# Why is $\text{Aut}(F)$ of the forgetful functor $F$ on $G$-sets isomorphic to $G$?

Here's something I've been trying to scratch out recently.

Let $G$ be a group and $\text{Set}(G)$ the category of $G$-sets. Let $F\colon\text{Set}(G)\to\text{Set}$ be the forgetful functor sending a $G$-set to its underlying set. Show that $\text{Aut}(F)$ is naturally isomorphic to $G$.

I've been trying to view $F$ as an object in the category of functors from $\text{Set}(G)\to\text{Set}$ where the morphisms are natural transformations. I think if $H$ is a natural transformation in $\text{Aut}(G)$, then for any $G$-set $X$, $H$ associates a morphism $H_X\colon F(X)\to F(X)$ such that for any morphism $f\colon X\to Y$ of $G$-sets, then $F(f)\circ H_Y=H_X\circ F(f)$?

I'm not really sure where I'm going with this. Given such an $H$, what is the natural $g\in G$ with which to associate it, and how would I get to this "natural isomorphism"?

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The more precise claim is that the following map $f$ from $Aut(F)$ to $G$ is a group isomorphism. If $a$ is in $Aut(F)$, then $a_G$ is an automorphism of $F(G)$. Put $f(a):=a_G(1)$. It suffices to check that $f$ is a group isomorphism. –  Pierre-Yves Gaillard Sep 22 '11 at 7:15
Thanks @Pierre-Yves. I'm curious to see how this works. Would you have time to expand this to an answer showing how $f$ is a group isomorphism? –  yunone Sep 25 '11 at 3:34

Define $u:\text{Aut}(F)\to G$ and $v:G\to\text{Aut}(F)$ by $$u(a):=a_G(1),\quad v(g)_X(x):=gx.$$ It suffices to show: (1) $u$ is a group morphism, (2) $u\circ v=\text{Id}_G$, (3) $v\circ u=\text{Id}_{\text{Aut}(F)}$.

(1) We have $u(ab)=(ab)_G(1)=a_G(b_G(1))$ and $u(a)u(b)=a_G(1)b_G(1)$. Define $f:G\to G$ by $f(g):=gb_G(1)$. As $f$ is a $G$-map, it commutes with $a_G$, and we get (1) by evaluating $a_G\circ f=f\circ a_G$ on $1$.

(2) We have $u(v(g))=v(g)_1=g$.

(3) We have $v(u(a))_X(x)=u(a)(x)=a_G(1)(x)$. It should be equal to $a_X(x)$.

Define $f:G\to X$ by $f(g):=gx$. Being a $G$-map, it satisfies $$a_X\circ F(f)=F(f)\circ a_G,$$ and it suffices to evaluate this equality on $1$.

EDIT. Here is a selfcontained version of Zhen Lin's answer.

Let $\mathcal C$ be a category, $\mathcal C'$ the opposite category, $\mathcal S$ the category of sets, and $\mathcal F$ the category whose objects are the functors from $\mathcal C$ to $\mathcal S$ and whose morphisms are the functorial morphisms.

It is straightforward to check the following statements.

The formula $$h(X):=\text{Hom}_{\mathcal C}(X,?)$$ defines a functorial morphism $$h:\mathcal C'\to \mathcal F.$$

Let $X$ be an object of $\mathcal C$. The formulas $$u(t):=t_X(\text{Id}_X),\quad v(a)_Y(f):=F(f)(a)$$ define functorial morphisms
$$u:\text{Hom}_{\mathcal F}(h(X),F)\to F(X),\quad v:F(X)\to \text{Hom}_{\mathcal F}(h(X),F)$$ which are functorial in $X$. Moreover

$u$ and $v$ are inverse isomorphisms.

