Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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"?

share|improve this question
1  
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

3 Answers 3

up vote 3 down vote accepted

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 $$\text{Aut}_{\mathcal F}(F)=\text{Aut}_{\mathcal C'}(G)=G.$$

share|improve this answer
    
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 $g$.

share|improve this answer
    
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 :).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.