Yoneda's lemma as a generalization of Cayley's theorem for groups

Question

$ <G, *> $ a group and $ G $ is seen as a category denoted by $ C_G $ that has a single object, say "@".

$H: C_G\to [C_G, Sets]$ is defined as:

In objects: $ @\in {C_G} $ we define an object $h _@$ of $[C_G, Sets]$ in

the following way: $H_@: C_G\longrightarrow {Sets} $

$ @\mapsto H _ @@ = Mor_ {C_G} (@, @) $

$ @ \xrightarrow [s {}] \, {@}\mapsto h _ @@ \xrightarrow [H_ @ s] \, {h _ @@} $ such that $ H_ @ s (t) = t * s $ with $ t: @ \longrightarrow {@} $, defined   so $ H _ @ $ is a contravariant functor.

In arrows: For every $ f: @ \longrightarrow {@} $ we define a natural transformation $ H_f: H _ @ \longrightarrow {H _ @} $.

Thus defined $ H $ is covariant functor.

The embedding of Yoneda tells us that $ H $ is fully faithful and injective in objects.

Denote for $ F $ to the image of the functor $ F $ then $ F = Im (H) $. Affirmation 1: $ F: C_G \to Sets $ is a faithful contravariant functor.

$ F $ is faithful yes and only if $ T $ is injective. $ T $ is defined as follows: the application $ T: Mor_{C_G} (@, @) \to Mor_{Sets} (F _@, F _@) $ defined as $ T (g): F _ @ \to F_@ $ such that $ T (g)_@: F_@@ \to F _ @@ $ with $ T (g)_@ (h) = h * g $. (This definition of $ T $ is because of how $ H $ is defined)

MY QUESTION IS THIS:

Why when $ F $ is the image of $ H $ does it have to $F _ @@ = Mor_{C_G} (@, @)$?

Let's see that $ T $ is injective:

Let $ f, g \in Mor_ {C_G} (@, @) $ such that $ T (g) = T (f) $

(I'm matching natural transformations, does this make sense?)

I think that equality means $ T (g) _@ = T (f) _@ $, then

$ T (g) _@ (h) = T (f)_@ (h) $ and this is $ h * g = h * f $ and since

these are elements

of the group then we take inverse of $ h $, then $ g = f $. Therefore $ T $ is injective.

Now let's look at the following:

$ G = Mor_{C_G} (@, @) $ and $ Mor_{Sets}(F _ @, F _ @) = \left\{{G \to G} \right \}$ is a group with the composition operation, then $ T: G \to \left \{{G \to G} \right \} $.  Missing to see that $ T $ is a homomorphism (this I got) and then I

already have a monomorphism osea $ G $ would be a subgroup of $ \left \{{G \to G} \right \} $.

It is necessary to see that the collection of applications $ {G \to G} $

are bijectives, for this last one what is the idea?

Thank you

Answer 1

In a general category, we have $H_a:=\hom(\_,a)$ is a contravariant functor to $\mathcal{Set}$, which sends an object $x$ to $\hom(x,a)$ and a morphism $f:x\to y\ $ to the function $\ \underset{y\to a}g\,\mapsto \underset{x\to a}{gf}$.
If $\alpha:a\to b$ is any morphism, it induces a natural transformation $H_a\to H_b$ by left composition.

So, now we have a single object and its contravariant hom functor $H_@=\hom(\_,@)$ which maps $@\mapsto\hom(@,@)=G\ $ and $\ s\mapsto \,\underset{G\to G}{(g\mapsto gs)}$.
Moreover, we consider here $H:C_G\to [{C_G}^{op},\,\mathcal{Set}]$ by the above natural transformations.
Note also, that if a category $C$ has only one object $@$, and $F,G:C\to D$ are functors, then a natural transformation $F\to G$ is (given by) a single morphism $F(@)\to G(@)$ of $D$, which has a special property (has to satisfy the commutative diagrams).

To your questions:

1. As I understand your description, you take $F$ to be the image of $H$, so it's simply $H_@$ as $C_G$ has only one object? That would also answer your question.
2. What you call $T$ is the morphism part of $H$, so $T$ preserves the group operation exactly because $H$ is a functor, i.e. preserves composition, which is defined as to be the group operation. And the Yoneda embedding is a functor. Verify it in the general case, then you should immediately see it for the one object case.
3. If a morphism $g$ is invertible (=isomorphism) in a category $\mathcal C$, and $F:\mathcal C\to\mathcal D$ is any functor, then $F(g)$ is invertible as well. Since all morphisms of $C_G$ are invertible, it follows that their images are invertible in $[{C_G}^{op},\mathcal{Set}]$, i.e. those are natural isomorphisms, so their single component is an invertible $\mathcal{Set}$-morphism.