In particular we have functorial isomorphisms $$\text{Hom}_{\mathcal F}(h(X),h(Y))=\text{Hom}_{\mathcal C}(Y,X)=\text{Hom}_{\mathcal C'}(X,Y),$$ $$\text{Aut}_{\mathcal F}(h(X))=\text{Aut}_{\mathcal C'}(X).$$

Now let $G$ be a group, $\mathcal C$ the category of $G$-sets, $F$ the forgetful functor. Then the formulas $$a_X(f):=f(1),\quad b_X(x)(g):=gx$$ define functorial morphisms
$$a:h(G)\to F,\quad b:F\to h(G).$$

Moreover $a$ and $b$ are inverse isomorphisms. This gives in particular canonical isomorphisms $$\text{Aut}_{\mathcal F}(F)=\text{Aut}_{\mathcal C'}(G)=G.$$

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Many thanks for adding this. –  yunone Sep 25 '11 at 6:00

This is a Yoneda-type argument. First, observe that $F \cong \textrm{Hom}_G(G, -)$, where $G$ is considered as a $G$-set by equipping it with the regular left action. So a natural transformation $F \Rightarrow F$ is also a natural transformation $\textrm{Hom}_G(G, -) \Rightarrow \textrm{Hom}_G(G, -)$, and by the Yoneda embedding, there is a natural bijection between such natural transformations and $\textrm{Hom}_G(G, G)$, which is just $G$ itself. (This implies all such natural transformations are in fact natural isomorphisms.)

More explicitly, let $\eta : F \Rightarrow F$ be a natural transformation, and let $g = \eta_G(e)$. Then $\eta_X(x) = g \cdot x$: indeed, if $f : G \to X$ be the $G$-equivariant map determined by $f(e) = x$, then $F(f) \circ \eta_G = \eta_X \circ F(f)$, so $\eta_X(x) = f(g) = g \cdot x$ (with some abuse of notation). Conversely, it is clear that this defines a unique natural transformation $\eta : F \Rightarrow F$ for each $g$.

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Thanks Zhen, I think I might have bit off more than I can chew with this question. I'm going to try to find a better book on category theory than the one I have now and then come back and try to understand this answer. –  yunone Sep 25 '11 at 4:43
Hmmm. I'm not sure there's that much category theory involved other than definitions, if you follow the approach in my second paragraph. The first paragraph is more conceptual and does require a good understanding of the Yoneda lemma though. –  Zhen Lin Sep 25 '11 at 5:12
1. $f$ is (freely and uniquely) determined by the image of $e$ because $f$ is a morphism of $G$-sets: so $f(g) = g \cdot f(e)$. 2. Because $F$ is the underlying-set functor we can basically just ignore it; we get $\eta_X(x) = f(g)$ by evaluating $F(f) \circ \eta_G = \eta_X \circ F(f)$ at $e$. 3. $\phi$ is an isomorphism because all these maps are natural (in a suitable sense). –  Zhen Lin Sep 25 '11 at 5:54
Thanks for the extra explanation. I think I'm getting more familiar with how these work. –  yunone Sep 25 '11 at 5:59

If $C$ is any algebraic category with forgetful functor $U : C \to \mathrm{Set}$ and left adjoint $F : \mathrm{Set} \to C$, then the free algebra on one generator $F(1)$ represents $U$. By the Yoneda lemma, it follows that $\mathrm{End}(U) \cong \mathrm{End}(F(1))$ as monoids. We also have a bijection $\mathrm{End}(F(1)) = \mathrm{Hom}_C(F(1),F(1)) \cong \mathrm{Hom}(1,U(F(1)) \cong U(F(1))$, which makes the underlying set of $F(1)$ a monoid such that this bijection becomes an isomorphism of monoids. Thus, we have $\mathrm{End}(U) \cong F(1)$ as monoids.

When $C=M-\mathrm{Set}$ for some monoid $M$, then $M$ is the free $M$-set on one generator and its monoid structure coincides with the given one.

When $C=R-\mathrm{Mod}$ for some ring $R$, then $R$ is the free $R$-module on one generator and its monoid structure is just the multiplicative structure.

When $C=R-\mathrm{CAlg}$ for some commutative ring $R$, the free commutative $R$-algebra on one generator is the polynomial algebra $R[T]$, and again the monoid structure is the multiplicative one.

You can add more categories as you like :).

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