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We have a notion of morphisms-between-morphisms in category theory as the morphisms in an arrow category $\mathcal{C}^\to$. If we take $\mathcal{C}$ to be $\mathcal{Cat}$, the category of categories, then we obtain a ``natural'' notion of morphisms-between-functors!

How close is the relation of this notion to that of natural transformations? Is there some form of equivalence?

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  • $\begingroup$ Your symbols are gone. $\endgroup$ – Pedro Tamaroff Feb 16 '16 at 1:41
  • $\begingroup$ Category of categories does not make any sense $\endgroup$ – Tsemo Aristide Feb 16 '16 at 1:42
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    $\begingroup$ let's not be so pedantic, what I meant was the category of small sets.. $\endgroup$ – Musa Al-hassy Feb 16 '16 at 2:06
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Let $\mathbf{2}$ be the category with two objects and one arrow between them. Then $\mathcal{C}^{\mathbf{2}}$ is what you are calling $\mathcal{C}^{\to}$. Now, a functor $\mathcal{C}\to\mathcal{D}^{\mathbf{2}}$ is, via the cartesian closed structure of $\mathcal{Cat}$ (at least for the category of small categories), equivalent to $\mathbf{2}\to\mathcal{D}^\mathcal{C}$. In other words, a functor into an arrow category is the same as an arrow in a functor category or a natural transformation.

However, what you actually talk about is $\mathcal{Cat}^{\mathbf{2}}$ which has as objects functors like $F : \mathcal{C}\to\mathcal{D}$ and $G : \mathcal{E}\to\mathcal{F}$ and the arrows (from $F$ to $G$ say) are pairs of functors $H : \mathcal{C}\to\mathcal{E}$ and $K : \mathcal{D}\to\mathcal{F}$ such that $G \circ H = K \circ F$. In other words, the arrows of an arrow category in general are just commutative squares of arrows of the category, and for $\mathcal{Cat}^\mathbf{2}$ are just commutative squares of functors. These are definitely not natural transformations even when restricted to endomorphisms.

Clearly natural transformations, like any functor, embed as both objects and as arrows $Id\to Id$. So $Nat(\mathcal{C},\mathcal{D})$ is $Hom(Id,Id)$ for $Id$ at say $\mathbf{2}\times\mathcal{C}\to\mathcal{D}$. Natural transformations are part of the 2-categorical structure of $\mathcal{Cat}$ and strict functors from $\mathbf{2}$ ignore that. Generalizing terminology, $\mathcal{C}^\mathbf{2}$ is the category of 1-cells in a bicategory $\mathcal{C}$. Like any category, it's 1-cells are represented by $\mathbf{2}\to\mathcal{C}^\mathbf{2}$ or $\mathbf{2}\times\mathbf{2}\to\mathcal{C}$. The 1-cells of $\mathcal{C}^\mathbf{2}$ are "1-cells between 1-cells" which need have no connection to 2-cells. $\mathcal{Cat}$ is relatively special in being able to reify its 2-cells as 1-cells. Looking at the 1-category of 1-cells and 1-cells between them in $\mathcal{Cat}$ is going to tell you something about the 2-cells only via this reification. Since this reification doesn't exist in arbitrary bi- or 2-categories, you aren't going to find a pretty, "natural" description.

If you consider laxer notions of functor though, you can find more connections in basically the same way $\mathcal{C}^\mathbf{2}$ works, i.e. we can consider 2-functors from a 2-category containing one 0-cell, one 1-cell, and one non-identity 2-cell.

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  • $\begingroup$ Yes, thank-you I am familiar with this alternate presentation of natural transformations as it can be thought of as generalisation of homotpoties. You have merely unfolded the definition of $\mathcal{Cat}^\to$ for me and have not indicated any sort of relationship to usual natural transformations, besides pointing out the obvious fact that pairs of functors are not natural transformations. What I am asking for is do you know of any useful relationship between the pair-of-functors approach derived from the arrow category and the usual natural transformation definition. That'd be helpful! $\endgroup$ – Musa Al-hassy Feb 16 '16 at 2:52
  • $\begingroup$ It is a good answer and I don't understand why it was downvoted. $\endgroup$ – Oskar Feb 16 '16 at 4:58
  • $\begingroup$ @Oskar I suspect Musa downvoted it because the original version didn't include the latter paragraphs. We'll see if the latter paragraphs address Musa's comment. Admittedly, it's hard to provide a definitive "no" to a question like "does some soft thing exist?". $\endgroup$ – Derek Elkins Feb 16 '16 at 5:37

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