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3.1. Schreier reflexive relations and graphs 31
Proposition 3.1.5. Any Schreier reflexive relation R is transitive, and we have q(yRz) · q(xRy) = q(xRz). It is an equivalence relation if and only if K[d0] is a group.
Proof. Suppose that xRy and yRz. Since the reflexive relation is a Schreier one, we have 1Rq(xRy) and 1Rq(yRz), whence 1R(q(yRz) · q(xRy)). From that we get xR(q(yRz) · q(xRy) · x), with q(yRz) · q(xRy) · x = q(yRz) · y = z thanks to the previous lemma, whence xRz. The uniqueness of the decomposition implies that q(yRz) · q(xRy) = q(xRz).
Suppose moreover that R is symmetric. Then, for any y such that 1Ry, we have that yR1. Whence q(yR1) · y = 1 for any y ∈ K[d0], which implies that K[d0] is a group. Conversely, suppose that K[d0] is a group and that we have xRy. Then we have q(xRy) · x = y with q(xRy) ∈ K[d0]. Whence x = q(xRy)−1 ·y with 1Rq(xRy)−1; so that yR(q(xRy)−1 ·y), which is yRx.
Proposition 3.1.6. Given a monoid X, the indiscrete equivalence relation ∇X given by:
p0 // X×Xoo s0 //X
p1
is a Schreier equivalence relation if and only if X is a group. In this case ∇X is actually homogeneous. The equivalence relation ∇X is explicitly given by: x1∇Xx2 forallx1,x2∈X.
Proof. If X is a group, it is an immediate consequence of Proposition 2.2.4; ac- tually the equivalence relation ∇X is even homogeneous. Conversely, suppose the indiscrete equivalence relation is a Schreier equivalence relation. Then, ac- cording to the previous proposition, K[p0] = X is a group.
Corollary 3.1.7. Given any monoid B, the three following conditions are equiv- alent:
(a) the monoid B is a group
(b) any split epimorphism with codomain B is homogeneous
(c) any split epimorphism with codomain B is a Schreier split epimorphism.
Proof. (a) ⇒ (b) is given by Proposition 2.2.4.
(b) ⇒ (c) holds since any homogeneous split epimorphism is a Schreier one. (c) ⇒ (a) Condition (c) implies that the indiscrete relation ∇B is Schreier and that consequently B is a group.
Corollary 3.1.8. Given a group B, any equivalence relation R on B in Mon is homogeneous, and R is itself a group.
From that, it is easy to build some examples:
