groupoid. Then (X1,X0,d1,s0), being isomorphic to a Schreier split epimorphism, is a Schreier one, too. This implies that K[d1] is a group; accord- ingly R[d1] is a Schreier equivalence relation. On the other hand, an internal groupoid being an internal category, the kernels K[d0] and K[d1] commute in X1, and consequently, according to the reflection of commutativity by the ker- nel functor(Theorem 2.4.6), the kernel equivalence relations R[d0] and R[d1] centralize each other.
3) ⇒ 4) Suppose that the kernel equivalence relations R[d0] and R[d1] central- ize each other. The preservation of commutativity by the kernel functor implies that the kernels K[d0] and K[d1] commute in X1.
4) ⇒ 1) Suppose that the kernels K[d0] and K[d1] commute in X1. First notice that when K[d0] is a group, any image of an element in K[d0] is invertible. To prove that the graph is underlying an internal category, we have to show that, for any pair (σ, α) ∈ X1 × K[d0], we have:
q(s0d1(σ) · α) · σ = σ · α