# Functor Category is Category

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## 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 $\map {\operatorname {Fun} } {\mathbf C, \mathbf D}$ is well-defined.

$\Box$

### Existence of identity morphisms (C2)

Let $F$ be an object of $\map {\operatorname {Fun} } {\mathbf C, \mathbf D}$, 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 $\operatorname {id}_{\map F A} : \map F A \to \map F A$.

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

The diagram:

$\xymatrix{ \map F A \ar[r]^{\operatorname {id}_{\map F A} } \ar[d]^{\map F f} & F(A) \ar[d]^{\map F f} \\ \map F B \ar[r]^{\operatorname {id}_{\map F B} } & \map F B }$

is commutative, since $\map F f = \operatorname {id}_{\map F B} \circ \map F f = \map F f \circ \operatorname {id}_{\map F A}$.

Hence $\operatorname {id}_F$ 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 $\alpha_A \circ \operatorname {id}_{\map F A} = \alpha_A$, so $\alpha \circ \operatorname {id}_F = \alpha$.

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 $\operatorname {id}_{\map F A} \circ \beta_A = \beta_A$, so $\operatorname {id}_F \circ \beta = \beta$.

$\Box$

### 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:

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

By associativity of composition in $\mathbf D$:

$\paren {\alpha_{3, A} \circ \alpha_{2, A} } \circ \alpha_{1, A} = \alpha_{3, A} \circ \paren {\alpha_{2, A} \circ \alpha_{1, A} }$

The associativity

$\paren {\alpha_3 \circ \alpha_2} \circ \alpha_1 = \alpha_3 \circ \paren {\alpha_2 \circ \alpha_1}$

follows.

$\Box$

$\blacksquare$