[proofplan] We verify directly that the specified functors and natural transformations satisfy the axioms of a category. The main point is that identity families are natural transformations and that [vertical composition of natural transformations](/theorems/3961) is again natural; both statements follow from the identity and associativity laws in $D$ together with naturality. Associativity and identity laws in $[C,D]$ are then checked componentwise at each object of $C$. Finally, the smallness of $C$ supplies the required size control for forming the families of components that define natural transformations. [/proofplan] [step:Define the candidate arrows between functors] Let $\operatorname{Ob}(C)$ denote the object set of $C$. Since $C$ is small, $\operatorname{Ob}(C)$ is a set and each hom-collection $\operatorname{Hom}_C(X,Y)$ is a set. For functors $F,G: C \to D$, define $\operatorname{Hom}_{[C,D]}(F,G)$ to be the collection of natural transformations $\eta: F \Rightarrow G$. Thus an element $\eta \in \operatorname{Hom}_{[C,D]}(F,G)$ is a family \begin{align*} \eta = (\eta_X)_{X \in \operatorname{Ob}(C)} \end{align*} such that, for each object $X \in \operatorname{Ob}(C)$, the component $\eta_X$ is a morphism \begin{align*} \eta_X: F(X) \to G(X) \end{align*} in $D$, and for every morphism $f: X \to Y$ in $C$ the naturality identity \begin{align*} G(f) \circ \eta_X = \eta_Y \circ F(f) \end{align*} holds in $D$. [/step] [step:Construct identity natural transformations] Let $F: C \to D$ be a functor. Define a family $\operatorname{id}_F = ((\operatorname{id}_F)_X)_{X \in \operatorname{Ob}(C)}$ by \begin{align*} (\operatorname{id}_F)_X := \operatorname{id}_{F(X)} \end{align*} for every object $X \in \operatorname{Ob}(C)$. We verify naturality. Let $f: X \to Y$ be a morphism in $C$. Since $F$ is a functor, $F(f): F(X) \to F(Y)$ is a morphism in $D$. The identity laws in $D$ give \begin{align*} F(f) \circ \operatorname{id}_{F(X)} = F(f) = \operatorname{id}_{F(Y)} \circ F(f). \end{align*} Therefore \begin{align*} F(f) \circ (\operatorname{id}_F)_X = (\operatorname{id}_F)_Y \circ F(f), \end{align*} so $\operatorname{id}_F: F \Rightarrow F$ is a natural transformation. This defines the identity morphism of $F$ in $[C,D]$. [/step] [step:Show vertical composition preserves naturality] Let $F,G,H: C \to D$ be functors. Let $\eta: F \Rightarrow G$ and $\theta: G \Rightarrow H$ be natural transformations. Define their vertical composite $\theta \circ \eta: F \Rightarrow H$ componentwise by \begin{align*} (\theta \circ \eta)_X := \theta_X \circ \eta_X \end{align*} for every object $X \in \operatorname{Ob}(C)$. We verify that this family is natural. Let $f: X \to Y$ be a morphism in $C$. Using associativity of composition in $D$, naturality of $\theta$, associativity again, and naturality of $\eta$, we compute \begin{align*} H(f) \circ (\theta \circ \eta)_X &= H(f) \circ (\theta_X \circ \eta_X) \\ &= (H(f) \circ \theta_X) \circ \eta_X \\ &= (\theta_Y \circ G(f)) \circ \eta_X \\ &= \theta_Y \circ (G(f) \circ \eta_X) \\ &= \theta_Y \circ (\eta_Y \circ F(f)) \\ &= (\theta_Y \circ \eta_Y) \circ F(f) \\ &= (\theta \circ \eta)_Y \circ F(f). \end{align*} Hence $\theta \circ \eta$ is a natural transformation $F \Rightarrow H$. [guided] We need to prove that composing natural transformations component by component gives another natural transformation. The proposed component at an object $X \in \operatorname{Ob}(C)$ is \begin{align*} (\theta \circ \eta)_X := \theta_X \circ \eta_X, \end{align*} which is defined because $\eta_X: F(X) \to G(X)$ and $\theta_X: G(X) \to H(X)$ are composable morphisms in $D$. The only condition to check is naturality. Take