Bourbaki

Corollary 3

Tag 01V3, permanent. Cite it as [Bourbaki, Tag 01V3].

Topologie algébrique, chapter II, § 5. Coégalisateur

Suppose that the groupoid G is simply transitive. Let K0K_0 be the subgroup of K generated by the stabilizers of the vertices of G. The mapping from K into (G/K)γ(o)(G/K)_{\gamma(o)} which to k∈Kk\in K associates γ(ck)\gamma (c_k) is a surjective group homomorphism, with kernel K0K_0.

If an element k∈Kk\in K fixes a vertex aa of G, the element g−1kgg^{-1}kg fixes the vertex g∗ag^*a; this implies that K0K_0 is a normal subgroup of K.

By proposition 5, the unique homomorphism λ:F(K)→\lambda : F(K)\rightarrow (G/K)γ(o)(G/K)_{\gamma(o)} such that λ([k])=γ(ck)\lambda ([k]) =\gamma (c_k) is surjective, and its kernel is the smallest normal subgroup of F(K) containing the elements [k][k], where kk is an element of K which fixes a vertex of G, and the elements [kh]−1[k][h][kh]^{-1}[k][h], where (k,h)∈K2(k, h)\in K^2. In particular, the mapping λ′:K→(G/K)γ(o)\lambda ': K\rightarrow (G/K)_{\gamma(o)}, defined by λ′(k)=γ(ck)=λ([k])\lambda '(k) =\gamma (c_k) =\lambda ([k]) for k∈Kk\in K, is a group homomorphism. One has λ=λ′∘p\lambda =\lambda '\circ p, where p:F(K)→Kp: F(K)\rightarrow K denotes the canonical surjective group homomorphism. Consequently, the homomorphism λ′\lambda ' is surjective and its kernel is the smallest normal subgroup of K which contains the elements kk, for kk fixing a vertex of G, that is to say K0K_0. This proves the proposition.

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.