Bourbaki

Proposition 5

Tag 01ZI, permanent. Cite it as [Bourbaki, Tag 01ZI].

Topologie algébrique, chapter III, § 4. Homotopie et revêtements

The mapping h(E,x)h_{(E,x)} of π1(B,b)\pi_1(B, b) into G which, to γ∈\gamma \in π1(B,b)\pi_1(B, b), associates the unique element gg of G such that x⋅g=x⋅γ−1x\cdot g=x\cdot \gamma^{-1}, is a group homomorphism, whose kernel is the subgroup p∗(π1(E,x))p_*(\pi_1(E, x)) of π1(B,b)\pi_1(B, b). If E is connected, this homomorphism is surjective.

For every g∈Gg\in G, one has h(E,x⋅g)=h_{(E,x\cdot g)}= Int(g−1g^{-1})∘h(E,x)\circ h_{(E,x)}.

Let E be a principal covering of B with group G, let xx be a point of EbE_b, and let us denote by pp the projection of E. For every g∈Gg\in G, the mapping y↦y⋅gy\mapsto y\cdot g is a B-automorphism of E. Hence, for every g∈Gg\in G, every y∈Eby\in E_b, and every γ∈π1(B,b)\gamma \in \pi_1(B, b), one has the relation (y⋅g)⋅γ=(y⋅γ)⋅g((y\cdot g)\cdot \gamma = (y\cdot \gamma )\cdot g(cf. III, p. 305, relation (4)). Consequently, for γ,δ∈π1(B,b)\gamma , \delta \in \pi_1(B, b), one has

x⋅h(E,x)(γδ)=x⋅δ−1γ−1 x\cdot h_{(E,x)}(\gamma \delta ) =x\cdot \delta^{-1}\gamma^{-1}
=(x⋅h(E,x)(δ))⋅γ−1 = (x\cdot h_{(E,x)}(\delta ))\cdot \gamma^{-1}
=(x⋅γ−1)⋅h(E,x)(δ) = (x\cdot \gamma^{-1})\cdot h_{(E,x)}(\delta )
=x⋅h(E,x)(γ)h(E,x)(δ) =x\cdot h_{(E,x)}(\gamma )h_{(E,x)}(\delta )

which proves that h(E,x)h_{(E,x)} is a group homomorphism.

Its kernel is the stabilizer of xx for the canonical operation of π1(B,b)\pi_1(B, b), hence is equal to p∗(π1(E,x))p_*(\pi_1(E, x)) by Theorem 1 of III, p. 305. The mapping g↦x⋅gg\mapsto x\cdot g is a bijection of G onto EbE_b. The image of the homomorphism h(E,x)h_{(E,x)} is therefore the set of g∈Gg\in G such that x⋅gx\cdot g belongs to the orbit of xx for this operation. If E is connected, π1(B,b)\pi_1(B, b) operates transitively on Eb(E_b(loc. cit.) and the homomorphism h(E,x)h_{(E,x)} is surjective.

Let g∈Gg\in G; for every γ∈π1(B,b)\gamma \in \pi_1(B, b), one has

x⋅gh(E,x⋅g)(γ)=(x⋅g)⋅γ−1=(x⋅γ−1)⋅g=x⋅h(E,x)(γ)g x\cdot gh_{(E,x\cdot g)}(\gamma ) = (x\cdot g)\cdot \gamma^{-1}= (x\cdot \gamma^{-1})\cdot g=x\cdot h_{(E,x)}(\gamma )g

whence h(E,x⋅g)(γ)=g−1h(E,x)(γ)gh_{(E,x\cdot g)}(\gamma ) =g^{-1}h_{(E,x)}(\gamma )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.

Cited by

Nothing.

Cites

Nothing.

Machine translation of the English, by gpt-5.4, against glossary version 34. Not checked by a person.