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\$$.