Set-sized groupoids, functors and composition conversion

A small category is a category whose objects and morphisms each form a set; each morphism carries its source and target. A groupoid is a category in which every morphism has a two-sided inverse. A groupoid is connected when it is nonempty and there is an arrow between every ordered pair of objects.

For arrows $a:x\to y$ and $b:y\to z$, write $ab=b\circ a$; hence composition is read from left to right. A covariant functor sends objects to objects and arrows to arrows with the corresponding sources and targets, preserving identities and composition. An automorphism is a functor from a category to itself with a two-sided inverse functor.

For an object $x$ of a small groupoid $\mathcal{G}$, its vertex group ${\mathcal{G}}_x$ is the set of arrows $x\to x$ with the left-to-right product and identity $1_x$. Given an arrow $a:x\to y$, the associated conjugation map $c_a:{\mathcal{G}}_x\to {\mathcal{G}}_y$ is $c_a(u)=a^{-1}ua$.

Category, object, morphism, domain, codomain, identity, composition, and hom-collection

Covariant functor, identity functor, composite functor, and contravariant functor

Isomorphism, groupoid, and connected category

Group and abelian group

The category axioms give associative composition and identities; in a groupoid every arrow has an inverse. Thus the endomorphisms at $x$ are closed under the indicated composition and form a group, with identity $1_x$ and inverses inherited from the groupoid. The left-to-right notation is a relabelling of the existing composite, so associativity and identities remain valid. The connectedness condition agrees with the zigzag definition because arrows in a groupoid can be reversed and composed.

For $a:x\to y$, the composite $a^{-1}ua$ is an endomorphism of $y$. Associativity and $a{a}^{-1}=1_x$ show $c_a(uv)=c_a(u)c_a(v)$; the map $v\mapsto av{a}^{-1}$ is its two-sided inverse. Hence $c_a$ is a group isomorphism (Group isomorphisms, automorphisms and the set Aut(G)). It depends on the specified arrow $a$; no global choice of connecting arrows is required.