Two questions on Yoneda's lemma

Question

I used to be okay with the Yoneda lemma because in the material I studied it was only used in order to speed up a bit the discussion of universal objects that came up (tensor products or spectra for instance), and for these the vague idea that the Yoneda lemma is about characterizing objects by the way they act or something was acceptable to me. Recently however I learned about putting structure on Hom-sets, for which you need almost its full power, namely that arrows $h_X \to h_Y$ all come from arrows $Y \to X$. This made me realize that I don't have a solid understanding of it. I have two questions.

First, a way to formulate the Yoneda lemma is to say that for an object $X$ of a category $C$ the functor $\mathbf{Set}^C \to \mathbf{Set}, F \mapsto F(X)$ of evaluation at $X$ is representable, by $h_X$ (this looks to me like the Riesz representation theorem but maybe I'm going off track here). This seems to me the cleanest way to put it but I haven't seen it anywhere. I wonder if this is a good way to think about it, if there's more to representable functors on category of functors, and if it is the case, where I could learn about it.

Secondly, trying to make sense of the proof, I read on this site that it should be seen as analogous to the way a map $A \to M$ of $A$-modules is determined by where it sends $1$, and to think of the naturality as a linearity. Now I agree that linearity can be expressed with commutative diagrams, and I get the vague idea, but I don't see the analogy precisely so I'm skeptical. In particular I think the crux of the proof is, given $\lambda : h_X \to F$ a natural transformation, to think of $f : X \to Y$ both as something on which $\lambda_Y$ is evaluated (since what we want to prove is that $\lambda_Y(f)=Ff(\xi)$ for a certain $\xi \in F(X)$, independant of $Y$), and a way to "pass" from $X$ to $Y$ to exploit naturality. I don't see the corresponding operation in the analogy with $A$-modules. Am I correct or is there a way to prove this fact about maps $A \to M$ directly with Yoneda, so as to make the analogy precise?

Answer 1

The analogy with module maps is quite clear : if you know what $1$ is sent to, then for an arbitrary $\lambda\in A$ you can think of it either as an element of the module, so it gets sent to $f(\lambda)$, either as a scalar acting on $1$, so it gets sent to $\lambda f(1)$.

This forces $f(\lambda) = \lambda f(1)$, and then you check that for any $m$, $f(\lambda) = \lambda m$ works.

It's the same as for the Yoneda lemma: you have $f: X\to Y$ as you noted, which can be seen as an element of $h_X(Y)$, where its image is then $\eta_Y(f)$, or as something "acting" on $h_X(X)$, so something to which naturality can be applied, where its image must then be $F(f)( \eta_X(id_X))$

So in both situations there's the idea of seeing something either as an element, or as an "agent" (I don't have specific references in mind, but you can see morphisms of $C$ as "acting" on $F$, for any functor $F:C\to D$ - to me the analogy with other types of action, e.g. group actions or modules over rings is clear but it could be that it's not for everyone)

As for your first query, yes, the Yoneda lemma can be seen as stating that $ev_X$ is represented by $h_X$ (well it's actually a bit better, because it says that specifically $id_X\in ev_X(h_X)$ is what makes the representation, but all in all, it's a representability statement). I don't remember seeing anything about representable functors on functor categories specifically, but there might be some stuff to be said (as they relate, e.g. to Kan extensions - that's something you could look up if you're interested)

Answer 2

I am going to sketch a proof here. I have no idea about the intuition behind Yoneda's lemma (I remember my professor saying "Yoneda came from the future" when we talked about Yoneda's lemma). But the proof is pretty easy to understand, once you get familiar with the involved concepts.

Statement: Let $F:\mathcal{C}\to\textbf{Set}$ be a functor and fix an object $C$ of $\mathcal C$. What we want to show is that the natural transformations from $\hom(C,-)$ to $F$ form a set (this is highly non trivial!)

So one has to begin a proof. Well, the only set known relatable to $C$ is $F(C)$. So it would be nice if we could embed the collection of all natural transformations $\hom(C,-)\to F$ inside $F(C)$, to get our result.

How can one do that? Suppose we are given $\eta:=(\eta_{X})_{X\in\text{ob}\mathcal C}$ a natural transformation from the $\hom(C,-)$ functor to $F$, thus for each $X\in\mathcal C$ $\eta_X$ is a mapping of the form $\eta_X:\hom(C,X)\to F(X)$. Where could we "send" $\eta$ to get an element of $F(C)$? Well, $\eta_C$ is a map $\eta_C:\hom(C,C)\to F(C)$, so if we compute $\eta_C$ to some homomorphism $C\xrightarrow{f}C$ of $\mathcal C$, then we will have an element of $F(C)$. Do we know any such homomorphisms? Of course, the identity morphism $1_C$. So $\eta_C(1_C)$ is indeed an element of the set $F(C)$. Moreover, in the general case, this could be the only morphism between $C$ and itself.

Thus the only reasonable map that one can define from the collection of all natural transformations $\hom(C,-)\to F$ is the following: $$\eta:=(\eta_X)_{X\in\text{ob}\mathcal C}\mapsto\eta_C(1_C)\in F(C)$$

Now the properties of natural transformations, naturality to be specific (naturality is the property with the commutative diagram) is used to show that this is a $1-1$ map (so we have embedded the natural transformations to $F(C)$, so we get that they form a set) but it also shows that it is a surjective map, so, as sets, we can identify $F(C)$ with all the natural transformations of $\hom(C,-)$ to $F$.

A final comment: As far as I understand (and I do not understand much about categories), the Yoneda lemma is a way to show that the set $F(C)$ pretty much contains all the info of how one could "bound" the behavior of the $\hom(C,-)$ functor with respect to $F$. It doesn't make much sense but that's the best I can understand for the nature of this result.

Also, I think your relation with the Riesz representation lemma is interesting, but keep in mind that the nature of the objects studied are much, much, different.