Bourbaki

Example 2

Tag 01Q7, permanent. Cite it as [Bourbaki, Tag 01Q7].

Topologie algébrique, chapter I, § 5. Revêtements principaux

Let (E,p)(E, p) be a principal fibre space with base B and group G; let (E′,p′)(E', p') be a principal fibre space with base B′B' and group G′G'. We define a law of operation on the right of the group G×G′G\times G' on the space E×E′E\times E' by putting (x,x′)⋅(g,g′)=(x⋅g,x′⋅g′)(x, x')\cdot (g, g') = (x\cdot g, x'\cdot g') for x∈E,x′∈E′,g∈Gx\in E,x'\in E',g\in G and g′∈G′g'\in G'; this operation is continuous. Let U be an open subset of B and f:U×G→p−1(U)f: U\times G\rightarrow \overset{-1}{p}(U) a trivialization of the U-space (p−1(U),pU)(\overset{-1}{p}(U), p_U);

let U′U' be an open subset of B′B' and f′:U′×G′→(−p′1)(U′)f': U'\times G'\rightarrow (^-{p'}^1)(U') a trivialization of the U′U'-space ((−p′1)(U′),pU′(^-{p'}^1)(U'), p'_U). The mapping ((b,b′),(g,g′))↦((b, b'),(g, g'))\mapsto (f(b,g),f′(b′,g′))(f(b, g), f'(b', g')) is a (U×U′)(U\times U')-isomorphism f′′f'' of (U×U′)×(G×G′)(U\times U')\times (G\times G')

onto p−1(U)×(−p′1)(U′)\overset{-1}{p}(U)\times (^-{p'}^1)(U') and one has

f′′((b,b′),(gh,g′h′))=(f(b,gh),f′(b′,g′h′))=f′′((b,b′),(g,g′))⋅(h,h′) f''((b, b'),(gh, g'h')) = (f(b, gh), f'(b', g'h')) =f''((b, b'),(g, g'))\cdot (h, h')

It follows that the (B×B′)(B\times B')-space E×E′E\times E', endowed with the law of operation of G×G′G\times G' defined above, is a principal fibre space, called the product of the principal fibre spaces E and E′E'.

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.