Bourbaki

Remark

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

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

Let bb and b′b' be two points of B,x,x a point of Eb,c∈E_b,c\in Λb,b′(B)\Lambda_{b,b'}(B) a path in B and c~\widetilde{c} the path in E originating at xx which lifts cc. For every s∈Is\in \mathbf{I}, let us denote by csc_s the path t↦c(st)t\mapsto c(st); the path in E originating at xx which lifts csc_s is the path t↦c~(st)t\mapsto \widetilde{c}(st) whose term is the point c~(s)\widetilde{c}(s). We have therefore c~(s)=x⋅cs\widetilde{c}(s) =x\cdot c_s for every s∈Is\in \mathbf{I}.

Let B′B' be a topological space and let p′:E′→B′p': E'\rightarrow B' be an étale and separated mapping satisfying the path lifting property. Let f:B→B′f: B\rightarrow B' and g:E→E′g: E\rightarrow E' be continuous mappings such that p′∘g=f∘pp'\circ g=f\circ p. For b,b′∈B,γ∈ϖb,b′(B)b,b'\in B,\gamma \in \varpi_{b,b'}(B) and x∈Ebx\in E_b, we have:

g(x⋅γ)=g(x)⋅f∗(γ)(3) g(x\cdot \gamma ) =g(x)\cdot f_*(\gamma ) \tag{3}

In fact, let cc be a path in B whose strict homotopy class is γ\gamma, and let us denote by c~\widetilde{c} the path in E originating at xx which lifts cc; the path g∘c~g\circ \widetilde{c} is then the lift originating at g(x)g(x) of f∘cf\circ c in E′E'.

In particular, for B=B′B = B' and f=f= IdB_B, we have

g(x⋅γ)=g(x)⋅γ(4) g(x\cdot \gamma ) =g(x)\cdot \gamma \tag{4}

Let p:E→Bp: E\rightarrow B and p′:E′→Bp': E'\rightarrow B be étale and separated mappings satisfying the path lifting property. Let bb be a point of B. If f:E→E′f: E\rightarrow E' is a B-morphism, the mapping fb:Eb→Eb′f_b: E_b\rightarrow E'_b is a π1(B,b)\pi_1(B, b)-morphism.

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.