Exponentiation Functor is Functor

Theorem
Let $\mathbf C$ be a Cartesian closed metacategory.

Let $A$ be an object of $\mathbf C$.

Let $\left({-}\right)^A: \mathbf C \to \mathbf C$ be the exponentiation functor.

Then $\left({-}\right)^A$ is a functor.

Proof
Let $B$ be an object of $\mathbf C$.

Let $\epsilon_B: B^A \times A \to B$ be the evaluation morphism at $B$.

Then, since:

it follows that $\left({\operatorname{id}_B}\right)^A = \operatorname{id}_{B^A}$.

Next, let $f: B \to C, g: C \to D$ be composable morphisms.

Then:

Thus, it is seen that $g^A f^A$ is the morphism satisfying the UMP for $\left({g f}\right)^A$.

That is to say:


 * $\left({g f}\right)^A = g^A f^A$

Hence $\left({-}\right)^A$ is a functor.