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20 Chapter 2. Schreier and homogeneous split epimorphisms in monoids
Proof. First notice that when g is surjective, so is f g and hence h. Let us denote by qf′ and qR the Schreier retractions associated with the two upper Schreier split epimorphisms. According to Proposition 2.3.1, we have p0qR = p0qf′ and p1qR = p1qf′ . Since the surjective homomorphisms are the quotients in Set of their kernel equivalence relations, there is a factorization qf in Set making the lower left hand side leftward square commute. The fact that h, g and K(g) are surjective allows to transfer the required conditions from qf′ to qf. The same facts hold for left homogeneous split epimorphisms by duality; consequently they hold for homogeneous split epimorphisms.
Corollary 2.3.6. Suppose that the right hand side square is a pullback, and h is a regular epimorphism.
k′
K[f′] // f //A′
K(g) ≃ g
oo s′ f′
//B′
h // // oos//
K[f] A B kf f
If the upper split epimorphism is a Schreier (resp. homogeneous) one, so is the lower one.
Proof. When the right hand side square is a pullback, the factorization K(g) is an isomorphism and consequently a regular epimorphism. When h is a regular epimorphism, so is g (because regular epimorphisms, i.e. surjective morphisms, are stable under pullbacks, see the Appendix). Moreover, all the commutative diagrams determined by the kernel equivalence relations R[g] and R[h] are pullbacks, and consequently the split epimorphism (R[g], R[h], R(f′), R(s′)) is a Schreier (resp. homogeneous) one. According to the previous proposition, so is (A,B,f,s)
Theorem 2.3.7. Consider a commutative diagram of split epimorphisms:
k // oo s K[f] A
// // B
f
K(u) u v
′ ooq′ ′oo s′ ′
K[f ] // A k′
// // B f′
where the lower row is a Schreier split sequence and where the map K(u) is an isomorphism. The following conditions are equivalent:
(a) the pair (u,f) is jointly monomorphic;