Bourbaki

Definition 7

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

Topologie algébrique, chapter II, § 3. Groupoïdes

Let G be a groupoid. One says that a subgroupoid H of G is normal if Som(H) = Som(G) and if for every pair (a,b)(a, b) of vertices of H and every arrow f∈Ga,bf\in G_{a,b}, one has Int(ff)(Hb)=Ha(H_b) = H_a.

Let G be a groupoid and let H be a subgroupoid of G whose set of vertices is equal to Som(G). To verify that H is distinguished, it is enough to prove that fHbf−1⊂HafH_bf^{-1}\subset H_a for every a∈a\in Som(G), every b∈b\in Som(G) and every f∈Ga,bf\in G_{a,b}. In fact, if this is so, one also has f−1Haf⊂Hbf^{-1}H_af\subset H_b for every arrow f∈Gb,af\in G_{b,a}, that is, Ha⊂fHbf−1H_a\subset fH_bf^{-1}, and consequently fHbf−1=HafH_bf^{-1}= H_a.

Let G be a groupoid and let H be a distinguished subgroupoid of G. For every vertex aa of G, the subgroup HaH_a of GaG_a is a distinguished subgroup of GaG_a.

Let φ:G→G′\varphi : G\rightarrow G' be a morphism of groupoids. The kernel of φ(\varphi (II, p. 166, example 5) is a distinguished subgroupoid of G. In fact, let ff be an arrow in G joining aa to bb and let g∈g\in Ker(φ\varphi )b_b; then one has φ(fgf−1)=φ(f)φ(g)φ(f−1)=φ(f)eφ(b)φ(f)−1=eφ(a)\varphi (f gf^{-1}) =\varphi (f)\varphi (g)\varphi (f^{-1}) =\varphi (f)e_{\varphi(b)}\varphi (f)^{-1}=e_{\varphi(a)}, whence the inclusion ffKer(φ\varphi )bf−1⊂_bf^{-1}\subset Ker(φ\varphi )a_a.

Let G be a groupoid. The groupoid G and the subgroupoid of G whose set of arrows is the set of identity elements of G are distinguished subgroupoids of G. The intersection of a family of distinguished subgroupoids of G is a distinguished subgroupoid of G. In particular, for every subset F⊂F\subset Fl(G), there exists a smallest distinguished subgroupoid of G whose set of arrows contains F. It is called the distinguished subgroupoid generated by F.

Let H be a distinguished subgroupoid of G. Let R\mathscr{R} be the equivalence relation in Fl(G) defined by R{f,g}\mathscr{R}\{f, g\} if and only if there exist arrows xx and yy in Fl(H) such that f=xgyf=xgy. If ff and gg are arrows of G equivalent modulo R\mathscr{R}, their origins (resp. their ends) belong to the same orbit of H. Put F′=F'= Fl(G)/R/\mathscr{R} and denote by o′o' and t′t' the mappings of F′F' into Orb(H) deduced from oo and tt by passing to the quotients. Denote by G/HG/H the quiver (Orb(H),F′,o′,t′,F', o', t').

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.