If N is a distinguished subgroup of Ga, the set of elements z∈Ω(Γ) such that c(z) is contained in N is a distinguished subset of Ω(Γ).
Let Z denote the set of elements z∈Ω(Γ) such that c(z)⊂N.
Let z∈ Fl(Γ). We have c(z,z)={ea} by definition, whence (z,z)∈Z.
Let (z1,...,zn)∈Z. By definition of c, the conjugacy class c(z1,...,zn) is equal to c(zn,z1,...,zn−1) and is formed by the inverses of the elements of c(zn,...,z1). This shows that Z satisfies condition (ii).
With the notation of condition (iii), one can choose elements u∈c(z),v∈c(z′) and w∈c(zz′) such that uv=w. If two of the elements u,v and w belong to N, the same is true of the third, therefore N satisfies (iii).
The notation being that of (iv), we have, for every integer i such that 1⩽i⩽n, the relation
φ(fi)h(bi)=h(ai)ψ(fi)
for h is a homotopy joining φ to ψ. This equality can also be written
φ(fi,εi)h(bi)εi=h(ai)εiψ(fi,εi)
Taking into account the relations ψ(fi,εi)=φ(fi+1,εi+1) for 1⩽i<n and ψ(fn,εn)=φ(f1,ε1), one deduces
so that the conjugacy classes c((a1,ε1),...,(an,εn)) and c((b1,ε1),...,(bn,εn)) are equal. This shows that Z satisfies condition (iv) and completes the proof of the proposition.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.