I came across the statement that Yoneda-lemma is a generalization of Cayley`s theorem which states, that every group is isomorphic to a group of permutations.

How exactly does generalizes Yoneda-lemma Cayley`s theorem? Can Cayley's theorem be deduced from Yoneda lemma, is it a generalization of a particular case of Yoneda, or is this instead, a philosophical statement?

To me, it seems that Yoneda embedding is more canonical than Cayley's theorem because in the latter you have to choose whether the group acts from the left or from the right on itself. But maybe this is an optical illusion.

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    $\begingroup$ Note, btw, that just as you can use left and/or right action in Cayley's theorem, you can use co-variant and contra-variant functor categories in Yoneda's lemma. $\endgroup$ – Mikael Vejdemo-Johansson Aug 9 '10 at 20:21
  • $\begingroup$ This seems relevant: mathoverflow.net/questions/12511/… $\endgroup$ – Baby Dragon Jan 30 '14 at 0:32

Just as Cayley's theorem states that every group is a subgroup of a symmetric group, Yoneda's lemma states that every (locally small) category $C$ embeds into a category of functors defined on $C$.

Specifically, Yoneda's lemma states that if $F:C\to Set$ is an arbitrary set-valued functor, then $F(A) = Nat(\hom(A,-), F)$, so that natural transformations from a hom-set functor are in bijective correspondence with the elements in the functor image.

For this to be a straight generalization, we may consider a group $G$ as a category $C_G$ (actually a groupoid) with a single object $\ast$, and each group element a morphism. Then, there is only a single object, so a set-valued functor is the same thing as a set $S$ with a map from $G$ to $Bij(S,S)$, the set of bijections from $S$ to $S$. We can see this by unrolling the definition of set-valued functor: $\ast$ goes to $S$, and each element of $G$ goes to some map $S\to S$; all of which maps have to be bijections since otherwise the group properties suffer.

Suppose now that we have some such $G$-set $S$, Yoneda's lemma tells us that its elements are bijective with natural transformations from $G$ to $S$; so what is a natural transformation here? Our two functors are $S$, that takes $\ast$ to $S$, and $\hom(\ast,-)$ that takes $\ast$ to $G$; both as sets. A natural transformation of these is a set-valued map from one image to the other, such that the 'obvious' square of induced maps from morphisms in $C_G$ commutes - thus, $Nat(\hom(\ast,-),S)$ is the collection of $G$-set maps from $G$ as a left $G$-representation to $S$ itself.

One of the things Yoneda brings is a bijection $Nat(\hom(a,-),\hom(b,-)) = \hom(b,a)$: the Yoneda embedding. Applied to the group situation, this tells us that $\hom_G(G,G)=G$. Certainly, left multiplication by an element is a group endomorphism; and what this tells us is that these are all there is. This bijection is exactly what is used in the proof of Cayley's theorem on the wikipedia page.

  • $\begingroup$ Thanks, I guess this is exactly what I was looking for. $\endgroup$ – Jan Aug 6 '10 at 13:47
  • $\begingroup$ What do you mean by "left multiplication by an element is a group endomorphism"? $\endgroup$ – Eric Auld Jun 14 '16 at 22:25
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    $\begingroup$ Fix an element $g\in G$. The function $\hat g:G\to G$ that maps $h$ to $gh$ will … not necessarily be a group homomorphism, because $\hat g(hk)=ghk$ which is not necessarily the same as $\hat g(h)\hat g(k)=ghgk$. I suspect that what I might have been looking for when I wrote this 6 years ago is the statement that there is an action of $G$ on itself; for instance $\bar g(h)=ghg^{-1}$. There is one of these for each element. $\endgroup$ – Mikael Vejdemo-Johansson Jun 16 '16 at 6:00

I can also add that both Yoneda and Cayley are results which follow from the general philosophy of investigating algebraic structures by letting them act on themselves.

1) If you let a group $G$ act on itself, you realize it as a subgroup of $\mathfrak{S}_n$.

2) If you let a ring with unit act on itself, you realize it as a subring of $E$, where $E$ is the underlying additive group.

3) Similarly if you let a $k$-algebra act on itself, you realize it as a subalgebra of $\mathcal{M}_n(k)$. In particular this gives the classical realization of $\mathbb{C}$ as a matrix algebra over $\mathbb{R}$ and of the quaternions as a matrix algebra over $\mathbb{C}$ or $\mathbb{R}$.

4) You can let a Lie algebra act on itself, but unfortunately this action need not be faithful (Lie algebra don't have units...). So you only obtain the easy first step of Ado's theorem about embedding Lie algebras into mtraix algebras.

5) If you let a category $\mathcal{C}$ act on itself, you obtain an embedding into $Fun(\mathcal{C}^{op}, Set)$, which is the content of Yoneda's lemma.

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    $\begingroup$ Thanks for this interesting perspective! Could you make 5) more precise and say what an action of a category is? $\endgroup$ – Jan Aug 6 '10 at 13:46
  • $\begingroup$ I don't think that there is such an established notion. I meant it in an intuitive sense. For groups, you associate to $g$ the map which sends $h$ to $gh$. So $g$ is determined by how it acts on other elements. For a category, you associate to $A$ the functor which sends $B$ to $Hom(B, A)$, and $A$ is determined by these $Hom$ sets. I will see if I can come up with a suitable notion of action which makes this precise. $\endgroup$ – Andrea Ferretti Aug 6 '10 at 14:18
  • $\begingroup$ @Jan: a presheaf is indeed sometimes called a module for a category, or an action of a category on a family of sets. This terminology is most often used in the study of bimodules between categories: functors $\mathcal{C}^\mathrm{op} \times \mathcal{D} \to \mathbf{Sets}$. (Also known as profunctors, distributors, and probably something else I’m forgetting.) $\endgroup$ – Peter LeFanu Lumsdaine Nov 21 '10 at 9:35

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