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I cannot understand if there is a way to understand the action of a functor between two functor categories, "abstractly".

I mean this: let $\mathcal{C}$, $\mathcal{D}$, $\mathcal{C}$', $\mathcal{D}$' be categories and consider the functor categories $Fnct(\mathcal{C},\mathcal{D})$ and $Fnct(\mathcal{C}',\mathcal{D}')$. Let $H : Fnct(\mathcal{C},\mathcal{D}) \longrightarrow Fnct(\mathcal{C}',\mathcal{D}')$ be a functor. How can I understand the action of the functor $H$ on objects and morphisms without any information about what categories $\mathcal{C}$, $\mathcal{D}$, $\mathcal{C}$', $\mathcal{D}$' actually are?

I found this question Functors Between Functor Categories, but it does not exactly answer to mine.

Thank you

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    $\begingroup$ Well, there is not much to understand with no additional information provided, $H$ just sends functors to functors and natural transformations to natural transformations such that identity transformation and composition of natural transformations is preserved. It might be good to figure out what such functors would be in concrete examples, such as if you take all categories to be sets or posets, grupoids, etc. $\endgroup$ – Ennar Feb 19 '16 at 20:09
  • $\begingroup$ Ennar speaks truth. Looking back at my question I can only agree with his suggestion. Looking forward to representable functors, the yoneda lemma, and formalizing limits with representability, you become comfortable with these kinds of functors, especially those of the type mentioned in the comments at the link. $\endgroup$ – Yeldarbskich Feb 19 '16 at 20:26
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    $\begingroup$ I agree that there's nothing to understand. There's nothing structural that distinguishes a functor category from an arbitrary category-indeed every category is isomorphic to a functor category. $\endgroup$ – Kevin Carlson Feb 19 '16 at 23:04
  • $\begingroup$ Pardon the revival of an old thread, but aren't functor categories structurally different due to size issues? Categories of functors between locally small categories can be large, and as such we may be unable to prove their existence. $\endgroup$ – Dawid K Feb 23 '18 at 18:36
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There is an important special case.

Let $\mathcal{C}$, $\mathcal{D}$, $\mathcal{C}'$, $\mathcal{D'}$ be categories, $T\colon\mathcal{C}'\to\mathcal{C}$ and $S\colon\mathcal{D}\to\mathcal{D'}$ be functors. Then we can define the functor $S^T\colon\mathcal{D}^{\mathcal{C}}\to\mathcal{D}'^{\mathcal{C}'}$ in the following way: $S^T(R)=S\circ T\circ R$ and $S^T(\alpha)(c)=S(\alpha(T(c)))$ for every functor $R\colon \mathcal{C}\to\mathcal{D}$, natural transformation $\alpha\in Arr(\mathcal{D}^{\mathcal{C}})$ and every object $c\in\mathcal{C}$. You can define this functor for every four categories and two functors between them.

For example, let $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$ be categories, $T\colon\mathcal{A}\to\mathcal{B}$ be a functor. Then $T^{I_C}\colon\mathcal{A}^{\mathcal{C}}\to\mathcal{B}^{\mathcal{C}}$ is called a direct image functor and is denoted by $T_*$. Also, the functor $I_C^T\colon\mathcal{C}^{\mathcal{B}}\to\mathcal{C}^{\mathcal{A}}$ is called an inverse image functor and is denoted by $T^*$.

But, it should be pointed out, that it is of course not true, that every functor $F\colon\mathcal{D}^{\mathcal{C}}\to\mathcal{D}'^{\mathcal{C}'}$ may be presented as $F=S^T$ for some $T\colon\mathcal{C}'\to\mathcal{C}$ and $S\colon\mathcal{D}\to\mathcal{D'}$ (it is not true even for sets), so the functor $(-)^{(-)}\colon\mathbf{Cat}^{op}\times\mathbf{Cat}\to\mathbf{Cat}$ is not full.

A one more special case. Let the first power is trivial, i.e. we consider functors of the form $\mathcal{A}\to\mathcal{C}^{\mathcal{B}}$. Then we may equivalently consider them as bifunctors $\mathcal{A}\times\mathcal{B}\to\mathcal{C}$, because of the exponential isomorphism $(\mathcal{C}^{\mathcal{B}})^{\mathcal{A}}\to\mathcal{C}^{\mathcal{A}\times\mathcal{B}}$.

So, as we can see, there is no (or I don't know) good ways to represent all functors between functor categories, but there are many interesting special cases when such ways exist.

Edit. After some thought, I decided to include the following presentation of all functors between functor categories (in spirit of the second special case): functor $\mathcal{D}^{\mathcal{C}}\to\mathcal{D}'^{\mathcal{C}'}$ is a functor $\mathcal{C}'\times \mathcal{D}^{\mathcal{C}}\to\mathcal{D}'$. But it's hard to call it a simplification.

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