Functor Category is Category

Theorem

Let $\mathbf C$ and $\mathbf D$ be categories.

Then the functor category $\mathrm{Fun}(\mathbf C, \mathbf D)$ is a category.

Proof

We check the metacategory axioms.

Composition (C1)

By Composite Of Natural Transformations is Natural Transformation composition of morphisms in $\mathrm{Fun}(\mathbf C, \mathbf D)$ is well-defined.

$\Box$

Existence of identity morphisms (C2)

Let $F$ be an object of $\mathrm{Fun}(\mathbf C, \mathbf D)$, that is a functor $F : \mathbf C \to \mathbf D$.

Define $\mathrm{id}_F : F \to F$ as the transformation, that assigns to every object $A$ of $\mathbf C$ the identity morphism $\mathrm{id}_{F(A)} : F(A) \to F(A)$.

Let $f : A \to B$ be a morphism in $\mathbf C$. The diagram

\begin{align*} \xymatrix{ F(A) \ar[r]^{\mathrm{id}_{F(A)}} \ar[d]^{F(f)} & F(A) \ar[d]^{F(f)} \\ F(B) \ar[r]^{\mathrm{id}_{F(B)}} & F(B) } \end{align*} is commutative, since $F(f) = \mathrm{id}_{F(B)} \circ F(f) = F(f) \circ \mathrm{id}_{F(A)}$.

Hence $\mathrm{id}_F$ is a natural transformation.

Let $G : \mathbf C \to \mathbf D$ be a functor.

Let $\alpha : F \to G$ be a natural transformation.

For every object $A$ of $\mathbf C$, we have $\alpha_A \circ \mathrm{id}_{F(A)} = \alpha_A$, so $\alpha \circ \mathrm{id}_{F} = \alpha$.

Let $H : \mathbf C \to \mathbf D$ be a functor.

Let $\beta : H \to F$ be a natural transformation.

For every object $A$ of $\mathbf C$, we have $\mathrm{id}_{F(A)} \circ \beta_A = \beta_A$, so $\mathrm{id}_{F} \circ \beta = \beta$.

$\Box$

Associativity of composition (C3)

Let $F_1,F_2,F_3,F_4 : \mathbf C \to \mathbf D$ be functors.

Let $\alpha_1 : F_1 \to F_2$, $\alpha_2 : F_2 \to F_3$, $\alpha_3 : F_3 \to F_4$ be natural transformations.

For every object $A$ of $\mathbf C$, we have morphisms

$\alpha_{1,A} : F_1(A) \to F_2(A)$
$\alpha_{2,A} : F_2(A) \to F_3(A)$
$\alpha_{3,A} : F_3(A) \to F_4(A)$

By associativity of composition in $\mathbf D$ $$(\alpha_{3,A} \circ \alpha_{2,A}) \circ \alpha_{1,A} = \alpha_{3,A} \circ (\alpha_{2,A} \circ \alpha_{1,A})$$ The associativity $$(\alpha_{3} \circ \alpha_{2}) \circ \alpha_{1} = \alpha_{3} \circ (\alpha_{2} \circ \alpha_{1})$$ follows.

$\Box$

$\blacksquare$