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18 Chapter 2. Schreier and homogeneous split epimorphisms in monoids
Suppose that qf and qg are the Schreier retractions of the Schreier split epi- morphisms (f,s) and (g,t), respectively. Let us set qgf(x) = qf(x)·sqg(f(x)) and show it satisfies the required identities. First we have:
qf (x) · sqg(f(x)) · stgf(x) = qf (x) · s(qg(f(x)) · tgf(x)) = qf (x) · sf(x) = x. Then, given any (α, y) ∈ K[gf] × Y , notice that f(α) is in K[g]. Hence:
qgf (α · st(y)) = qf (α · st(y)) · sqg(f(α · st(y))) =
= qf (α) · qf (sf(α) · qf (st(y))) · sqg(f(α) · t(y)) =
= qf(α)·qf(sf(α)·1)·sf(α) = qf(α)·qf(sf(α))·sf(α) = qf(α)·1·sf(α) = α.
Suppose now that the composite is a Schreier split epimorphism with Schreier retraction qgf . Let us set qg (w) = f qgf (s(w)) and show it satisfies the required identities. First we have:
qg(w) · tg(w) = fqgf (s(w)) · tg(w) = fqgf (s(w)) · fstg(w) =
= f (qgf (s(w)) · stg(w)) = f (qgf (s(w)) · stgf s(w)) = f (s(w)) = w.
Now, given any (α, y) ∈ K[g] × Y , we have s(α) ∈ K[gf] and get:
qg(α · t(y)) = fqgf (s(α · t(y))) = fqgf (s(α) · st(y)) = f(s(α)) = α.
The results concerning the homogeneous split epimorphisms are obtained du- ally.
Proposition 2.3.3. Schreier split epimorphisms are stable under products, i.e. the product of two Schreier split epimorphisms is a Schreier one. The same is true for homogeneous split epimorphisms.
Proof. Consider the two Schreier split exact sequences
K[f]
oo q oo s // A
kf
// // B
// // B′,
oo s×s′ f ×f ′
and
K[f′] Their term by term product
// A′
oo q′ k′
oo q×q′ K[f] × K[f′]
k×k′
oo s′ f′
// A × A′
// // B × B′,
clearly satisfies the conditions of Proposition 2.1.4. Dually we have the same result for left homogeneous split epimorphisms, and hence for homogeneous split epimorphisms.