# The Unapologetic Mathematician

## Enriched Categorical Constructions

We’re going to need to talk about enriched functors with more than one variable, so we’re going to need an enriched analogue of the product of two categories.

Remember that the product $\mathcal{C}\times\mathcal{D}$ of two categories has the product of the object-classes as its objects, and it has pairs of morphisms for its morphisms. That is, the hom-set $\hom_{\mathcal{C}\times\mathcal{D}}((C_1,D_1),(C_2,D_2))$ is the product $\hom_\mathcal{C}(C_1,C_2)\times\hom_\mathcal{D}(D_1,D_2)$. Of course, in the enriched setting we no longer have hom-sets to work with.

So we’ll keep the same definition for the objects of our product category, but we’ll replace the definition of the hom-objects:
$\hom_{\mathcal{C}\times\mathcal{D}}((C_1,D_1),(C_2,D_2))=\hom_\mathcal{C}(C_1,C_2)\otimes\hom_\mathcal{D}(D_1,D_2)$
Now we can use the associativity and commutativity of our monoidal category $\mathcal{V}$ (remember we’re assuming it’s symmetric now) to move around factors like this:
$(\hom_\mathcal{C}(C_2,C_3)\otimes\hom_\mathcal{D}(D_2,D_3))\otimes(\hom_\mathcal{C}(C_1,C_2)\otimes\hom_\mathcal{D}(D_1,D_2))\rightarrow$
$(\hom_\mathcal{C}(C_2,C_3)\otimes\hom_\mathcal{C}(C_1,C_2))\otimes(\hom_\mathcal{D}(D_2,D_3)\otimes\hom_\mathcal{D}(D_1,D_2))$
at which point we can use the composition in each category to give a composition of the original pairs.

To get an identity, we use $\mathbf{1}\cong\mathbf{1}\otimes\mathbf{1}$ and then hit the left copy of $\mathbf{1}$ with the identity morphism for the object $C\in\mathcal{C}$ and the right copy with the identity morphism for $D\in\mathcal{D}$.

What about the opposite category? Well, it works pretty much the same as before. We just define $\hom_{\mathcal{C}^\mathrm{op}}(A,B)=\hom_\mathcal{C}(B,A)$. For an identity, we just use the same $\mathbf{1}\rightarrow\hom_\mathcal{C}(C,C)=\hom_{\mathcal{C}^\mathrm{op}}(C,C)$ as before.

Actually, these same constructions apply to functors. If we have functors $F:\mathcal{C}\rightarrow\mathcal{C}'$ and $G:\mathcal{D}\rightarrow\mathcal{D}'$, we can assemble them into a functor $F\otimes G:\mathcal{C}\otimes\mathcal{D}\rightarrow\mathcal{C}'\otimes\mathcal{D}'$. Just define $F\otimes G(C\otimes D)=F(C)\otimes G(D)$, and use a similar definition for the morphisms. Also, given $F:\mathcal{C}\rightarrow\mathcal{D}$ we get a functor $F^\mathrm{op}:\mathcal{C}^\mathrm{op}\rightarrow\mathcal{D}^\mathrm{op}$.

Now we know we have a 2-category $\mathcal{V}\mathbf{-Cat}$ of categories enriched over $\mathcal{V}$. Since a 2-category is a category enriched over categories, we can pass to the underlying category $\mathcal{V}\mathbf{-Cat}_0$ of enriched categories. It turns out that all the foregoing discussion gives this category some nice, familiar structure.

The product of two enriched categories turns out to be weakly associative. Also, remember from our discussion of the underlying category that we have a $\mathcal{V}$-category $\mathcal{I}$. This behaves like a weak identity for the product. That is, when we equip $\mathcal{V}\mathbf{-Cat}_0$ with this product and identified object, it turns out to be a monoidal category! Even better, it’s symmetric — $\mathcal{C}\otimes\mathcal{D}\cong\mathcal{D}\otimes\mathcal{C}$. And what is the opposite category but a duality on this category?

So now we can define contravariant enriched functors, as well as functors of more than one variable. As usual, you should go back and try to think of these definitions in terms of ordinary categories ($\mathbf{Set}$-categories) as well as $\mathbf{Ab}$-categories.

Incidentally, if you want to run ahead a bit, try working out how natural transformations fit into the picture. It turns out that the 2-category $\mathcal{V}\mathbf{-Cat}$ is an example of an even deeper structure I haven’t defined yet: it’s a symmetric monoidal 2-category with duals.

[UPDATE]: Excuse me.. I should have said that $\mathcal{V}\mathbf{-Cat}$ is a symmetric monoidal 2-category with a duality involution rather than “with duals”, and similarly for the underlying category. I blame my inattention on being stuck around the house all day waiting for repairmen to come by to put the dishwasher they left in my living room last Friday into the dishwasher-sized hole they put in my kitchen. Basically, the “duality involution” means that the opposite of the opposite $\mathcal{V}$-category is the original $\mathcal{V}$-category back again, and that the opposite of a tensor product is the tensor product of the opposites.

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Posted by John Armstrong | Category theory

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