Bourbaki

Proposition 6

Tag 0204, permanent. Cite it as [Bourbaki, Tag 0204].

Topologie algébrique, chapter III, § 5. Homotopie et revêtements (cas des espaces localement connexes par arcs)

Let B be a topological space and let bb be a point of B. Endow the group π1(B,b)\pi_1(B, b) with the admissible topology.

a) For every discrete group G, every principal covering E of B with group G, and every x∈Ebx\in E_b, the homomorphism h(E,x)h_{(E,x)} is a continuous homomorphism of topological groups.

b) For every covering E of B, the canonical action of π1(B,b)\pi_1(B, b) on the fibre EbE_b is admissible.

It is enough to prove the second assertion. Let K be the set of elements k∈π1(B,b)k\in \pi_1(B, b) such that x⋅k=xx\cdot k=x for every x∈Ebx\in E_b; let us prove that K is an admissible subgroup of π1(B,b)\pi_1(B, b).

Let b′b' be a point of B and let V′V' be a neighbourhood of b′b' in B over which the covering E is trivializable. Let γ\gamma be the class of a path cc joining bb to b′b' in B and let c′c' be the unique path in E with origin xx which lifts cc. For every δ∈π1(V′,b′)\delta \in \pi_1(V', b') and every x′∈Eb′x'\in E_{b'}, one has x′⋅δ=x′x'\cdot \delta =x'. Consequently, for x∈Ebx\in E_b and δ∈π1(V′,b′)\delta \in \pi_1(V', b'), one has

x⋅γδγ−1=((x⋅γ)⋅δ)⋅γ−1=(x⋅γ)⋅γ−1=x x\cdot \gamma \delta \gamma^{-1}= ((x\cdot \gamma )\cdot \delta )\cdot \gamma^{-1}= (x\cdot \gamma )\cdot \gamma^{-1}=x

so that γδγ−1\gamma \delta \gamma^{-1} belongs to K. In other words, K is an admissible subgroup, whence 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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Cites

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