Bourbaki

Example 5

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

Topologie algébrique, chapter II, § 3. Groupoïdes

Let φ:G→G′\varphi : G\rightarrow G' be a morphism of groupoids. Let H be the subquiver of G with set of vertices Som(G) and whose set of arrows consists of the f∈f\in Fl(G) such that φ(f)\varphi (f) is an identity element of G′G'. For every a∈a\in Som(G),ea∈Ha,a,e_a\in H_{a,a}. For every f∈f\in Fl(H), φ(f−1)=φ(f)−1\varphi (f^{-1}) =\varphi (f)^{-1}, hence f−1∈f^{-1}\in Fl(H). Further, if ff and gg are composable arrows of H, one has φ(fg)=φ(f)φ(g)=et(φ(f))eo(φ(g))=\varphi (f g) =\varphi (f)\varphi (g) =e_{t(\varphi(f))}e_{o(\varphi(g))}= et(φ(f))e_{t(\varphi(f))}, hence fgf g is an arrow of H. This shows that H is a subgroupoid of G. It is called the kernel of φ\varphi and is denoted by Ker(φ\varphi ).

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.