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34 Chapter 3. Schreier internal relations, categories and groupoids
but q0q1(x) = q0(x), because, thanks to Proposition 2.1.5, we have q0(x) = q0(q1(x) · s0d1(x)) =
= q0q1(x) · q0(s0d0q1(x) · q0s0d1(x)) = q0q1(x) · 1 = q0q1(x); since q0(x) = x (because x ∈ K[d0]), we get
1 = d1(x) · d0q1(x).
Conversely, suppose that d1(x) has an inverse for any x ∈ K[d0]. We have to define a map q1 : X1 K[d1] satisfying the properties of the Schreier retraction of (d1,s0). This map can be defined as:
q1(x) = q0(x) · s0(d1q0(x))−1,
where q0 is the Schreier retraction of (d0, s0). First of all, q1(x) ∈ K[d1] for any
x ∈ K[d0], since:
d1q1(x) = d1q0(x) · d1s0(d1q0(x))−1 = d1q0(x) · (d1q0(x))−1 = 1.
Moreover, x = q1(x) · s0d1(x) for any x ∈ X1, because:
q1(x) · s0d1(x) = q0(x) · s0(d1q0(x))−1 · s0d1(x) =
= q0(x) · s0(d1q0(x))−1 · s0d1(q0(x) · s0d0(x)) = = q0(x) · s0(d1q0(x))−1 · s0d1q0(x) · s0d0(x) =
= q0(x) · s0d0(x) = x.
Finally, we have that q1(x · s0(b)) = x for any x ∈ K[d1] and any b ∈ X0,
because:
where the second equality holds because
q0(x · s0(b)) = q0(x) · q0(s0d0(x) · q0s0(b)) = q0(x)
q1(x · s0(b)) = q0(x · s0(b)) · s0(d1q0(x · s0(b)))−1 =
= q0(x) · s0(d1q0(x))−1 = q0(x) · s0(d1q0(x))−1 · s0d1(x) = = q0(x) · s0(d1q0(x))−1 · s0d1q0(x) · s0d1s0d0(x) =
= q0(x) · s0d0(x) = x,
thanks to Proposition 2.1.5. This concludes the proof.