Christopher wrote:

> Well \\( C^{\text{op}} \times D \\) is a cost enriched category if we use some operator to turn the two costs into one.

Right! I guess that's another thing I didn't explain: how to make the product of two \\(\mathcal{V}\\)-enriched categories into another \\(\mathcal{V}\\)-enriched category! Ugh, I was getting way ahead of myself. In general the answer is just what you guessed: if \\(\mathcal{A}\\) and \\(\mathcal{B}\\) are \\(\mathcal{V}\\)-enriched categories, there's a \\(\mathcal{V}\\)-enriched category whose objects are pairs consisting of an object in \\(\mathcal{A}\\) and one in \\(\mathcal{B}\\), and where we define

\[ (\mathcal{A}\times \mathcal{B})( (a,b), (a',b') ) = \mathcal{A}(a,a') \otimes \mathcal{B}(b,b') \]

where \\(\otimes\\) is the multiplication (or 'tensor product') in \\(\mathcal{V}\\). Remember, \\(\mathcal{V}\\) is a monoidal preorder with multiplication \\(\otimes\\) and unit object \\(I\\).

> Well \\( C^{\text{op}} \times D \\) is a cost enriched category if we use some operator to turn the two costs into one.

Right! I guess that's another thing I didn't explain: how to make the product of two \\(\mathcal{V}\\)-enriched categories into another \\(\mathcal{V}\\)-enriched category! Ugh, I was getting way ahead of myself. In general the answer is just what you guessed: if \\(\mathcal{A}\\) and \\(\mathcal{B}\\) are \\(\mathcal{V}\\)-enriched categories, there's a \\(\mathcal{V}\\)-enriched category whose objects are pairs consisting of an object in \\(\mathcal{A}\\) and one in \\(\mathcal{B}\\), and where we define

\[ (\mathcal{A}\times \mathcal{B})( (a,b), (a',b') ) = \mathcal{A}(a,a') \otimes \mathcal{B}(b,b') \]

where \\(\otimes\\) is the multiplication (or 'tensor product') in \\(\mathcal{V}\\). Remember, \\(\mathcal{V}\\) is a monoidal preorder with multiplication \\(\otimes\\) and unit object \\(I\\).