# Yoneda lemma as a generalisation of Cayley's theorem

I have seen answers in questions asking the same question. They have first described what is Yoneda lemma and then deduced Cayley's theorem from that. I am not asking for that.

I am planing to explain Yoneda lemma for a group of students who know some group theory some basic definitions in category theory.

I don't want to prove Yoneda lemma and say that as a special case we can get Cayley's theorem. I want to recall Cayley's theorem and then say that it can be generalised to some extent and then state and prove Yoneda lemma.

Any suggestions are welcome.

• What’s the difference really? I mean, you could state Cayley, say “When interpreting this group as a category, you could also state Cayley like this: …”, and then only use category theory parlance to state Cayley. And finally you could say: “And this is also true for all categories, not just when $\mathcal C$ is a group.” Or do you want to prove Yoneda by reducing to Cayley? – k.stm Jul 18 '17 at 20:40
• @k.stm i want to state and/or prove Cayley and say that it can be generalised to some other theorem and then prove Yoneda. I am not trying to prove yoneda by reducing to Cayley – user87543 Jul 18 '17 at 20:43
• It seems futile to "try to state and prove" Yoneda: one of the bad things about this lemma is people sometimes learn it informally, and then fail to see it is quite simple. Why not state it clearly, prove it, and relate it to Cayley so people don't see it as something strange? – Pedro Tamaroff Jul 18 '17 at 20:51
• @k.stm I saw your edited comment now. Your "when interpreting ... like this :" seems to be a good idea. Please feel free to elobarate if you have anything more to say. – user87543 Jul 18 '17 at 21:01
• @PraphullaKoushik Here's my thoughts on how to present the idea: start by reframing Cayley's theorem as "we can completely recover a group's structure if we look at the homomorphisms from that group. In fact, we can do so in a systematic (i.e. 'natural') way". Then, you can present Yoneda's lemma as a generalization of this idea. – Omnomnomnom Jul 18 '17 at 21:35

$\newcommand{\Sym}{\operatorname{Sym}}$ $\newcommand{\Ob}{\operatorname{Ob}}$ $\newcommand{\Ar}{\operatorname{Ar}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\End}{\operatorname{End}}$ $\newcommand{\Nat}{\operatorname{Nat}}$ $\newcommand{\cod}{\operatorname{cod}}$ $\newcommand{\dom}{\operatorname{dom}}$ $\newcommand{\id}{\operatorname{id}}$ $\newcommand{\com}{\operatorname{com}}$ $\newcommand{\Set}{\mathrm{Set}}$

First of all, I personally define categories to be structures $C = (\Ob C, \Ar C, \dom, \cod, \id, ∘)$, where $\Ob C$ is the class of objects, $\Ar C$ is the class of arrows and

• $\dom \colon \Ar C → \Ob C$ is the domain.
• $\cod \colon \Ar C → \Ob C$ is the codomain.
• $\id \Ob C → \Ar C$ is the identity selection.
• $∘ \colon \Ar C × \Ar C \dashrightarrow \Ar C$ the arrow composition.

Let $G$ be a group. Cayley says:

There is an injective group homomorphism $G → \Sym G$, where $\Sym G$ is the group of bijections $G → G$.

Proof. The injection is given by $G → \Sym G,~g ↦ (g·~)$ where $(g·~)$ is the left multiplication by $g$. It’s a homomorphism since $∀g,h ∈ G \colon (gh·~) = (g·~)∘(h·~)$. It clearly is injective because $(g·~) = \id_G ⇒ g·1 = 1$, so its kernel is trivial.

The proof also shows that the image of each $g ∈ G$, as a left multiplication, respects the right multiplication by $h ∈ G$ since $∀x ∈ G\colon g(xh) = (gx)h$. So we can specify Cayley:

There is an injective group homomorphism $G → \Nat G$, where $\Nat G$ is the subgroup of bijections on $G$ respecting the right action of $G$, i.e. $\Nat G = \{σ ∈ \Sym G;~σ(gh) = σ(g)h\quad∀g,h ∈ G\}$.

We will soon see why we have named this subgroup $\Nat G$.

Now, let’s interpret $G$ as a category in the following way:

Let $C = (\Ob C, \Ar C, \dom, \cod, \id, ∘)$ be the category with $\Ob C = \{\star\}$, $\Ar C = G$ with $\cod = \dom = \star$ (constantly) and $\id \star = 1$ and $∘$ is the group multiplication $G × G → G$.

Now consider the contravariant hom-functor $h_\star = \Hom_C (–,\star) \colon C → \Set$. The only object it can accept is $\star$ and $\Hom_C (\star,\star) = \Ar C = G$. Any natural transformation $h_\star → h_\star$ is therefore given by a sole arrow $\Hom_C (\star,\star) → \Hom_C(\star,\star)$, i.e. by a map $G → G$ (since in general it’s given by a family of arrows – one for each object in $C$ – but here there’s only one object). This map $σ\colon G → G$ also respects the right action of $G$ – this comes from the naturality: For $g ∈ G$ and $h ∈ H$ we have $$σ(gh) = (σ∘h_\star(h))(g) = (h_\star(h)∘σ)(g) = σ(g)h,$$ and vice versa: By using the very same equation we see that very such map $σ$ respecting the right action of $G$ automatically yields a natural transformation $h_\star → h_\star$.

Thus, $\Nat (h_\star, h_\star) \cong \Nat G$ as monoids. And $G = \Ar C = \Hom_C (\star,\star)$. And we can restate Cayley as

There is an injective monoid homomorphism $\Hom_C(\star,\star) → \Nat (h_\star, h_\star)$.

Because $\star$ is the only object of $C$ and $\Nat (h_\star,h_\star) = \Hom_{\Set^{C^\mathrm{op}}} (h_\star,h_\star) ⊆ \Ar \Set^{C^\mathrm{op}}$, one can enlarge such a monoid homomorphism to a functor by sending the object $\star$ to the object $h_\star$, so Cayley can be further restated as

There is a faithful functor $C → \Set^{C^\mathrm{op}}$.

It turns out, the last statement even holds true when $C$ is any category, not just a category that is a group in disguise. And even more is true: This functor is in fact full.

Then I would state some version of Yoneda …

• I am afraid this would mean $G$ is isomorphic to End(G) – user87543 Jul 18 '17 at 21:59
• @PraphullaKoushik Did I miss something? How so? – k.stm Jul 18 '17 at 22:00
• The version that I have of Yoneda lemma is that set of all morphism between objects is isomorphic to set of natural transformations. I did not check but I was thinking every natural transformation is inveruble – user87543 Jul 18 '17 at 22:03
• @PraphullaKoushik Er, of course. : / And the naturality is not trivial because we still have endomorphisms. I will fix this, but not now – because I’m tired and it’s bed time here, so I will delete the answer for now. Thanks for spotting the mistake. – k.stm Jul 18 '17 at 22:06
• @PraphullaKoushik Okay, I have tried to correct it now after all. See if it’s okay now. It should be okay, but then again: I’m quite tired now … – k.stm Jul 18 '17 at 22:34