Page 91 - Textos de Matemática Vol. 45

7.3. Special Schreier extensions with abelian kernel 83 converse:
1) Suppose that K(u) is a monomorphism and that u(x1) = u(x2). Then f(x1) = f′u(x1) = f′u(x2) = f(x2),
which means that x1R[f]x2. Since R[f] is a Schreier equivalence relation, de- noting by q its associated Schreier retraction, we get q(x1R[f]x2) · x1 = x2 thanks to Lemma 3.1.4. Then
uq(x1R[f]x2) · u(x1) = u(x2) = u(x1),
and the uniqueness condition for the equivalence relation R[f′] (which is also a Schreier one) implies that uq(x1R[f]x2) = 1, and since K(u) (which is nothing but the restriction of u to the kernels) is a monomorphism, we have q(x1R[f]x2) = 1. Since the kernel of q, in the category of pointed sets, is s0 : X → R[f], we have that the pair (x1, x2) belongs to the image of s0, which means that x1 = x2.
2) Suppose that K(u) is a surjective homomorphism. Let x′ be in X′. There exists x ∈ X such that f(x) = f′(x′); so we get f′u(x) = f(x) = f′(x′), which means that u(x)R[f′]x′. Denoting by q′ the Schreier retraction of the Schreier equivalence relation R[f′], we obtain f′(x′) = q′(u(x)R[f′]x′) · u(x), again thanks to Lemma 3.1.4. But q′(u(x)R[f′]x′) belongs to K[f′] and so, since K(u) is surjective, there exists t ∈ K[f] such that q′(u(x)R[f′]x′) = u(t). Then we have f′(x′) = u(t) · u(x) = u(t · x).
7.3 Special Schreier extensions with abelian kernel
Definition 7.3.1. A special Schreier (resp. homogeneous) extension with abelian kernel is an exact sequence where f is a special Schreier (resp. special homoge- neous) surjection and K[f] is an abelian group A:
A=K[f]//k //Xf////Y
Thanks to Proposition 3.1.12, we have that, when a special Schreier surjection is split, it is a Schreier split epimorphism. According to the property of the Schreier split extension classifier, the split special Schreier surjections with kernel the abelian group A are in one-to-one correspondence with the structures of Y -modules: Y → End(A).
The kernel K[f] is abelian if and only if the kernel inclusion kf : K[f] X cooperates with itself, and, according to Proposition 4.3.6, this happens if and only if the Schreier equivalence relation R[f] is such that [R[f],R[f]] = 0. Consider then the associated double centralizing relation and its levelwise