# A functor $\mathcal G\to \mathbf{Set}$ is the same as a left $G$-set

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?

• 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 $\mathcal{G}$. Jun 19 '19 at 23:46
• 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)? Jun 19 '19 at 23:51
• 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. Jun 20 '19 at 3:18
• @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. Jun 20 '19 at 4:18
• 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 $G = \operatorname{Hom}_\mathcal{G}(\star, \star)$. Jun 20 '19 at 4:20

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.

• 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 $\mathcal C\to\mathbf {Set}$. Jun 20 '19 at 0:34
• 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. Jun 20 '19 at 0:36
• 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. Jun 20 '19 at 4:44
• 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. Jun 20 '19 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 $$\mathscr M\to \mathbf{Set}$$.