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Let $Cat$ be the category of categories, then its morphism category consists of functors as objects and morphism between functors as morphisms. If we restrict to the case where the two functors $\mathcal{F},\mathcal{G}:A\to B$, then these morphisms between $\mathcal{F},\mathcal{G}$ are the natural transformations. What is the name of a general morphism between two functors $\mathcal{F}:A\to B,\mathcal{G}:C\to D$?

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What makes you think there is such a thing? –  Zhen Lin May 29 '14 at 10:14
@ZhenLin If there are a pair of functors $\mathcal{H}:A\to C$ and $\mathcal{I}:B\to D$ such that $\mathcal{G}\mathcal{H}= \mathcal{I}\mathcal{F}$, it's a morphism between $\mathcal{F}$ and $\mathcal{G}$ right? –  mez May 29 '14 at 11:29
That's a possible definition. In some situations it is more natural to require that $GH$ and $IF$ are merely isomorphic (and such an isomorphism belongs to the data of a morphism between $F$ and $G$). I would just call it a morphism from $F$ to $G$. But if you use that concept somewhere, better explain it to the readers. –  Martin Brandenburg May 29 '14 at 12:19
The first example one usually encounters of a "category of morphisms" is the arrow category, which would correspond to this definition‌​. For $F$ and $G$ parallel functors, however, I don't think that's equivalent to the natural transformations between them... –  askyle May 29 '14 at 15:16

2 Answers 2

up vote 3 down vote accepted

$\require{AMScd}$ I think you are mistaking between two distinct notions (see also the comments).

The first notion is the arrow category, defined as follow. Let $\mathscr C$ be category. The category $\operatorname{Arr}(\mathscr C)$ (also denoted $\mathscr C^{\mathbf 2}$ or $\mathscr C^\rightarrow$) is the category whose

  • objects are the arrow $f$ of $\mathscr C$,
  • morphisms $(f\colon a \to b) \to (g \colon c \to d)$ are the commutative square $$ \begin{CD} a @>f>> b \\ @VVV @VVV \\ c @>>g> d , \end{CD}$$
  • composition is the concatenation of such squares.

You can of course apply that definition with $\mathscr C = \mathsf{Cat}$.

The second notion is the enrichment of $\mathsf {Cat}$ over itself. That is, the category $\mathsf{Cat}$ has the property that, for any two objects $A$ and $B$, the hom-set $\hom_{\mathsf{Cat}}(A,B)$ actually carries a category structure in such a way that the composition $$ \hom_{\mathsf{Cat}}(B,C) \times \hom_{\mathsf{Cat}}(A,B) \to \hom_{\mathsf{Cat}}(A,C) $$ is a functor. The short way to say it is : $\mathsf{Cat}$ is enriched over the (cartesian closed) monoidal category $(\mathsf{Cat},\times,\mathbf 1)$ (where $\mathbf 1$ is the final category).

The two notions are very distinct and not to be confused !

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There is a more general notion that might be relevant here: the double category $\Bbb{CAT}$ of categories, functors and profunctors, where a profunctor $\mathcal A\not\to\mathcal B$ can be defined as a category $\mathcal U$ equipped with a functor $U:\mathcal U\to 2$ where $2=\,[x\to y]$ is the category of one arrow and $\mathcal A=H^{-1}(x)$ and $\mathcal B=H^{-1}(y)$ [see e.g. in nLab].

Let the horizontal arrows of $\Bbb{CAT}$ be the profunctors, the vertical arrows the functors, and let the cells -what you are looking for- be the functors $T:\mathcal U\to\mathcal V$ for which $T|_\mathcal A=F$ and $T|_\mathcal C=G$: $$ \mathcal A \overset{\mathcal U}{\not\to}\mathcal C \\ \ F\downarrow\ T\ \downarrow G \\ \mathcal B \underset{\mathcal V}{\not\to}\mathcal D $$

You can verify that if $\mathcal U$ and $\mathcal V$ are both the identity profunctors, then we get back exactly the natural transformations $F\leadsto G$.

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