Page 63 - Textos de Matemática Vol. 45

5.5. The case of commutative monoids 55 double centralizing relation (lower left hand side part of the diagram below),
see Section 4.4:
XXX
σ0k k ιX
p1 // φ¯
R ×M S oo σ // S // Hol(X)
p0
OO 0 OO (dR0 p0,p)
(p,dS1 p1) dS0 dS1 θX dR1 //
R oo
OO
ρX
dR0
// M φ // End(X)
All the commutative squares of the lower part are pullbacks, again see Section 4.4; accordingly both leftward whole lower rectangles are pullbacks. We are going to show that the two maps φdR0 and φdR1 classify the same split extension (R×M S, R, p0, s0, σ0k) with kernel X. According to the property of the Schreier split extension classifier, they will be equal, and consequently we shall get a factorization, which means R ⊂ R[φ]. So let us denote by σ0 : S R ×M S the above horizontal map. Then necessarily σ0k is a kernel of p0 : R ×M S → R, so that we get p1σ0k = k and (p,dS1 p1)σ0k = k since σ0 is a section of any map of the pair ((p, dS1 p1), p1). So, both φdR0 and φdR1 classify the split extension with
kernel X in question.
5.5 The case of commutative monoids

Proposition 5.5.1. Given a Schreier split epimorphism (A, B, f, s), whose cor- responding split sequence is the following:
K[f]oo q //Aoo s //B, kf
A is a commutative monoid if and only if K[f] and B are commutative and A is isomorphic to the direct product K[f] × B.
Proof. Obviously, if K[f] and B are commutative, then also K[f]×B is. Hence, if A is isomorphic to K[f] × B, it is commutative.
Conversely, if A is commutative, then also K[f] and B are, since they are submonoids of A. It remains to prove that A is isomorphic to K[f] × B. For this, consider the map ψ: A → K[f] × B sending an element a ∈ A to the pair (q(a), f(a)). It is a bijection, whose inverse is the map φ: K[f] × B → A