Bourbaki

Theorem 2

Tag 01R3, permanent. Cite it as [Bourbaki, Tag 01R3].

Topologie algébrique, chapter I, § 5. Revêtements principaux

Let B be a connected topological space, let E be a covering of B, connected and nonempty. The following properties are equivalent:

(i) The covering E is Galois;

(ii) There exists a discrete topological group G and a continuous right operation of G on E making E into a principal covering with group G;

(iii) The covering E×BEE\times_BE of E defined by the projection pr1_1 is trivializable.

When these conditions are satisfied, the canonical mapping G→G\rightarrow AutB(E)∘_B(E)^{\circ} defined by the operation of G is a group isomorphism.

If, moreover, the space B is locally connected, the preceding properties are equivalent to the following property:

(i′') There exists a point bb of B such that the operation of the group AutB(E)_B(E) on the fibre EbE_b is transitive.

The implication (i)⇒\Rightarrow(ii) results from proposition 5 and the implication (ii)⇒\Rightarrow(iii) from remark 1 of I, p. 95. Let us prove (iii)⇒\Rightarrow(i). Let us denote by p:E→Bp: E\rightarrow B the projection of the covering E. Since B is connected, this covering has a degree, and this degree is not zero since E is not empty. Let bb be a point of B. Let us prove that the operation of AutB(E)_B(E) on the fibre EbE_b is transitive. From the above, this fibre is not empty. Let xx and x′x' be points of EbE_b. The B-morphisms h:E→Eh: E\rightarrow E such that h(x)=x′h(x) =x' correspond bijectively to the continuous sections ss of the mapping pr1:E×BE→E_1: E\times_BE\rightarrow E such that s(x)=(x,x′)s(x) = (x, x') (I, p. 9, prop. 3). Under assumption (iii), such a section exists (I, p. 70, cor. 2) and is unique (I, p. 34, cor. 1 of prop. 11) since the space E is connected. Thus, for every pair (x,x′)∈E×BE(x, x')\in E\times_BE, there exists a unique B-morphism h:E→Eh: E\rightarrow E such that h(x)=x′h(x) =x'. If h′:E→Eh': E\rightarrow E is the unique B-morphism such that h′(x′)=xh'(x') =x, one has h′(h(x))=xh'(h(x)) =x and h(h′(x′))=x′h(h'(x')) =x', whence h′∘h=h'\circ h= IdE_E and h∘h′=h\circ h'= IdE′_{E'}. This proves that hh is a B-automorphism of E. Hence the covering E is galoisian.

We have proved that conditions (i), (ii), (iii) are equivalent. Suppose them satisfied. Let δ:G→\delta : G\rightarrow AutB(E)_B(E) be the mapping which to g∈Gg\in G associates the B-automorphism x↦x⋅gx\mapsto x\cdot g of E. The mapping δ\delta is a group homomorphism of G into AutB(E)∘_B(E)^{\circ}. Since G operates freely in E, the mapping δ\delta is injective. Let h∈h\in AutB(E)_B(E), let x∈x\in E and let gg be the unique element of G such that x⋅g=h(x)x\cdot g=h(x). The B-morphisms hh and δ(g)\delta (g) coincide at xx, hence everywhere, since the space E is connected (I, p. 34, cor. 1 of prop. 11), which proves that δ\delta is an isomorphism.

Now suppose the space B locally connected and let us prove, under assumption (i′'), that the covering E is galoisian. Let E′E' be the quotient space of E by the right operation of AutB(E)∘_B(E)^{\circ} and let us denote by q:E→E′q: E\rightarrow E' the canonical mapping. It makes E into a surjective covering of E′(E'(I, p. 98, example 2); the mapping p:E→Bp: E\rightarrow B defines by passing to the quotient a continuous mapping p′:E′→Bp': E'\rightarrow B such that p′∘q=pp'\circ q=p. Since the space B is locally connected, the B-space (E′,p′)(E', p') is a covering (I, p. 81, prop. 7). Under assumption (i′'), there exists a point bb of B such that the fibre Eb′E'_b has exactly one element; since the space B is connected, the same is then true for every point bb of B (I, p. 74, prop. 4), which proves that E is a galoisian covering of B.

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.