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7.3. Special Schreier extensions with abelian kernel 85
Proof. Let q be the Schreier retraction of the Schreier split epimorphism (d0,s0). Since [R[f], R[f]] = 0, for any xR[f]x′ and any t ∈ A we have q(x′R[f](x′ · t)) · x = x · t thanks to Proposition 4.3.3. Hence
φ(x′)(t) = q(x′R[f](x′ · t)) and φ(x)(t) = q(xR[f](x · t))
(see the construction of φ given in Theorem 5.1.2). If we have (x,x′) ∈ R[f], then we get q(x′R[f](x′ · t)) · x = x · t. On the other hand, by the uniqueness in the Schreier condition, q(xR[f](x · t)) is the only element in A such that q(xR[f](x · t)) · x = x · t. Accordingly φ(x′)(t) = φ(x)(t), and consequently φ(x) = φ(x′).
Since f is the quotient of R[f], according to the previous proposition there is a factorization φ˜: Y → End(A) such that φ = φ˜f. We have φ˜(y)(a) = q(xR[f](x · a)) for any x such that f(x) = y. Consider now the following diagram:
R[f] ×X R[f] // R[f]
Q
Hol(A) OO
φ¯ oop1//f¯//// //
OO (d0 p0 ,p)
d0
OO //
// X
d1 (1) ψ ////
OO Y
φˇ
(2) θA
p0 R[f]
σ
ρA End(A)
(p,d1p1)
oo d1
// f φ˜ GG
d0
Let us check now that φ¯ coequalizes the upper horizontal relation. Given any triple xR[f]x′R[f]x′′ belonging to R[f]×XR[f], we have (recalling the definition of φ¯ given in Theorem 5.1.2) that
φ¯p1(xR[f]x′R[f]x′′) = φ¯(x′, x′′) = (q(x′R[f]x′′), φ(x′)). On the other hand, we have that:
φ¯(d0p0,p)(xR[f]x′R[f]x′′) = φ¯(x,p(xR[f]x′R[f]x′′)) = = (q(xR[f]p(xR[f]x′R[f]x′′)), φ(x)).
According to the explicit expression of p given in Proposition 4.3.3, this is equal to
(q(xR[f](q(x′R[f]x′′) · x), φ(x))) = (q(x′R[f]x′′), φ(x)),
where the last equality comes from Lemma 3.1.4. Observing that φ(x) = φ(x′), because (x,x′) is in R[f], we conclude the proof that φ¯ coequalizes the upper
φ