Editing Groupoid (section)

=== Comparing the definitions ===
The algebraic and category-theoretic definitions are equivalent, as we now show. Given a groupoid in the category-theoretic sense, let ''G'' be the [[disjoint union]] of all of the sets ''G''(''x'',''y'') (i.e. the sets of morphisms from ''x'' to ''y''). Then <math>\mathrm{comp}</math> and <math>\mathrm{inv}</math> become partial operations on ''G'', and <math>\mathrm{inv}</math> will in fact be defined everywhere. We define ∗ to be <math>\mathrm{comp}</math> and <sup>−1</sup> to be {{tmath|1= \mathrm{inv} }}, which gives a groupoid in the algebraic sense. Explicit reference to ''G''<sub>0</sub> (and hence to {{tmath|1= \mathrm{id} }}) can be dropped.

Conversely, given a groupoid ''G'' in the algebraic sense, define an equivalence relation <math>\sim</math> on its elements by
<math>a \sim b</math> iff ''a'' ∗ ''a''<sup>−1</sup> = ''b'' ∗ ''b''<sup>−1</sup>. Let ''G''<sub>0</sub> be the set of equivalence classes of {{tmath|1= \sim }}, i.e. {{tmath|1= G_0:=G/\!\!\sim }}. Denote  ''a'' ∗ ''a''<sup>−1</sup> by <math>1_x</math> if <math>a\in G</math> with {{tmath|1= x\in G_0 }}.

Now define <math>G(x, y)</math> as the set of all elements ''f'' such that <math>1_x*f*1_y</math> exists. Given <math>f \in G(x,y)</math> and {{tmath|1= g \in G(y, z) }}, their composite is defined as {{tmath|1= gf:=f*g \in G(x,z) }}. To see that this is well defined, observe that since <math>(1_x*f)*1_y</math> and <math>1_y*(g*1_z)</math> exist, so does {{tmath|1= (1_x*f*1_y)*(g*1_z)=f*g }}. The identity morphism on ''x'' is then {{tmath|1= 1_x }}, and the category-theoretic inverse of ''f'' is ''f''<sup>−1</sup>.

Sets in the definitions above may be replaced with [[class (set theory)|class]]es, as is generally the case in category theory.