The mapping h(E,x) of π1(B,b) into G which, to γ∈ π1(B,b), associates the unique element g of G such that x⋅g=x⋅γ−1, is a group homomorphism, whose kernel is the subgroup p∗(π1(E,x)) of π1(B,b). If E is connected, this homomorphism is surjective.
For every g∈G, one has h(E,x⋅g)= Int(g−1)∘h(E,x).
Let E be a principal covering of B with group G, let x be a point of Eb, and let us denote by p the projection of E. For every g∈G, the mapping y↦y⋅g is a B-automorphism of E. Hence, for every g∈G, every y∈Eb, and every γ∈π1(B,b), one has the relation (y⋅g)⋅γ=(y⋅γ)⋅g(cf. III, p. 305, relation (4)). Consequently, for γ,δ∈π1(B,b), one has
x⋅h(E,x)(γδ)=x⋅δ−1γ−1
=(x⋅h(E,x)(δ))⋅γ−1
=(x⋅γ−1)⋅h(E,x)(δ)
=x⋅h(E,x)(γ)h(E,x)(δ)
which proves that h(E,x) is a group homomorphism.
Its kernel is the stabilizer of x for the canonical operation of π1(B,b), hence is equal to p∗(π1(E,x)) by Theorem 1 of III, p. 305. The mapping g↦x⋅g is a bijection of G onto Eb. The image of the homomorphism h(E,x) is therefore the set of g∈G such that x⋅g belongs to the orbit of x for this operation. If E is connected, π1(B,b) operates transitively on Eb(loc. cit.) and the homomorphism h(E,x) is surjective.
Let g∈G; for every γ∈π1(B,b), one has
x⋅gh(E,x⋅g)(γ)=(x⋅g)⋅γ−1=(x⋅γ−1)⋅g=x⋅h(E,x)(γ)g
whence h(E,x⋅g)(γ)=g−1h(E,x)(γ)g. This 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.
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