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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 \begin{equation} \prod_{A\in \operatorname{Ob}(\mathcal{C})} \hom(S(A),T(A)) \end{equation}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. 
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 $[S]\ne [T]$ iff $S_1\ne T_1$ , 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. 
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