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I am trying to understand the proof of the following theorem from Qiaochu Yuan's answer

Theorem: Let $G$ be a group and let $P : G\text{-FinSet} \to \text{Set}$ be the forgetful functor from the category of finite $G$-sets to sets. Then $\text{Aut}(P)$ is naturally isomorphic to the profinite completion $\widehat{G}$.

Proof. Here is a sketch. If we replaced "finite sets" with "sets" above, the automorphism group of the forgetful functor would just be $G$, and this would be because the forgetful functor in this case is representable by $G$ itself (being acted on by left multiplication).

If $G$ is infinite the forgetful functor is no longer representable. However, it can be written as a filtered colimit of representable functors; it is the filtered colimit over the functors represented by the finite quotients $G/G_i$ of $G$ (this is the key step). Together with the Yoneda lemma, it follows that

$$\text{Hom}(P, P) \cong \text{Hom}(\text{colim}_i G/G_i, P) \cong \lim_i \text{Hom}(G/G_i, P) \cong \lim_i P(G/G_i) \cong \lim_i G/G_i$$

and although this is only an identification as sets, as written, you can show that it is also an identification as groups. $\Box$

I wonder why $P$ can be written as a filtered colimit of representable functors. In particular, why $P=\text{colim}_i G/G_i$? Also, why is $\text{Hom}(P, P)=\text{Aut}(P)$? I think $\text{Hom}(P, P)$ might not only have isomorphisms. Thanks in advance.

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    $\begingroup$ 1. Check the blog post I linked to, it explains this sort of thing in more detail. 2. The argument computes the endomorphisms and shows that they are all automorphisms, which means it also computes the automorphisms. $\endgroup$ Oct 12, 2023 at 21:54
  • $\begingroup$ @QiaochuYuan thanks for the reply, but I think your blog only gave a bunch of statements without proof, I am still unsure why we have that inverse limit. Also for the 2, could you please explain why it computes the endomorphism? $\endgroup$
    – Kat
    Oct 12, 2023 at 23:31
  • $\begingroup$ @QiaochuYuan also, when you say $G$-set, do you mean transitive $G$-set? $\endgroup$
    – Kat
    Oct 13, 2023 at 1:07
  • $\begingroup$ I mean arbitrary (finite) $G$-set. $\endgroup$ Oct 13, 2023 at 1:10

1 Answer 1

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If $H \subseteq G$ is a subgroup and $X$ is a $G$-set (meaning: left $G$-set), then $G$-maps $G/H \to X$ correspond naturally to elements $x \in X$ with $H \subseteq G_x$. (You can easily write down a bijection.) This describes the hom-functor $\hom(G/H,-)$ on the category of $G$-sets. Notice that it is (isomorphic) to a subfunctor of the forgetful functor $U$, since what we get above is a subset of (the underlying set of) $X$. This means that when we take colimits of such functors, we are actually just taking a union inside the partial order of subfunctors. So, the canonical map

$\mathrm{colim}_{H \subseteq G \text{ finite index}} \hom(G/H,X) \to U(X)$

is injective, and the image consists of those elements $x \in X$ such that there is a finite index subgroup $H$ with $H \subseteq G_x$. Equivalently, $G_x$ must be of finite index.

Let $X$ be a finite $G$-set. Then for every $x \in X$ the stabilizer $G_x$ is of finite index (the index is the size of the orbit). Hence,

$\mathrm{colim}_{H \subseteq G \text{ finite index}} \hom(G/H,X) \to U(X)$

is bijective. We get an isomorphism of functors

$\mathrm{colim}_{H \subseteq G \text{ finite index}} \hom(G/H,-) \xrightarrow{\cong} U,$

where $U : \mathbf{FinSet}_G \to \mathbf{Set}$ is the forgetful functor.

We want the same with normal subgroups. Well, if $x \in X$, there is always a finite index normal subgroup $N \subseteq G_x$, namely $N = \bigcap_{g \in G} g G_x g^{-1} = \bigcap_{y \in Gx} G_{y}$.

I just write $N$ in the following to indicate a normal subgroup. Therefore, we get a bijection: $$\hom(U,U) \cong \hom\bigl(\mathrm{colim}_{N \subseteq G \text{ finite index}} \hom(G/N,-),U\bigr)\\ \cong \lim_{N \subseteq G \text{ finite index}} \hom\bigl(\hom(G/N,-),U\bigr)$$ The next step is to use the Yoneda Lemma. It implies that the natural map $$\hom\bigl(\hom(G/N,-),U\bigr) \to U(G/N)$$ is an isomorphism. Thus, we get $$\hom(U,U) \cong \lim_{N \subseteq G \text{ finite index}} U(G/N).$$ The next step is to upgrade this to an isomorphism of monoids $$\mathrm{End}(U) \cong \lim_{H \subseteq G \text{ finite index}} G/N.$$ This is not completely formal, the above argument is not symmetric! To see this, one just has to go through the proof and find an explicit formula for the bijection. It maps a natural transformation $\eta : U \to U$ to the compatible family of elements $\eta_{G/N}([1])$, where $[1] \in G/N$ is the canonical element. Conversely, if a compatible family $([g^N] \in G/N)_{N}$ is given, $\eta_X : U(X) \to U(X)$ is defined as follows: If $x \in X$, then $\eta_X(x) = [g^{G_x} \cdot x]$. For $X = G/N$, which also has a right $G/N$-action, this shows that $\eta_X$ is actually compatible with the right action*.

Using this, if $\eta,\mu : U \to U$, then $\eta \circ \mu$ is mapped to the family of elements $$\eta_{G/N}(\mu_{G/N}([1])) = \eta_{G/N}([1] \cdot \mu_{G/N}([1])) = \eta_{G/N}([1]) \cdot \mu_{G/N}([1]).$$ And clearly $\mathrm{id} : U \to U$ is mapped to the family $([1])$, which is $1$. This proves the isomorphism of monoids.

A projective limit of monoids, which happen to be groups, is again a monoid which happens to be a group. (This is basically because the compatibility condition transfers to inverse elements.) So the projective limit above is a group, and hence the monoid $\mathrm{End}(U)$ is a group. So it must coincide with its group of units, which is $\mathrm{Aut}(U)$. Also, an isomorphism of monoids between monoids which happen to be groups is actually an isomorphism of groups (it is well-known that we get preservation of inverses for free). So we indeed get an isomorphism of groups $$\mathrm{Aut}(U) \cong \lim_{N \subseteq G \text{ finite index}} G/N.$$ Instead of $\mathbf{FinSet}_G$ we can also the the category of $G$-sets whose stabilizers are of finite index (equivalently, all orbits must be finite).

*Remember that, even though $\eta$ is indexed by left $G$-sets, each map only operates between their underlying sets. The map has no reason to be a left $G$-map. It is a pure "coincidence" (as always, more general categorical arguments reveal that it is no coincidence) that sometimes these maps happen to be right $G$-maps, though. The natural transformations that consist of left $G$-maps are quite boring, this is $\mathrm{End}(\mathrm{id}_{\mathbf{Set}_G}) \cong Z(G)$, the center of $G$.

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