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# functor category

Let $\mathcal{C},\mathcal{D}$ be categories. Consider the class $O$ of all covariant functors $T:\mathcal{C}\to\mathcal{D}$, and the class $M$ of all natural transformations $\tau:S\dot{\to}T$ for every pair $S,T:\mathcal{C}\to\mathcal{D}$ of functors. Write $\mathcal{D}^{\mathcal{C}}$ for the pair $(O,M)$.

For each pair of functors $S,T:\mathcal{C}\to\mathcal{D}$, write $\hom(S,T)$ the class of all natural transformations from $S$ to $T$. If $\tau$ is in both $\hom(S,T)$ and $\hom(U,V)$, then $S=U$ and $T=V$.

Using the composition of natural transformations, we have a mapping

$\bullet:\hom(R,S)\times\hom(S,T)\to\hom(R,T),$ |

for every triple $R,S,T:\mathcal{C}\to\mathcal{D}$. Since composition of natural transformations is associative, the associativity of $\bullet$ applies.

In addition, for each $S:\mathcal{C}\to\mathcal{D}$, we have the identity natural transformation $1_{S}\in\hom(S,S)$. For every $\tau\in\hom(S,T)$ and every $\eta\in\hom(T,S)$, we have $\tau\bullet 1_{S}=\tau$ and $1_{S}\bullet\eta=\eta$.

From the discussion above, we are ready to call $\mathcal{D}^{\mathcal{C}}$ a category. However, unless $\hom(S,T)$ is a set for every pair of functors in $O$, $\mathcal{D}^{\mathcal{C}}$ is not a category. When $\mathcal{D}^{\mathcal{C}}$ is a category, we call it the *category of functors* from $\mathcal{C}$ to $\mathcal{D}$, or simply a *functor category*.

That $\mathcal{D}^{\mathcal{C}}$ is a functor category depends on various restrictions being placed on the “sizes” of $\mathcal{C}$ and $\mathcal{D}$:

###### Proposition 1.

If $\mathcal{C}$ is $\mathcal{U}$-small, then $\mathcal{D}^{\mathcal{C}}$ is a category.

###### Proof.

Suppose $\mathcal{C}$ is $\mathcal{U}$-small. Consider the class $\hom(S,T)$. Each $\tau\in\hom(S,T)$ is determined by the collection of morphisms $S(A)\to T(A)$ for each object $A$ in $\mathcal{C}$. This means that, for each $A$ in $\mathcal{C}$, $\hom(S(A),T(A))$ contains the image of every $\tau\in\hom(S,T)$ under $A$. So the class of all these natural transformations is a subclass of the product

$\prod_{{A\in\operatorname{Ob}(\mathcal{C})}}\hom(S(A),T(A))$ | (1) |

Since $\operatorname{Ob}(\mathcal{C})$, as well as each $\hom(S(A),T(A))$ is a set, so is the product (1). Hence $\hom(S,T)$, being a subclass of (1), is a set, or that $\mathcal{D}^{{\mathcal{C}}}$ is a category. ∎

###### Proposition 2.

If in addition $\mathcal{D}$ is a $\mathcal{U}$-category, then so is $\mathcal{D}^{\mathcal{C}}$.

###### Proof.

$\mathcal{D}$ being a $\mathcal{U}$-category means that $\hom(S(A),T(A))$ is $\mathcal{U}$-small, for every object $A$ in $\mathcal{C}$. Since $\operatorname{Ob}(\mathcal{C})$ is also $\mathcal{U}$-small (assumption in Proposition 1), the product (1) above is $\mathcal{U}$-small. Consequently, $\hom(S,T)$, being a subclass of (1), is $\mathcal{U}$-small. This shows that $\mathcal{D}^{{\mathcal{C}}}$ is a $\mathcal{U}$-category. ∎

###### Proposition 3.

If $\mathcal{D}$ is furthermore $\mathcal{U}$-small, so is $\mathcal{D}^{\mathcal{C}}$.

###### Proof.

We want to show that the class $\mathcal{M}$ of all functors from $\mathcal{C}$ to $\mathcal{D}$ is $\mathcal{U}$-small. A functor $S:\mathcal{C}\to\mathcal{D}$ can be broken up into two components: a function $S_{1}:\operatorname{Ob}(\mathcal{C})\to\operatorname{Ob}(\mathcal{D})$, and a function $S_{2}:\operatorname{Mor}(\mathcal{C})\to\operatorname{Mor}(\mathcal{D})$, so that $S_{2}(A\to B)=S_{1}(A)\to S_{1}(B)$.

Define a binary relation $\sim$ on $\mathcal{M}$ so that $S\sim T$ iff they have the same first component: $S_{1}=T_{1}$. It is easy to see that $\sim$ is an equivalence relation on $\mathcal{M}$. Let $[S]$ be the equivalence class containing the functor $S$. For every morphism $A\to B$, its image under the second component of every functor in $[S]$ lies in $\hom(S_{1}(A),S_{1}(B))$. So the size of $[S]$ can not exceed the size of

$\prod_{{A,B\in\operatorname{Ob}(\mathcal{C})}}\hom(S_{1}(A),S_{1}(B))$ |

Since $\operatorname{Ob}(\mathcal{C})$ is $\mathcal{U}$-small (assumption in Prop 1), so is $\operatorname{Ob}(\mathcal{C})\times\operatorname{Ob}(\mathcal{C})$. Furthermore, since each $\hom(S_{1}(A),S_{1}(B))$ is $\mathcal{U}$-small (assumption in Prop 2), $[S]$ is $\mathcal{U}$-small as well.

Next, let us estimate the size of the class $\mathcal{M}/\sim$ of equivalence classes in $\mathcal{M}$. First, note that for every functor $S:\mathcal{C}\to\mathcal{D}$, its first component is a function from the *set* $\operatorname{Ob}(\mathcal{C})$ to the *set* $\operatorname{Ob}(\mathcal{D})$ by assumption. As $[S]\neq[T]$ iff $S_{1}\neq T_{1}$, the size can not exceed

$|\operatorname{Ob}(\mathcal{D})^{{\operatorname{Ob}(\mathcal{C})}}|$ |

the cardinality of the set of all functions from $\operatorname{Ob}(\mathcal{C})$ to $\operatorname{Ob}(\mathcal{D})$. By assumption, $\operatorname{Ob}(\mathcal{D})$ is $\mathcal{U}$-small, so is $\operatorname{Ob}(\mathcal{D})^{{\operatorname{Ob}(\mathcal{C})}}$. As a result, $\mathcal{M}/\sim$ is $\mathcal{U}$-small. Together with the fact that $[S]$ is $\mathcal{U}$-small for each functor $S$, we have that $\mathcal{M}$ itself must be $\mathcal{U}$-small, which completes the proof. ∎

## Mathematics Subject Classification

18-00*no label found*18A25

*no label found*18A05

*no label found*

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