Oh, I think I see what I did wrong.

I made a functor from \\(\mathcal{C} \to \mathbf{Set}\\) instead of \\(\mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\).

On a side note, since \\(\mathcal{C}^{op} \times \mathcal{C}\\) is a category, we must also have a functor,

\\[

(\mathcal{C}^{op} \times \mathcal{C})^{op} \times \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\\\

=\mathcal{C} \times \mathcal{C}^{op} \times \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}.

\\]

We can keep doing this construction ad infinitum.

Also, on another side note, since \\(\mathrm{hom} : \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\) is a functor to \\(\mathbf{Set}\\), \\(\mathrm{hom}\\) counts as a database instance, however it is one that comes automatic with every category.

I made a functor from \\(\mathcal{C} \to \mathbf{Set}\\) instead of \\(\mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\).

On a side note, since \\(\mathcal{C}^{op} \times \mathcal{C}\\) is a category, we must also have a functor,

\\[

(\mathcal{C}^{op} \times \mathcal{C})^{op} \times \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\\\

=\mathcal{C} \times \mathcal{C}^{op} \times \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}.

\\]

We can keep doing this construction ad infinitum.

Also, on another side note, since \\(\mathrm{hom} : \mathcal{C}^{op} \times \mathcal{C} \to \mathbf{Set}\\) is a functor to \\(\mathbf{Set}\\), \\(\mathrm{hom}\\) counts as a database instance, however it is one that comes automatic with every category.