I'm trying to understand the first part of Example 1.2.8 from here: https://arxiv.org/pdf/1612.09375.pdf
Let $Ob(\mathcal G)=\{\star\}$. A functor $F:\mathcal G\to \mathbf{Set}$ consists of:
- An assignment $F: Ob(\mathcal G)\to Ob(\mathbf{Set}),\star\mapsto S_\star$. This is indeed "the same as" choosing a set (I guess formally this means that the class of such assignments is in bijection with the class of sets.)
- An assignment $F: \mathcal G(\star,\star)\to\mathbf {Set}(S_\star,S_\star)$ satisfying $F(f\circ g)=F(f)\circ F(g)$ and $F(1_\star)=1_{S_\star}$ for all $f,g:\star\to\star$. Since $\mathcal G(\star,\star)$ is bijective to the set of elements of the monoid $G$ and since $\circ$ in the category corresponds to $\cdot$ in the monoid, the above can be written as $F:G\to \mathbf {Set}(\star,\star)$ subject to $F(f\cdot g)=F(f)\circ F(g)$, $F(1_\star)=1_{S_\star}$.
How does one get from the above that $F:\mathcal G\to\mathbf{Set}$ consists of a set $S$ together with, for each $g\in G$, a function $F(g):S\to S$, satisfying the functoriality axioms, as claimed in the text linked above?
Best Answer
As Daniel says in the comments, the claim is nothing more than 'unpacking' the definition of functor in this particular case.
The first realization one has to have is that a groupoid $\mathcal{G}$ that has only one object $*$ "is a group". That is, the arrows $G = \mathcal{G}(*,*)$ for a group and determine $\mathcal{G}$ (recall that for any category one could forget the objects and just work with arrows, as the former are represented by identities).
Now, to be formal, consider the category $G\mathsf{Set}$ of $G$-sets toghether to functions that commute with the $G$-actions. We can think of the objects here as pairs $(X,\rho)$ where $\rho : G \to S(X)$ is the action.
Now, as per your bullet points we can define the functor
$$ \begin{align} \mathcal{\Gamma} :\mathsf{Set}&^\mathcal{G} \to G\mathsf{Set}\\ & F \longmapsto (F* , \rho_F) \\ & \downarrow_{\eta}\ \mapsto \quad \downarrow_{\eta_*}\\ & F' \mapsto (F'*,\rho_{F'}) \end{align} $$
where $\rho_F(g)(x) = F(g)(x)$ and $\eta_* : F* \to F'*$ is the $*$-component of the natural transformation $\eta$.
You can check that this is not only an equivalence of categories but a category isomorphism, with the inverse sending $(X,\rho)$ to the functor that maps $* \mapsto X$ and $ * \xrightarrow{g} * $ to $\rho(g) : X \to X$. Likewise, a $G$-function $h$ from $(X,\rho)$ to $(X',\rho')$ gives rise to a natural transformation whose only component is $h$ itself.