Clarification of the statement of the Yoneda lemma

Question

Here is one version of the Yoneda lemma: Yoneda (Previously, for any category $\mathcal C$ and any $X\in \text{Ob} \mathcal C$, the guy $h_X$ was defined as the functor $\operatorname {Hom}_{\mathcal C}(-,X):\mathcal C^{op}\to \mathcal{Set}$. And $\mathcal {PreSh(C)}=\mathcal {Set}^{\mathcal C^{op}}$)

Here is another verison (Lemma A.4.6, p.21). There, $\text y$ is a functor $$\text y: \mathcal C\to \mathcal {Set}^{\mathcal C^{op}}$$ (see p.20).

My questions are:

1. Are these two statements equivalent, and if so, why? In particular, I don't see why $h_A$ in the first statement is the same functor as $\text y$ in the second statement.

2. In the first statement, what is meant by "functorial in $F$ and $A$"?

3. In the second statement, why does the first assertion mean that $\text{Nat} (\text yA, F)\simeq FA$, and what is $\simeq$? (As far as I understand, the first assertion says that there are natural transformations between the two functors whose compositions are identities, but I don't see how the statement $\text{Nat} (\text yA,F)\simeq FA$ follows, nor do I understand what it means)

Answer 1

1. They're the same statement, but note that $h_X$ is a functor $C^{op}\to\mathbf{Set}$ and $y$ is a functor $C\to\mathbf{Set}^{C^{op}}$; they're not stating they're the same functor. In this case, $h_X$ is the same presheaf as $y(X)$.

2. The more usual language for what it's calling "functorial in $A$ and $F$" is "natural in $A$ and $F$", because it's describing a natural transformation. The operations taking $(A,F)$ to $F(A)$ and $\mathrm{Hom}(h_A,F)$ are the object parts of a pair of functors $C^{op}\times\mathbf{Set}^{C^{op}}\to\mathbf{Set}$, and the theorem is saying that the bijection it's displaying is a natural isomorphim. The reason you might say this explicitly is it's possible to have things that look like natural transformations, but are only actually natural if you fix one argument and treat the functors as functors only in one variable. This is just saying explicitly that this is a natural transformation that works as the $A$ argument varies, the $F$ argument varies, or both vary.

3. Here $\simeq$ means "naturally isomorphic"; it's the same statement as in the first link, just a slight difference in notation. $\mathrm{Nat}(F,G)$ is the same thing as $\mathrm{Hom}_{PreSh(C)}(F,G)$; $yA$ is the same as $h_A$.