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3.1. Schreier reflexive relations and graphs 33 Conversely, suppose that (X, Y, f, s) is a Schreier split epimorphism and
K [f ] is a group. Define qp0 (x, x′ ) = (1, qf (x′ ) · qf (x)−1 ). We can check that qp0(x,x′)·s0p0(x,x′) = (1,qf(x′)·qf(x)−1)·(x,x) = (x,qf(x′)·qf(x)−1 ·x) =
= (x, qf (x′ ) · sf (x)) = (x, qf (x′ ) · sf (x′ )) = (x, x′ ),
qp0 ((1, k) · s0(x)) = qp0 ((1, k) · (x, x)) = qp0 (x, k · x) = (1, qf (k · x) · qf (x)−1) =
= (1, qf (k) · qf (sf(k) · qf (x)) · qf (x)−1) = (1, qf (k) · qf (x) · qf (x)−1) = (1, k). If moreover (X,Y,f,s) is left homogeneous, in the same way we can define
q¯ (x, x′ ) = (1, q¯ (x)−1 · q¯ (x′ )) which satisfies the desired conditions for a left p0 ff
homogeneous split epimorphism. Accordingly R[f] is a homogeneous equiva- lence relation.
We conclude this section with a result which will be useful later on.
and
Proposition 3.1.13. Let
d0 // X1oos0 //X0
d1
be a Schreier reflexive graph. The split epimorphism (d1,s0) is also a Schreier one if and only if, for any x ∈ K[d0], the element d1(x) is invertible in X0. In particular, when K[d0] is a group, the split epimorphism (d1,s0) is also a Schreier one.
Proof. Suppose that (d1, s0) is a Schreier split epimorphism. Let us denote by q0 and q1 the Schreier retractions associated with (d0, s0) and (d1, s0), respectively. Then, for any x ∈ K[d0], the inverse of d1(x) is d0q1(x). Indeed:
and hence
and moreover hence
x = q1(x) · s0d1(x)
1 = d0(x) = d0q1(x) · d0s0d1(x) = d0q1(x) · d1(x), q1(x) = q0q1(x) · s0d0q1(x),
1 = d1q1(x) = d1q0q1(x) · d1s0d0q1(x) = d1q0q1(x) · d0q1(x),