§ Perspectives on Yoneda

We can try to gain intuition for Yoneda by considering a finite category where we view arrows as directed paths.
The "interpretation" of a path is taken by going from edges to labels and then concatenating all edge labels. We "interpret" the label id_x as "" (the empty string), and we "interpret" all other arrows a as some unique string associated to each arrow. Composition of arrows becomes concatenation of strings. This obeys all the axioms of a category. We are basically of a category as a free monoid.
Let's being our consideration of the covariant functor Hom(O, -): C → Set. Note that the objects of this category are sets of arrows Hom(O, P). To every arrow a: P → Q we associate the set function a': Hom(O, P) → Hom(O, Q). a'(op) = pq . op.
Now, to apply Yoneda, we need another covariant functor G: S → Set. We now need to show that the set of natural transformations η: F → G are in bijection with the set G[O] ∈ Set.
We do this by the following consideration. Recall that for the natural transformation, we have the commuting diagram:

x -p-> y

Hom(o, x) -p'-> Hom(o, y)
|                |
ηx               ηy
|                |
v                v
G[x] -G[p] --> G[y]

∀ x y ∈ C,
  o2x ∈ Hom(o, x),
  p ∈ Hom(x, y),
    G[p](ηx(o2x)) = ηy(p'(o2x))

Which on using the definition of p' becomes:

∀ x y ∈ C,
  o2x ∈ Hom(o, x),
  p ∈ Hom(x, y),
  G[p](ηx(o2x)) = ηy(o2x . p')

Now pick the magic x = o and o2x = o2o = id_o. This gives:

x = o, y ∈ C
o2x = id_o

∀ y ∈ C,
  p ∈ Hom(o, y),
  G[p](ηo(id_o)) = ηy(id_o . p')
  G[p](ηo(id_o)) = ηy(p') [By identity arrow]
  [assume we fix ηo(id_o) ∈ G[o] ]
  ηy(p') = G[p](ηo(id_o)) [ηy is now forced.
                          everything on the RHS is known]

Hence, we learn how to map every other arrow ηy(p). If we know how to map the arrows, we can map the objects in the hom-sets as images of the arrows, since we know what ηo[id_o] maps to. Concretely:

§ Images ηo(q) for q ∈ Hom(o, o) after ηo(id_o) is fixed:

We have the relation id_o . q = q. So we get that the arrow q': Hom(o, o) → Hom(o, o) takes id_o to q. By the structure of the natural transformation, we have that:

∀ x y ∈ C,
  o2x ∈ Hom(o, x),
  p ∈ Hom(x, y),
    G[p](ηx(o2x)) = ηy(p'(o2x))

• Pick x = y = o, o2x = id_o, p = q. This gives:

    G[q](ηo(id_o)) = ηo(q'(id_o))
    G[q](ηo(id_o)) = ηo(q' . id_o)
    G[q](ηo(id_o)) = ηo(q')
    ηo(q') = G[q](ηo(id_o))

Hence, we've deduced ηo(q'), so we know what element q gets mapped to. The same process works with any arrow!

§ A shift in perspective: Yoneda as partial monoid.

Since we're considering the sets $Hom(o, -)$, note that we can always pre-compose any element of $Hom(o, o)$ to every $Hom(o, p)$. More-over, if we know the value of $id_o$, then we have the equation that $id_o \circ a = a$. since $id_o$ is the identity. Moreover, $id_o$ is the only identity arrow we possess across all $Hom(o, -)$: We can only access the identity arrow inside $Hom(o, o)$. For all other $Hom(o, p)$ where $p \neq o$, we do not have the identity arrow $id_o$ or $id_p$. So we have a sort of partial monoid , where we have a unique identity element $id_o$, and arrows that compose partially based on domain/codomain conditions.
From this perspective, we can read the commutative diagram laws as a sort of "Cayley's theorem". We have as elements the elements of the set $Hom(o, -)$. For every arrow $a: p \rightarrow q$, we have the action $Hom(o, p) \xrightarrow{a} Hom(o, q)$.
From this perspective, it is trivial to see that:

• Every monoid can be embedded into its action space (Cayley's theorem).
• This mapping of yoneda from $Hom(o, -)$ to arbitrary sets is like a "forgetful" functor from a monoid into a semigroup.
• If our monoid is "well represented" by a semigroup, then once we know what the identity maps to, we can discover all of the other elements by using the relation $f(ex) = f(e) f(x)$. The only "arbitrariness" introduced by forgetting the monoid structure is the loss of the unique identity. NOTE : This is handwavy, since the data given by a natural transformation is somehow "different", in a way that I'm not sure how to make precise.