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88 Chapter 7. Special Schreier and special homogeneous surjections such that μh¯ = l. Furthermore, the surjection f′ is special homogeneous as soon
as f is.
Proof. The fact that h is a morphism of Y -modules means that ψ(y)(h(a)) = hq(xR[f](x · a)) for any x such that f(x) = y. Consider now the following pullback in Mon:
f∗ ////K[h] φY kh 1Y

R[f] f¯ A φY
It produces a relation S on X defined by xSz if and only if xR[f]z and q(xR[f]z) ∈ K[h]. This relation S is clearly reflexive. The kernel of d0 : S → X is the abelian group K[h] and the reflexive relation S is a Schreier one with the Schreier retraction qS : S → K[h] defined by qS(xSz) = q(xR[f]z). Accord- ingly S is a Schreier equivalence relation on X in Mon. Let us define X′ as the quotient monoid of this equivalence relation S:
d0// h¯
Soo s0 //X ////X′.
d1
Since the kernel of d0 : S → X is the abelian group K[h], K[h] is also the kernel of h¯. On the other hand, since S is included in R[f], h¯ coequalizes p0 and p1, hence there exists a unique surjective homomorphism f′ : X′ Y making the following lower right hand side square commutative:
K(h¯)
(0,kf ) R(h¯)
////Y
S
j
// //
K[k(h¯)] //
//K[h] //1
//R[f]oo s0 //X p1
k¯ A//
kh¯ p0// f
K[f′] // (0,kf′)
p′0// h¯ //R[f′]oo s′0 //X′
////Y.
p′ f′ 1
Completing the diagram above with the kernels of the vertical maps produces the upper horizontal kernel diagram, which shows that the kernel of K(h¯) is, up to isomorphisms, the abelian group K[h]; consequently the abelian group K[f′] is, up to isomorphisms, the abelian group A′, and makes the map k′ : A′ X′, defined by k′(a′) = h¯(a) for any a such that a′ = h(a), a kernel for f′. We have to show now that f′ is a special Schreier surjection. If it is the case, the associated Schreier retraction q′ is necessarily defined by q′(x¯R[f′]z¯) =