Page 97 - Textos de Matemática Vol. 45

7.5. Baer sums in Mon 89
hq(xR[f]z) for any x,z ∈ X such that h¯(x) = x¯ and h¯(z) = z¯. Let us check that this definition does not depend on the choice of x and z. Suppose that h¯(x) = h¯(x′) and h¯(z) = h¯(z′). Then xSx′ and zSz′, and this implies that xR[f]x′ and zR[f]z′. Then, by Proposition 3.1.5, we get
q(x′R[f]z′) · q(xR[f]x′) = q(xR[f]z′) = q(zR[f]z′) · q(xR[f]z). Hence
hq(x′R[f]z′) · hq(xR[f]x′) = hq(zR[f]z′) · hq(xR[f]z).
But xSx′ and zSz′ imply also that q(xR[f]x′) ∈ K[h] and q(zR[f]z′) ∈ K[h], and so hq(xR[f]z) = hq(x′R[f]z′).
Let us now determine the direction of the special Schreier extension
A′ // k′ //X′ f′ //// Y.
It is determined by the map φ′ : Y → End(A′) defined by φ′(y)(a′) = q′(h¯(x)R[f′](h¯(x) · h¯(a))),
where f(x) = y and h(a) = a′. Hence
φ′(y)(a′) = hq(xR[f](x · a)) = ψ(y)(h(a)) = ψ(y)(a′),
and consequently φ′ = ψ. Accordingly, the direction of this special Schreier extension is (Y ψ A′, Y, pψY , ιψY ), as desired.
Suppose now that we have the morphisms l of special Schreier extensions and m of left Y -modules such that mh = K(l). Showing that l factors through h¯ is equivalent to showing that l coequalizes the equivalence relation S. So, suppose that we have xSz. Then xR[f]z and q(xR[f]z) ∈ K(h). Since f′′l = f, we get l(x)R[f′′]l(z), hence q′′(l(x)R[f′′]l(z)) · l(x) = l(z) by Lemma 3.1.4. Moreover
q′′(l(x)R[f′′]l(z)) = lq(xR[f]z) = mhq(xR[f]z) = m(1) = 1.
So we have l(x) = l(z), and hence a factorization μ: X′ → X′′ such that μh¯ = l, so that K(μ)h = K(l) = mh. Since h is surjective, we get K(μ) = m. The last point of the proposition is just Corollary 7.3.4.
7.5 Baer sums in Mon
We shall denote by SExtφ(Y,A) (resp. HExtφ(Y,A)) the set of isomorphic classes of special Schreier (resp. homogeneous) extensions with codomain Y , abelian kernel A and direction the Schreier split epimorphism (A φY, Y, pφY , ιφY )