Bourbaki

Proposition 2

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

Topologie algébrique, chapter III, § 5. Homotopie et revêtements (cas des espaces localement connexes par arcs)

Let E and E′E' be coverings of B.

a) Let f:E→E′f: E\rightarrow E' be a B-morphism. The mapping fb:Eb→Eb′f_b: E_b\rightarrow E'_b derived from ff is compatible with the operations of π1(B,b)\pi_1(B, b) on EbE_b and Eb′E'_b respectively. It is bijective if and only if ff is an isomorphism.

b) The mapping f↦fbf\mapsto f_b is a bijection of the set CB(E;E′)\mathscr{C}_B(E; E') of B-morphisms from E into E′E' onto the set Fπ1(B,b)(Eb;Eb′)\mathscr{F}_{\pi_1(B,b)}(E_b; E'_b) of π1(B,b)\pi_1(B, b)-morphisms from EbE_b into Eb′E'_b.

If ff is a B-morphism from E into E′E', the mapping fb:Eb→Eb′f_b: E_b\rightarrow E'_b is a π1(B,b)\pi_1(B, b)-morphism (cf. III, p. 305). Moreover, two B-morphisms from E into E′E' which coincide on the fibre EbE_b are equal (I, p. 80, cor. 3 of prop. 6).

Let φ:Eb→Eb′\varphi : E_b\rightarrow E'_b be a morphism of π1(B,b)\pi_1(B, b)-sets; let us prove that there exists a B-morphism ff of E into E′E' such that fb=φf_b=\varphi. We may suppose that the spaces E and E′E' are connected and non-empty, so that EbE_b and Eb′E'_b are homogeneous π1(B,b)\pi_1(B, b)-sets. Let xx be a point of EbE_b; its stabilizer G is the image of the mapping π1(p,x)\pi_1(p, x) (III, p. 305, Theorem 1). Since the mapping φ\varphi is a π1(B,b)\pi_1(B, b)-morphism, the group G fixes the point x′=φ(x)x'=\varphi (x) of Eb′E'_b, and is therefore contained in the image of the mapping π1(p′,x′)\pi_1(p', x'). By Prop. 1 of III, p. 308, there exists a B-morphism ff of E into E′E' such that f(x)=x′f(x) =x'. The mappings fbf_b and φ\varphi are π1(B,b)\pi_1(B, b)-morphisms of the homogeneous π1(B,b)\pi_1(B, b)-set EbE_b into Eb′E'_b which agree at the point xx; they are therefore equal.

If f:E→E′f: E\rightarrow E' is a B-isomorphism, the mapping fb:Eb→Eb′f_b: E_b\rightarrow E'_b is bijective. Conversely, suppose that the mapping fbf_b is bijective, and let g:E′→g: E'\rightarrow E be a B-morphism such that gb=(fb)−1g_b= (f_b)^{-1}. The B-morphism g∘f:E→Eg\circ f: E\rightarrow E induces on EbE_b the identical mapping; it therefore follows from the proposition that g∘f=g\circ f= IdE_E. Analogously, f∘g=f\circ g= IdE′_{E'}, which proves that ff is a B-isomorphism.

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.