Functor from $BG \to \text{Gpd}$ induces functors for each $g \in G$.

abstract-algebracategory-theoryfunctorsgroupoids

I'm reading Groupoids and Stuff. Page 7.

Definition 1.3.1. A group action of a a finite group $G$ on a groupoid $X$ is a functor $A : B G \to \text{Gpd}$ such that $A(I) = X$ where $I$ is the unique object of $BG$ (the category with one object and morphisms elements of $G$.)

This by definition means that each element $g \in G$ induces a functor $A(g) : X \to X$.

I'm having trouble seeing this. Can you explain?

Best Answer

A functor takes a morphism $f : x \to y$ to a morphism $\mathcal{F}(f) : f(x) \to f(y)$.

In $BG$, the only object is $I$ and the only morphisms are the group elements $g : I \to I$. The functor $A$ takes one of these morphisms to a morphism $A(g) : A(I) \to A(I)$. Since $A(I) = X$ we have $A(g) : X \to X$.