Page 50 - Textos de Matemática Vol. 45

42 Chapter 3. Schreier internal relations, categories and groupoids Proposition 3.3.6. Consider the following commutative diagram
d′0 //
′′ X1′oos0 X
// 0 d′1
uv d0 //
X, X1 oo s // 0
0
d1
where the two rows are Schreier reflexive graphs, u is a regular epimorphism, and the kernels of d0 and d′0 are groups. If the upper row is an internal groupoid, then so is the lower one.
Proof. Thanks to Proposition 3.3.5, we only have to show that k: K[d0] → X1 and l: K[d1] → X1 cooperate, where k is the kernel of d0 and l is the kernel of
k′ : K[d′0] → X1′ and l′ : K[d′1] → X1′ cooperate, where k′ is the kernel of d′0 and l′ is the kernel of d′1. This means that
x′·y′ =y′·x′ forall x′ ∈K[d′0], y′ ∈K[d′1].
Consider the following diagrams, where i and j are induced by u via the uni-
versal property of the kernels:
d′0 // ′′
d1. Since X1′ oo s0 // X0 is a Schreier internal groupoid, we already know that d′1
s′ ′ k′ //′oo0
′
s′
′ l′ //′oo0 ′
K[d0] X1
d′0
// X0
K[d1] X1
// X0
d′1 iuvjuv
// oo s0
K[d0] k X1
d0

// X0,
K[d1]
// oo s0
l
X1
// X0.
Since u is a regular epimorphism, i.e. it is surjective, also i is. Indeed, there exists a set-theoretical map r : X1 → X1′ such that ur = 1X1 . Then the map q′rk: K[d0] → K[d′0] (where q′ is the map given by the Schreier condition for the split epimorphism d′0) is such that
iq′rk = qurk = qk = 1K[d0],
where q is the map given by the Schreier condition for the split epimorphism d0, and this proves that i is surjective. Similarly we prove that j is surjective, too. Now, given x ∈ K[d0] and y ∈ K[d1], there exist x′ ∈ K[d′0] and y′ ∈ K[d′1] such that i(x′) = x and j(y′) = y. Then we have:
x·y=i(x′)·j(y′)=u(x′ ·y′)=u(y′ ·x′)=j(y′)·i(x′)=y·x,
and this proves that k and l cooperate.
d1