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?

  • $\begingroup$ It's just the definition of functor: the first bullet tells you what the functor does on objects of $\mathcal{G}$ and the second bullet tells you what the functor does on morphisms of $\mathcal{G}$. $\endgroup$ Jun 19, 2019 at 23:46
  • $\begingroup$ Right, that's how I got them. But how to use my bullets to obtain what is claimed in the text (and what is in italics in my question)? $\endgroup$
    – user557
    Jun 19, 2019 at 23:51
  • $\begingroup$ The data of a functor $F: \mathcal{G} \to \mathbf{Set}$ is a map on objects $\operatorname{Ob}(\mathcal{G}) \to \operatorname{Ob}(\mathbf{Set})$ (in this case, simply specifying the target set $F(\star)$), along with a map of hom-sets $F_{\star, \star}: \operatorname{Hom}_{\mathcal{G}}(\star, \star) \to \operatorname{Hom}_{\mathbf{Set}}(F(\star), F(\star))$. This data needs to satisfy some axioms about how composition and identity behave. Note that for any element of $G$, $F_{\star, \star}(g)$ is a function from $F(\star)$ to itself. $\endgroup$
    – Joppy
    Jun 20, 2019 at 3:18
  • $\begingroup$ @Joppy My problem is that I don't understand how the author gets e.g. $(g'g)\cdot s=g'\cdot(g\cdot s)$ from the functoriality condition that I wrote in the question. $\endgroup$
    – user557
    Jun 20, 2019 at 4:18
  • 1
    $\begingroup$ Let $g, h$ be elements of the monoid $G$. Then $F_{\star, \star}(g)$ and $F_{\star, \star}(h)$ are functions from $F(\star)$ to itself. One of the conditions to be a functor gives that $F_{\star, \star}(g) \circ F_{\star, \star}(h) = F_{\star, \star}(g \circ h)$, where the first is a composition of functions, and the second is a composition inside the monoid $G = \operatorname{Hom}_\mathcal{G}(\star, \star)$. $\endgroup$
    – Joppy
    Jun 20, 2019 at 4:20

2 Answers 2


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.

  • $\begingroup$ Thank you, but it doesn't quite fit into the framework of the cited text: the text does not assume familiarity with natural transformations (at this point), nor does it assume familiarity with $G$-sets: as far as I understand, they are defined in this example as our functor $\mathcal C\to\mathbf {Set}$. $\endgroup$
    – user557
    Jun 20, 2019 at 0:34
  • $\begingroup$ Ok, forget that $\Gamma$ is a category isomorphism. In particular it is bijective on objects, which is what you wanted. Every $G$-set is an instance of such a functor. If $G$-sets are defined this way, then there is not much else to say other than what you yourself have written in the post. $\endgroup$
    – qualcuno
    Jun 20, 2019 at 0:36
  • $\begingroup$ I’m not sure that an answer involving groupoids, the category of $G$-sets, and natural transformations is helpful given that the original question was essentially about following through the definition of a functor. $\endgroup$
    – Joppy
    Jun 20, 2019 at 4:44
  • $\begingroup$ I mean OP's bullet points seemed as he was familiar with these things, and wanted to know why this construction translated to having a $G$-set for some group $G$. Context was behind a link. But yes, I (a posteriori) agree that this is not as helpful as it could be. $\endgroup$
    – qualcuno
    Jun 20, 2019 at 5:11

Let $M$ be a monoid regarded as a one-object category $\mathscr M$ with unique object $\star$.

We first show that any functor $F: \mathscr M\to\mathbf{Set}$ gives rise to a left $M$-set (which is, by definition, a pair $(S,\cdot)$, where $S$ is a set and $\cdot$ is a left action of $M$, i.e., a map $$M\times S\to M,\\(m,s)\mapsto m\cdot s $$ such that $(m_1m_2)\cdot s=m_1\cdot(m_2\cdot s)$ and $e\cdot s=s$, where $e$ is the identity of $M$.)

Let $S=F(\star)$ and define the map $M\times S\to S$, written $(m,s)\mapsto m\cdot s$, by $m\cdot s=F(m)(s)$. (Here we identify the elements of $M$ with the arrows of $\mathscr M$ and use one and the same letter $m$ to denote them.) We need to check that the axioms of action hold. Well, since $F$ is a functor, we have $F(1_\star)=1_S$ and $F(m_1\circ m_2)=F(m_1)\circ F(m_2)$. Evaluating both sides of each equation at $s\in S$, we get, respectively, $F(1_\star)(s)=1_S(s)$ and $(m_1\circ m_2)(s)=F(m_1)(F(m_2)(s))$ or, equivalently, $1_\star\cdot s=s$ and $(m_1\circ m_2)\cdot s=m_1\cdot (m_2\cdot s) $. Since $\circ$ corresponds to multiplication in $M$ and $1_\star$ corresponds to $e$, this translates to $e\cdot s=s$ and $(m_1m_2)\cdot s=m_1\cdot(m_2\cdot s)$. In this way, $F$ gives rise to a left $M$-set.

Conversely, consider a left $M$-set $(S,\cdot)$. Define the functor $F:\mathscr M\to \mathbf{Set}$ as follows. Define the image of the unique object $\star$ by $F(\star)=S$. If $m:\star\to \star$ is an arrow in $\mathscr M$, define $F(m): S\to S$ by $F(m)(s)=m\cdot s$. Let us prove functoriality: $$F(m_1\circ m_2)=(m_1\circ m_2)\cdot s=(m_1m_2)\cdot s=m_1\cdot (m_2\cdot s)=\\ m_1\cdot F(m_2)(s)=F(m_1)(F(m_2)(s))=(F(m_1)\circ F(m_2))(s).$$ The second requirement for $F$ being a functor is checked similarly. This shows that to every left $M$-set there corresponds a functor $\mathscr M\to \mathbf{Set}$.


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