No, there is no such thing as a functor \\(\mathbf{op}: \mathcal{C} \to \mathcal{C}^{\mathrm{op}}\\). Functors can't 'flip arrows around', since

a functor \\(F\\) maps a morphism \\(f : x \to y\\) to a morphism \\(F(f) : F(x) \to F(y)\\).

This is a great example of a level slip! What we really have is a functor

\[ \mathbf{op} : \mathbf{Cat} \to \mathbf{Cat} \]

This functor sends each object of \\(\mathbf{Cat} \\) - that is, each category \\(\mathcal{C}\\) - to its opposite:

\[ \mathbf{op}(\mathcal{C}) = \mathcal{C}^{\mathrm{op}} \]

This functor send each morphism of \\(\mathbf{Cat} \\) - that is, each functor \\(F: \mathcal{C} \to \mathcal{D} \\) - to some obvious thing.

**Puzzle.** What is this obvious thing?

a functor \\(F\\) maps a morphism \\(f : x \to y\\) to a morphism \\(F(f) : F(x) \to F(y)\\).

This is a great example of a level slip! What we really have is a functor

\[ \mathbf{op} : \mathbf{Cat} \to \mathbf{Cat} \]

This functor sends each object of \\(\mathbf{Cat} \\) - that is, each category \\(\mathcal{C}\\) - to its opposite:

\[ \mathbf{op}(\mathcal{C}) = \mathcal{C}^{\mathrm{op}} \]

This functor send each morphism of \\(\mathbf{Cat} \\) - that is, each functor \\(F: \mathcal{C} \to \mathcal{D} \\) - to some obvious thing.

**Puzzle.** What is this obvious thing?