Bourbaki

Example 3

Tag 01R8, permanent. Cite it as [Bourbaki, Tag 01R8].

Topologie algébrique, chapter I, § 5. Revêtements principaux

Analogously, let E be a topological space and let G be a topological group operating continuously on the right in E. If the topological space G is endowed with the operation of G by left translations, the space E×GGE\times^GG is identified with E.

Let E and F be topological spaces and let G be a topological group operating continuously on the right in E and on the left in F. Let B be a topological space and let p:E→Bp: E\rightarrow B be a continuous mapping such that p(x⋅g)=p(x)p(x\cdot g) =p(x) for x∈Ex\in E and g∈Gg\in G. The mapping p∘p\circ pr1:E×F→B_1: E\times F\rightarrow B defines, by passing to the quotient, a continuous mapping pF:E×GF→Bp^F: E\times^GF\rightarrow B and the canonical mapping π:E×F→E×GF\pi : E\times F\rightarrow E\times^GF is a B-morphism.

Let B′B' be a topological space and let h:B′→Bh: B'\rightarrow B be a continuous mapping. The group G operates continuously on the right in B′×BEB'\times_BE by the operation law ((b′,x),g)↦(b′,x⋅g)((b', x), g)\mapsto (b', x\cdot g). By composing the canonical isomorphism ((b′,x),y)↦(b′,(x,y))((b', x), y)\mapsto (b',(x, y)) of (B′×BE)×F(B'\times_BE)\times F onto B′×B(E×F)B'\times_B(E\times F) and the mapping IdB′×π_{B'}\times \pi of B′×B(E×F)B'\times_B(E\times F) into B′×B(E×GF)B'\times_B(E\times^GF), one obtains a continuous mapping λ0:(B′×BE)×F→B′×B(E×GF)\lambda_0: (B'\times_BE)\times F\rightarrow B'\times_B(E\times^GF). It defines, by passing to the quotient, a continuous mapping

λ:(B′×BE)×GF→B′×B(E×GF) \lambda : (B'\times_BE)\times^GF\rightarrow B'\times_B(E\times^GF)

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.