Cayley's Theorem and the Yoneda Lemma [closed]

Hi, from wiki, I know that yoneda lemma is the generalization of Cayley's theorem. But I am not quite understand the intuition behind that. Anyone can help me with that? Cheers!

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closed as off topic by Ryan Budney, J.C. Ottem, Andres Caicedo, Zev Chonoles, Dan PetersenMay 2 '11 at 6:47

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which theorem of Cayley are you thinking of? his collected mathematical works make a dozen volumes or so. –  Abdelmalek Abdesselam May 1 '11 at 23:13
@Abdelmalek: according to wiki, the theorem that says that a group G is isomorphic to a subgroup of Sym(G). –  Thierry Zell May 1 '11 at 23:28
Is "yoneda" a person's name? –  timur May 1 '11 at 23:50
@timur: yes. en.wikipedia.org/wiki/Nobuo_Yoneda –  Qiaochu Yuan May 2 '11 at 0:28
The way I understand timur's comment is that Yoneda's name should begin with a capital letter. But given that his own monicker has none, I am unsure… –  Jonathan Chiche May 2 '11 at 6:20

The way I view Yoneda's lemma as a generalization of Cayley's theorem is by viewing a group as a category with one object $A$. To see how a category with one element is a group, let the points of the group be morphisms $f:A\rightarrow A$. By the definition of a category, we must have an identity morphism $id_A:A\rightarrow A$, and this is the identity of the group. Similarly, morphisms are associative, so $f\circ (g\circ h) = (f\circ g)\circ h$. So composition is the group operation. As for invertibility, let's just force our morphisms to be invertible by assuming they are all isomorphisms. Ok, so now the collection of morphisms forms a group in the sense we learn in undergrad.

Cayley's Theorem says we can embed a group $G$ into $Sym(G)$, the set of bijections $f:G\rightarrow G$. This means we have an isomorphism $G\cong H$ for some $H\leq Sym(G)$. If we're going to generalize this to category theory we need to understand maps that take morphisms to morphisms (those are our group elements, after all). Functors do this, so a generalization of Cayley's Theorem needs to say something about embedding a category $\mathcal{C}$ into a category of functors going out of $\mathcal{C}$. If I want a category of functors, I need to know what maps are between functors. They are natural transformations.

Now, Yoneda's Lemma says we have $\mathrm{Nat}(h^A,F) \cong F(A)$ where $h^A = Hom(A, − )$ and $F$ is any functor from $\mathcal{C}$ to the category of Sets. If we set $F$ to be the functor $h^A$ then Yoneda is telling us $\mathrm{Nat}(h^A,h^A) \cong Hom(A,A)$. But $Hom(A,A)$ is exactly our group (elements of our group are exactly morphisms from $A$ to $A$), so now we see our group as isomorphic to $\mathrm{Nat}(h^A,h^A)$ sitting in a larger functor category (the category of all functors from $\mathcal{C}$ to Set).

Some notational points: this category which $\mathcal{C}$ is equivalent to is called the category of representable functors with maps between functors given by natural transformations. Yoneda's lemma also discusses how it relates to the larger functor category it sits in, because a natural transformation $Φ: h^A \rightarrow F$ is sent to $Φ_A(id_A)$ in $F(A)$. I've ignored an issue of covariance vs. contravariance because all I need is $\mathrm{Nat}(h^A,h^A) \cong Hom(A,A)$, although the more general fact is that $\mathrm{Nat}(h^A,h^B) \cong Hom(B,A)$.

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Let $G$ be a finite group of order $n$ and consider the category $\mathscr G$ with a single object $\{ \bullet \}$ and whose morphisms consist of the elements of $G$, i.e., $\mathrm{Hom}_{\mathscr G}(\bullet, \bullet)=G$.

Let $h^{\bullet}=\mathrm{Hom}_{\mathscr G}(\bullet , {\_} )$ be the Yoneda functor. Then according to Yoneda's lemma $$\mathrm{Nat}(h^{\bullet},h^\bullet)\simeq \mathrm{Hom}_{\mathscr G}(\bullet,\bullet),$$ which implies that $h^\bullet$ is a faithful functor. In other words, $$h^{\bullet} : \mathscr G \to \mathscr Set$$ maps $\bullet$ to the set $\mathrm{Hom}_{\mathscr G}(\bullet,\bullet)\simeq_{\mathscr Set} |G|$ (where $|G|$ is the set of elements of $G$) and also embeds the group (!) $\mathrm{Hom}_{\mathscr G}(\bullet,\bullet)\simeq_{\mathscr Groups} G$ into $\mathrm{Hom}_{\mathscr Set}(|G|,|G|)\simeq S_n$, which gives you Cayley's Theorem.

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\bullet makes for a more readable category object :) –  Mariano Suárez-Alvarez May 2 '11 at 6:01
@Mariano: thanks for the advice! –  Sándor Kovács May 2 '11 at 6:29

One can define the notion of a right category action on a set. This involves assigning a domain (an object of the category) to each element of the set, and a partial multiplication of elements by arrows defined whenever the domain of an element is the codomain of an arrow. The prototypical example is a category acting by composition on its set of arrows.

The category of right $\mathbf{C}$-sets winds up being equivalent to the category of functors $\mathbf{C} \to \mathbf{Set}$. The reverse equivalence applied to Yoneda of an object $A$ is essentially the set of arrows with codomain $A$, with $\mathbf{C}$ acting by composition.

I believe there is also an equivalence of left-right $\mathbf{C}$-sets to the category of functors $\mathbf{C}^\circ \times \mathbf{C} \to \mathbf{Set}$. The action of $\mathbf{C}$ on itself corresponds to $\text{Hom}_{\mathbf{C}}(-,-)$, which in turn is the adjoint transpose of the Yoneda embedding.

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