Natural transformations from a small category to a locally small one as limits in $\operatorname{Set}$?

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I'm working through the following exercise,

Exercise 3.2.iv. Generalize Exercise 3.2.iii to show that for any small category $J$, any locally small category $C$, and any parallel pair of functors $F,G: J \to C$, the set $\hom(F,G)$ of natural transformations from $F$ to $G$ can be defined as a small limit in $\cat{Set}$. (Hint: The diagram whose limit is $\hom(F,G)$ is indexed by a category $J^§$ whose objects are morphisms in $J$ and which has morphisms $1_x \to f$ and $1_y \to f$ for every $f : x \to y$ in $J$.)

from Emily Riehl's Category Theory in Context. It is shown in this chapter that given a presheaf $H: J \to \cat{Set}$ with $J$ small, then $\lim_JH$ always exists and is isomorphic (in $\cat{Set}$) to $\operatorname{Cone}(1,H)$, the set of cones with summit $1$ over $H$.

More concretely, the elements of $\lim_J H = \operatorname{Cone}(1,H)$ are of the form $(\lambda_j)_{j\in ob(J)} \in \prod_{j\in ob(J)}Fj$ with $\lambda_{j'} = Hf(\lambda_j)$ for each morphism $f:j \to j'$ (here we abuse notation and identify each morphism $1 \to \lambda_j$ of the cone with its image).

In this case, then, I would like to construct a functor $H : J^§ \to \cat{Set}$ so that $\hom(F,G) = \lim_{J^§}H$, i.e. I want a functor so that tuples $(\lambda_j)_{j \in J^§}$ such that $H\phi(\lambda_j) = \lambda_j'$ correspond to natural transformations from $F$ to $G$. I also know that natural transformations can be thought as tuples $(\mu_j)_{j \in J}$ of morphisms $\mu_j : Fj \to Gj$ so that

$$Gf\mu_j = \mu_{j'}Ff \quad (\forall f: j \to j'). \tag{1}$$

This makes me think that $H$ sould take objects $(f : x \to y) \in J^§$ to $\hom(Fx,Gy)$ so that a cone over $H$ with summit $1$ consists of $\lambda_f \in \hom(Fdom(f), Gcod(f))$ which could act, when $f$ is an identity, as the components of $\mu: F \Rightarrow G$.

However, I'm not so sure how to finish the construction of $H$ in order to guarantee that $(1)$ holds, and so that the correspondence is bijective. Any ideas?

• For the posterity (and because it has caused me confusion): by definition, the category $J^\S$ has, as objects, morphisms in $J$, $$\operatorname{ob}J^\S:=\operatorname{mor}J.$$ For each morphism $f:x\to y$ in $J$, in the category $J^\S$ there is exactly one morphism $1_x\to f$ and exactly one morphism $1_y\to f$, and $J^\S$ has no additional non-identity morphisms. Commented Oct 4, 2022 at 17:09

Your instincts are good. The functor $H$ does indeed take $f:x\to y$ to $\hom(Fx,Gy)$ and a morphism $id_y\to f$ to $$\hom(Ff,Gy):\hom(Fy,Gy)\to\hom(Fx,Gy)$$ and $id_x\to f$ to $$\hom(Fx,Gf):\hom(Fx,Gx)\to\hom(Fx,Gy).$$ A cone over this diagram from $1$ (call the components of this cone $\alpha_f$) means that, for all $(f:x\to y)\in J$, $$\hom(Fx,Gf)\circ\alpha_{id_X}=\alpha_f=\hom(Ff,Gy)\circ\alpha_{id_y}.$$ Denote by $\beta_x$ the value of $\alpha_{id_x}$ in $\hom(Fx,Gx)$. This equality says that $$Gf\circ\beta_x=\beta_y\circ Ff,$$ which is exactly the $(1)$ you're going for; so every such cone has an associated natural transformation.
The converse, that every natural transformation has an associated cone of this form (and that these operations are inverse to each other) is easy to check. Just remember that if you have $\mu:F\to G$, we just need $\alpha_{id_x}$ to pick out $\mu_x$, and for $f:x\to y$, $\alpha_f$ just needs to pick out $Gf\circ\mu_x$. And since cones $1\to H$ correspond to elements of $\lim H$, this is just $\hom(F,G)$ as we wanted.