# Reconstructions of Groups From Category of $G-\mathbf{Sets}$; Construction of a Group Homomorphism [duplicate]

I try to come up with a proof of the following statement, but I find it a little difficult. I hope that I can get some help from someone on this site. I think this is what they give a proof of, on Ncatlab - Tannakian Duality (at the section $$G-\mathbf{Sets}$$). But I can't really follow that proof: https://ncatlab.org/nlab/show/Tannaka+duality#ForPermutationRepresentations.

Statement. Let $$F:G-\mathbf{Sets}\to\mathbf{Sets}$$ be the forgetful functor, where $$G-\mathbf{Sets}$$ is the category of sets equipped with a group action by the group $$G$$. I'm trying to understand the proof of the following fact $$\operatorname{Aut}(F)\cong G.$$

### What I have done

I have managed to construct a map $$\varphi:G\to\operatorname{Aut}(F)$$ This was done by the following rule $$\varphi(g)=\eta^g$$, where $$\eta_S^g:S\to S$$ is defined by $$\eta_S^g(s)=s\cdot g$$. It is straightforward to check that this gives a natural transformation from $$F$$ to $$F$$ and that it also is a group homomorphism.

However, the other way is more problematic for me. I want to find a map $$\psi:\operatorname{Aut}(F)\to G.$$ That is, given a natural transformation $$\eta$$, I want to assign it to a group element $$g\in G$$.


### Concerns and Questions


Remark 1. I remember a professor told me that the morphism $$\eta_G$$ is totally understood by what it does to the identity element $$e\in G$$ (from which I should be able to understand how to construct the group homomorphism), $$e\mapsto \eta_G(e).$$

I don't really understand what the above means. I think I have misunderstood something about the forgetful functor. When I think about the forgetful functor $$F:A\to B$$, I think that the functor forgets everything that is present in $$A$$, but isn't present in $$B$$. In our case, it forgets the structure of group actions. And so, in particular, I cannot use the property of being a $$G$$-equivariant map. Only the properties of being a set-theoretic map.

Question 1.

If $$\eta_G(e)=s$$, and if I would like to make sense of what the professor told me, I think I would reason something as follows $$\eta_G(g)=\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)=s\eta_G(g).$$ where I in the second equality used the property of being a group homomorphism. But on the other hand, if I want to treat it as a group homomorphism, then I think I had to do it to start with. That is, $$\eta_G$$ must map identities to identities (in order to be consistent in my reasoning). So I think my argument fails.

My question is: What does he mean?

I don't think what I did above makes any sense. But I think I have seen others using the properties of the morphisms in the category $$A$$, after having applied the forgetful functor, hence my reasoning. Once again, I am not really sure what I am doing. So I may very well be wrong.

Question 2. How does this tell me where to map a natural transformation?

Given a $$\eta\in\operatorname{Aut}(F)$$, where do I map it? Do I map it as follows $$\eta\mapsto \eta_G(e)?$$ Doing so, do I know that I have exhaustively told where to map every natural transformation?

Question 3. I guess I also, somehow, have to use the commutative diagram in the definition of the natural transformation when I construct the group homomorphism, which I haven't done? I guess my suggestion above is not the correct way to do it. Do you have any ideas how I can construct the map?

I would be really happy I could have any help from someone on this site to understand this better. Because I am really lost, and confused.

Best wishes,

Joel

I am going to use left $$G$$-sets, not right.

Question 1 & 3.

You cannot write $$\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)$$, we are not assuming $$\eta_G:G\to G$$ is a group homomorphism, only that it is a morphism of $$G$$-sets. You can ue that to say $$\eta_G(g\cdot e)=g\cdot\eta_G(e)$$ though (which you'd reverse the order of if you insist on right group actions).

$$\require{AMScd} \begin{CD} G @>{\eta_G}>> G \\ @VVV @VVV \\ Y @>{\eta_Y}>> Y \end{CD}$$

Here, we can let the map $$G\to Y$$ be the evaluation-at-$$y$$ map $$g\mapsto gy$$ where $$y\in Y$$ is fixed (note the evaluation map is also useful in establishing the orbit-stabilizer theorem - its fibers are cosets of $$y$$'s stabilizer). Then we chase the diagram starting from $$e\in G$$ in the top left.

If we follow the upper-right path, we get $$e\mapsto \eta_G(e)\mapsto \eta_G(e)y$$. In the lower-left path, $$e\mapsto y\mapsto \eta_Y(y)$$. Therefore we may equate $$\eta_Y(g):=\eta_G(e)y$$. That is, every automorphism $$\eta$$ applied to a $$G$$-set $$Y$$ is just applying a particular group element $$\eta_G(e)\in G$$.

Qusetion 3.

Yes, $$\eta\mapsto \eta_G(e)$$. This applies for all $$\eta\in\mathrm{Aut}\,F$$.

• Thanks for your answer! Just to check: Shouldn't it be $\eta_G(y)=\eta_G(e)y$? Or maybe I am misunderstanding something? – Joel Dec 30 '20 at 10:54
• @Joel No. It is $\eta_Y(y)=\eta_G(e)y$. Look at the morphims in the diagram, we are applying $\eta_Y$! And $\eta_G(y)$ doesn't even make sense, as $\eta_G$ is a function on $G$ whereas $y$ is an element of $Y$. – runway44 Dec 30 '20 at 11:03
• Ok, thanks! I am not 100% what the following expression tells me $\eta_Y(y)=\eta_G(e)y$, though. Does it tell me that letting the natural transformation act on a $G$-Set $Y$ is the same as fixing an element $y$ in $Y$ and then letting it act on the identity element of the group instead? In which sense is this important when constructing the map $\eta\mapsto \eta_G(e)$? You are probably saying exactly that in "every automorphism $\eta$ applied to a $G$-set $Y$ is just applying a particular group element $\eta_G(e)\in G$". But I am a little dumb and don't see it. – Joel Dec 30 '20 at 11:24
• It tells you that letting the natural transformation act on a $G$-set $Y$ is the same as applying the group element $\eta_G(e)$ to it. That means you can evaluate $\eta_Y(y)$ for any $y\in Y$ by applying the group action, i.e. $\eta_G(e)y$. We don't have elements of $Y$ act on $G$, we know that $G$ acts on $Y$. I only fix $y$ in my argument to show what $\eta_Y$ does to an arbitrary element $y\in Y$. – runway44 Dec 30 '20 at 11:27
• Ah, thank you very much. I got it now! – Joel Dec 30 '20 at 11:33