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 $g\in G$.

The natural transformation $\eta$ is defined by the following commutative diagram $\require{AMScd}$ $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} F(X) & \ra{\eta_X} & F(X) \\ \da{F(f)} & & \da{F(f)} \\ F(Y) & \ra{\eta_Y} & F(Y) & \\ \end{array} $$ where $\eta_X$ is a morphism in $\mathbf{Sets}$ and $f:X \to Y$ is a morphism in the category $G-\mathbf{Sets}$. Since $F$ is just the forgetful functor, the above diagram reduces to $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X & \ra{\eta_X} & X \\ \da{f} & & \da{f} \\ Y & \ra{\eta_Y} & Y & \\ \end{array} $$

Concerns and Questions

In the definition of natural transformation - I have that - given any $G-\text{Set}$ $X$, $\eta_X:F(X)\to F(X)$ is a morphism. A natural $G-\text{Set}$ is simply to take $X=G$ and to let it act on itself through the group structure: $$\varphi: G\times G\to G \\ (g,s)\mapsto g\cdot s.$$ So the commutative diagram now becomes $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} G & \ra{\eta_G} & G \\ \da{f} & & \da{f} \\ Y & \ra{\eta_Y} & Y & \\ \end{array} $$

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,



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).

Consider your commutative diagram again:

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

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 $\eta\in\mathrm{Aut}\,F$.

  • $\begingroup$ Thanks for your answer! Just to check: Shouldn't it be $\eta_G(y)=\eta_G(e)y$? Or maybe I am misunderstanding something? $\endgroup$ – Joel Dec 30 '20 at 10:54
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    $\begingroup$ @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 $Y$. $\endgroup$ – runway44 Dec 30 '20 at 11:03
  • $\begingroup$ 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. $\endgroup$ – Joel Dec 30 '20 at 11:24
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    $\begingroup$ 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 $y\in Y$. $\endgroup$ – runway44 Dec 30 '20 at 11:27
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    $\begingroup$ Ah, thank you very much. I got it now! $\endgroup$ – Joel Dec 30 '20 at 11:33

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