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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 $F$.

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    $\begingroup$ So you actually want a $2$-functor? $\endgroup$ Commented Jul 22, 2020 at 16:41
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    $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2020 at 16:52
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    $\begingroup$ @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? $\endgroup$
    – alf262
    Commented Jul 22, 2020 at 16:59
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    $\begingroup$ @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. $\endgroup$ Commented Jul 22, 2020 at 17:00
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    $\begingroup$ Indeed, my bad, I was not attentive enough. $\endgroup$ Commented Jul 22, 2020 at 17:05

1 Answer 1

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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 $$\require{AMScd} \begin{CD} [\mathcal C,\mathcal D] @>{[\mathcal C,F]}>> [\mathcal C,\mathcal D'];\\ @V{[c,\mathcal D]}VV @V{[c,\mathcal D']}VV \\ [*,\mathcal D]@>[*,F]>> [*,\mathcal D']; \end{CD}$$

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