# Functor between small functor categories

Let $$\textbf{Cat}$$ denote the category of small categories and functors between them. Fix $$\mathcal{C}\in\textbf{Cat}$$. I want to construct a functor $$[\mathcal{C},-]:\textbf{Cat}\rightarrow\textbf{Cat}$$ analogous to the hom-functors.

Obviously, $$[\mathcal{C},-](\mathcal{D}):=[\mathcal{C},\mathcal{D}]$$ for all $$\mathcal{D}\in\textbf{Cat}$$.

Let $$F:\mathcal{D}\rightarrow\mathcal{D}'$$ be a functor between two small categories. Then $$[\mathcal{C},-](F):=[\mathcal{C},F]$$ has to be a functor from $$[\mathcal{C},\mathcal{D}]$$ to $$[\mathcal{C},\mathcal{D'}]$$. For a functor $$G:\mathcal{C}\rightarrow\mathcal{D}$$, we can set $$[\mathcal{C},F](G):=F\circ G$$. Now, let $$\alpha:G\Rightarrow H$$ be a natural transformation between $$G,H:\mathcal{C}\rightarrow\mathcal{D}$$. Should I define $$[\mathcal{C},F](\alpha):=F*\alpha$$? ($$*$$ denotes the Godement product)

Edit:

Note that $$F*\alpha$$ actually denotes $$1_F*\alpha$$, where $$1_F$$ is the identity natural transformation on $$F$$.

• So you actually want a $2$-functor? Commented Jul 22, 2020 at 16:41
• If you just want a functor, you don't have to deal with natural transformations. It's only if you want to respect the structure of $2$-category. Commented Jul 22, 2020 at 16:52
• @CaptainLama For a given functor $F:\mathcal{D}\rightarrow\mathcal{D}'$ between two categories small categories, $[\mathcal{C},F]$ would have to be a functor between the functor categories $[\mathcal{C},\mathcal{D}]$ and $[\mathcal{C},\mathcal{D}']$, right? In that case, I have to define $[\mathcal{C},F]$ for objects of $[\mathcal{C},\mathcal{D}]$ and morphisms of $[\mathcal{C},\mathcal{D}]$. The latter are natural transformations, no? So, to specify $[\mathcal{C},F]$, I would have to give a rule mapping natural transformations to natural transformations. What is my mistake? Commented Jul 22, 2020 at 16:59
• @alf262 You're perfectly right. There was some confusion between $[\mathcal C,\alpha]$, which you're not trying to define, and $[\mathcal C,F](\alpha)$, which you are. Commented Jul 22, 2020 at 17:00
• Indeed, my bad, I was not attentive enough. Commented Jul 22, 2020 at 17:05

Well, if $$\mathcal C=*$$ is the terminal category, then you presumably want $$[*,-]$$ to be naturally isomorphic to the identity functor via evaluation at the only object of $$*$$. In this case a morphism $$\alpha$$ in $$[*,\mathcal D]$$ is identified with a morphism in $$\mathcal D$$, and the Godement product $$F*\alpha$$ is identified with $$F(\alpha)$$, so this looks good.
To conclude for a general $$\mathcal C$$, you just have to insist that you want $$[\mathcal C,F]$$ to be natural in $$\mathcal C$$. Then for any object of $$\mathcal C$$, viewed as a functor $$c:*\to \mathcal C$$, and any morphism $$\alpha$$ in $$[\mathcal C,\mathcal D]$$, we find $$[\mathcal C,F](\alpha)_{c}=F(\alpha_c)$$, as desired. The relevant naturality square here, to be clear, is $$\require{AMScd} \begin{CD} [\mathcal C,\mathcal D] @>{[\mathcal C,F]}>> [\mathcal C,\mathcal D'];\\ @V{[c,\mathcal D]}VV @V{[c,\mathcal D']}VV \\ [*,\mathcal D]@>[*,F]>> [*,\mathcal D']; \end{CD}$$