Bourbaki

Example 2

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

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

Let H be a discrete topological group, let f:G→Hf: G\rightarrow H be a group morphism. Endow H with the left operation of G given by g⋅h=f(g)hg\cdot h=f(g)h, for g∈Gg\in G and h∈Hh\in H. Let E′=E×GHE'= E\times^GH be the associated covering; it is a principal covering of group H (I, p. 107, Example 6). Let us denote by q:E×H→E×GHq: E\times H\rightarrow E\times^GH the canonical mapping and put x′=q(x,e)x'=q(x, e). We have h(E′,x′)=f∘h(E,x)h_{(E',x')}=f\circ h_{(E,x)}.

In fact, let cc be a loop at bb in B, let γ\gamma be its class in π1(B,b)\pi_1(B, b) and let g=h(E,x)g=h_{(E,x)}; one has therefore x⋅γ=x⋅g−1x\cdot \gamma =x\cdot g^{-1}. Let c~\widetilde{c} be a path with origin xx in E; then t↦q(c~(t),e)t\mapsto q(\widetilde{c}(t), e) is a path with origin x′x' in E′E' which lifts the path p∘c~p\circ \widetilde{c} in B, so that

x′⋅γ=q(x⋅γ,e)=q(x⋅g−1,e)=q(x,f(g)−1)=q(x,e)⋅f(g)−1 x'\cdot \gamma =q(x\cdot \gamma , e) =q(x\cdot g^{-1}, e) =q(x, f(g)^{-1}) =q(x, e)\cdot f(g)^{-1}

which proves that h(E′,x′)(γ)=f(g)h_{(E',x')}(\gamma ) =f(g).

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