a morphism $f: X \to Y$ in $C$. We must prove \begin{align*} H(f) \circ (\theta \circ \eta)_X = (\theta \circ \eta)_Y \circ F(f). \end{align*} Starting from the left-hand side and substituting the definition of the composite component gives \begin{align*} H(f) \circ (\theta \circ \eta)_X = H(f) \circ (\theta_X \circ \eta_X). \end{align*} Associativity of composition in $D$ allows us to regroup: \begin{align*} H(f) \circ (\theta_X \circ \eta_X) = (H(f) \circ \theta_X) \circ \eta_X. \end{align*} Now use the naturality of $\theta: G \Rightarrow H$, applied to $f: X \to Y$: \begin{align*} H(f) \circ \theta_X = \theta_Y \circ G(f). \end{align*} Substituting this identity gives \begin{align*} (H(f) \circ \theta_X) \circ \eta_X = (\theta_Y \circ G(f)) \circ \eta_X. \end{align*} Regroup again by associativity in $D$: \begin{align*} (\theta_Y \circ G(f)) \circ \eta_X = \theta_Y \circ (G(f) \circ \eta_X). \end{align*} Now use the naturality of $\eta: F \Rightarrow G$, applied to the same morphism $f$: \begin{align*} G(f) \circ \eta_X = \eta_Y \circ F(f). \end{align*} Therefore \begin{align*} \theta_Y \circ (G(f) \circ \eta_X) = \theta_Y \circ (\eta_Y \circ F(f)). \end{align*} A final use of associativity in $D$ gives \begin{align*} \theta_Y \circ (\eta_Y \circ F(f)) = (\theta_Y \circ \eta_Y) \circ F(f) = (\theta \circ \eta)_Y \circ F(f). \end{align*} Thus the naturality square for $\theta \circ \eta$ commutes for every morphism $f: X \to Y$ in $C$, so $\theta \circ \eta: F \Rightarrow H$ is a natural transformation. [/guided] [/step] [step:Verify associativity of composition componentwise] Let $F,G,H,K: C \to D$ be functors. Let \begin{align*} \eta &: F \Rightarrow G, & \theta &: G \Rightarrow H, & \rho &: H \Rightarrow K \end{align*} be natural transformations. For every object $X \in \operatorname{Ob}(C)$, the definition of vertical composition and associativity of composition in $D$ give \begin{align*} ((\rho \circ \theta) \circ \eta)_X &= (\rho_X \circ \theta_X) \circ \eta_X \\ &= \rho_X \circ (\theta_X \circ \eta_X) \\ &= (\rho \circ (\theta \circ \eta))_X. \end{align*} Since natural transformations are equal exactly when their components agree at every object of $C$, it follows that \begin{align*} (\rho \circ \theta) \circ \eta = \rho \circ (\theta \circ \eta). \end{align*} Thus composition in $[C,D]$ is associative. [/step] [step:Verify the identity laws componentwise] Let $F,G: C \to D$ be functors, and let $\eta: F \Rightarrow G$ be a natural transformation. For every object $X \in \operatorname{Ob}(C)$, the identity laws in $D$ give \begin{align*} (\eta \circ \operatorname{id}_F)_X &= \eta_X \circ \operatorname{id}_{F(X)} = \eta_X, \\ (\operatorname{id}_G \circ \eta)_X &= \operatorname{id}_{G(X)} \circ \eta_X = \eta_X. \end{align*} Since equality of natural transformations is componentwise equality, we obtain \begin{align*} \eta \circ \operatorname{id}_F &= \eta, & \operatorname{id}_G \circ \eta &= \eta. \end{align*} Therefore the identity morphisms constructed above are left and right identities for vertical composition. [/step] [step:Conclude that the functor category exists] The preceding steps define objects, morphisms, identity morphisms, and composition for $[C,D]$, and verify associativity and the identity laws. The smallness of $C$ ensures that the component families \begin{align*} (\eta_X)_{X \in \operatorname{Ob}(C)} \end{align*} are indexed by a set, so the natural transformations used as morphisms are legitimate arrows in the ambient size convention. Hence the specified data form a category, denoted $[C,D]$. This proves that the functor category from the small category $C$ to the category $D$ exists. [/step]