Bourbaki

Corollary 1

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

Topologie algébrique, chapter I, § 5. Revêtements principaux

Let G be a topological group operating continuously on the right in a topological space E. The following conditions are equivalent:

(i) The orbit space E/GE/G is Hausdorff and the space E, endowed with the canonical mapping p:E→E/Gp: E\rightarrow E/G, is a principal fibre space with group G;

(ii) The group G operates properly (TG, III, p. 27) and freely in E, and every point of E/GE/G possesses a neighbourhood over which there exists a continuous section of the mapping pp.

Put B=E/GB = E/G. Under each of the hypotheses (i) and (ii), the group G operates continuously and freely in E, so that the mapping θ:(x,g)↦(x,x⋅g)\theta : (x, g)\mapsto (x, x\cdot g) is a continuous bijection of E×GE\times G onto E×BEE\times_BE. Let us place ourselves under this hypothesis and denote by φ:E×BE→G\varphi : E\times_BE\rightarrow G the mapping pr2∘θ−1_2\circ \theta^{-1}. Having regard to formula (1), in order that θ\theta be a homeomorphism, it is necessary and sufficient that the mapping φ\varphi be continuous. On the other hand, since the equivalence relation defined by G is open (TG, III, p. 10, lemma 2), in order that the space E/GE/G be Hausdorff, it is necessary and sufficient that the graph E×BEE\times_BE of this equivalence relation be closed in E×EE\times E (TG, I, p. 55, prop. 8). Finally (TG, III, p. 31, prop. 6), in order that G operate properly in E, it is necessary and sufficient that E×BEE\times_BE be closed in E×EE\times E and that the mapping φ\varphi be continuous. The equivalence of conditions (i) and (ii) then follows from prop. 3.

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.