# 2-functor and CAT

Let $$\cal K$$ be a small category. Let $$\cal A$$ be a subcategory of $$\mathbf {CAT}$$ and $$U:{\cal A}\hookrightarrow{\mathbf {CAT}}$$ the underlying functor. Now how is $$U^{\cal K}$$ naturally defined as a $$2$$-functor

$$U^{\cal K}:\cal A\hookrightarrow{\mathbf {CAT\ } } ?$$ I understand that on an object $$A$$ in $$\cal A$$, $$U^{\cal K}(A)$$ is the class of all functors $$F:{\cal K}\to A$$. Is this correct?

What about $$U^\cal K$$ on functors and natural transformations? Note that I'm a beginner to the $$2$$-category stuff.

Let me first start with a precision, to be sure we agree : $$\textbf{CAT}$$ is a two category, whose objects are all small categories, morphisms are functors, and $$2$$-cells are natural transformations, and $$\mathcal{A}$$ is a sub-$$2$$-category of $$\textbf{CAT}$$.
Now you want to define a functor $$U^{\mathcal{K}} : \mathcal{A} \to \textbf{CAT}$$, and you already know that for every object $$A$$ of $$\mathcal{A}$$ (by definition, $$A$$ is a small category), you have $$U^{\mathcal{K}}(A):=[\mathcal{K},A]$$ (the category of functors from $$\mathcal{K}$$ to $$A$$).
• For any functor $$G : A \to B$$ is $$\mathcal{A}$$, define $$U^\mathcal{K} (G) : [\mathcal{K},A] \to [\mathcal{K},B]$$ to be the composition $$G \circ \_$$ You can check that this defines indeed a functor between the two functor categories. Indeed, any object $$F$$ of $$[\mathcal{K},A]$$ (note, $$F : \mathcal{K} \to A$$ is just a functor) is sent to the functor $$G\circ F : \mathcal{K} \to B$$. Moreover, a functor $$\alpha : F \Rightarrow F'$$ in $$[\mathcal{K},A]$$ (note that $$\alpha$$ is just a natural transformation) is sent to the natural transformation $$(G\circ \alpha)_K = G(\alpha_K) : G(F(K)) \to G(F'(K))$$. You can check that this is indeed a natural transformation (and by the way, this is called the left whiskering of $$\alpha$$ by $$G$$)
• For any natural transformation $$\beta : G \to G'$$ in $$\mathcal{A}$$ define $$U^{\mathcal{K}}(\beta) : U^\mathcal{K}(G) \Rightarrow U^\mathcal{K}(G')$$ to be the natural transformation such that for all $$F : \mathcal{K} \to A$$, $$U^\mathcal{K}(\beta)_F$$ is the natural transformation $$\beta\circ F : G\circ F \Rightarrow G'\circ F$$ (It gets a little confusing here, since we are defining a natural transformation whose components are themselves natural transformation, so we have to be careful). The definition of $$\beta\circ F$$ is essentially the dual to the previous one and is called the right whiskering. Now to prove that this is well defined, we have to show that $$U^\mathcal{K}(\beta)$$ is indeed a natural transformation. Unwrapping all the definitions, it turns our that this is exactly given by the exchange law (which is an axiom of $$2$$-categories)
• So how looks the components of $\beta \circ F$ like? At which components should we calculate it when we have already used $F$ so the components won't be $F$'s... – user175304 Jul 12 '19 at 15:36
• Given that $\beta : G\circ F \Rightarrow G'\circ F$, there is not so many choices. The only thing that makes sense is to pose $(\beta\circ F)_X = \beta_{F(X)} : G(F(X)) \to G'(F(X))$ – Thibaut Benjamin Jul 12 '19 at 15:54