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My job is to prove Cayley's theorem through the Yoneda Lemma, I would greatly appreciate your input as I find category theory rather difficult. This is my homework, so I appreciate any critiques. (The bold part is the part that is in the statement of the problem).

Let $(G,\cdot,e)$ be a group seen as a category with exactly one object.

Right from this part I am confused, so I will let the category be $G$, letting $*$ be the only object of $G$ and abusing notation I will let $G$ also be the hom set $G(*,*)$.

$1)$ Show that every functor $F:G\rightarrow \textbf{Set}$ defines a group action for $G$ on $F*$.

Again I am confused, Here is what I did: I showed every element of $G$ maps to a bijection on $F*$ and that if $g_1,g_2\in G$ then $F(g_1\cdot g_2)=F(g_1)\circ F(g_2)$. Notice we have $f(g_1\cdot g_2)=F(g_1)\circ F(g_2)$ simply because $F$ is a functor, moreover the domain and codomain of $F(f_1)$ must be $F*$ because $G$ is a category with only one object, $*$. Finally $F(f_1)$ is a bijection because it has an inverse, namely $F(f^{-1})$ To see this notice $F(f)\circ F(f^{-1})=F(f\cdot f^{-1})=F(f^{-1}\cdot f)=F(1_*)=1_{F*}$, where all of the equalities are obtained from the fact $F$ is a functor. So this shows $F$ is indeed a homomorphism from $G$ to $S_{F*}$, in other words an action of $G$ on $F*$

$2)$ Suppose $F*=G$ and prove every natural transformation $\tau:G(*,\_ )\rightarrow F$ defines a bijection on the group $G$

I got lost here, I realized the component on $*$ of a natural transformation is a function $G\rightarrow G$ but could not see it s bijective.

$3)$ Use Yoneda's lemma to prove $G\hookrightarrow S_G$

We have $G\cong Nat((*,_)$ and by $2)$ we can send a natural isomorphism $\kappa$ to the bijection on $G$ $\kappa_*$.

Please help me complete the proof.

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The statement in (2) is not correct. For instance, $F$ could be the functor that sends $*$ to $G$ and sends every element $g\in G$ to the identity map on $G$. There is then a natural transformation $\tau:G(*,-)\to F$ whose map on objects is any constant map from $G(*,*)=G$ to $F*=G$.

However, you do not need the general statement of (2) to get the final result; you only need it in the case that $F=G(*,-)$. Actually, it's probably easier to just skip showing the "bijection" part of (2) and prove it only after using Yoneda's lemma as you do in part (3).

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  • $\begingroup$ Okay, now I am confused (I've been like this for a couple of hours though). So then I should send $g\in G$ to its respective natural transformation $\kappa:(*,\_)\rightarrow \kappa:(*,\_)$ and then send $\kappa$ to the bijection $\kappa_*$? If so how do I prove $\kappa_*$ is a bijection? $\endgroup$ – Jorge Fernández Hidalgo Sep 22 '15 at 1:51
  • $\begingroup$ Well, $g$ has an inverse in $G$... $\endgroup$ – Eric Wofsey Sep 22 '15 at 1:52
  • $\begingroup$ Sorry, how can I use this fact to show $\kappa_*$ is a bijection? Am I correct in saying $\kappa_*$ is a function from $G$ to $G$? Am I supposed to use the naturality square? I tried but I don't see how. $\endgroup$ – Jorge Fernández Hidalgo Sep 22 '15 at 2:04
  • $\begingroup$ All I get is if $g\in G$ and $\mathcal G$ is premultiplying by $g$ then $\mathcal G \kappa_*\mathcal G^{-1}=\kappa_*$ $\endgroup$ – Jorge Fernández Hidalgo Sep 22 '15 at 2:06
  • $\begingroup$ Use the fact that the Yoneda correspondence between maps $*\to *$ in the category $G$ and natural transformations $G(*,-)\to G(*,-)$ preserves composition. $\endgroup$ – Eric Wofsey Sep 22 '15 at 2:10
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You might find helpful the section "Yoneda meets Cayley" in the freely available Notes on Basic Category Theory where I try to explain the connection clearly. (A little too long, perhaps, to reproduce here, sorry: but then if, as you say, you are finding category theory a little difficult, you may find the gently-paced surrounding chapters useful.)

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