# The Unapologetic Mathematician

## Enriched Categories II

So we have the basic data of a category $\mathcal{C}$ enriched over a monoidal category $\mathcal{V}$. Of course, what I left out were the relations that have to hold. And they’re just the same as those from categories, but now written in terms of $\mathcal{V}$ instead of $\mathbf{Set}$: associativity and identity relations, as encoded in the following commutative diagrams:

Notice how these are very similar to the axioms for a monoidal category or a monoid object. And this shouldn’t be unexpected by now, since we know that a monoid is just a (small) category with only one object. In fact, if we only have one object in a $\mathcal{V}$-enriched category we get back exactly a monoid object in $\mathcal{V}$!

Now, often we’re thinking of our hom-objects as “hom-sets with additional structure”. There should be a nice way to forget that extra structure and recover just a regular category again. To an extent this is true, but for some monoidal categories $\mathcal{V}$ the “underlying set” functor isn’t really an underlying set at all. For now, though, let’s look at a familiar category of “sets with extra structure” and see how we get the underlying set out of the category itself.

Again, the good example to always refer back to for enriched categories is $\mathbf{Ab}$, the category of abelian groups with tensor product as the monoidal structure. We recall that the functor giving the free abelian group on a set is left adjoint to the forgetful functor from abelian groups to sets. That is, $\hom_\mathbf{Ab}(F(S),A)\cong\hom_\mathbf{Set}(S,U(A))$. We also know that we can consider an element of the underlying set $U(A)$ of an abelian group as a function from a one-point set into $U(A)$. That is, $\hom_\mathbf{Set}(\{*\},U(A))\cong U(A)$. Putting these together, we see that $U(A)\cong\hom_\mathbf{Ab}(\mathbb{Z},A)$, since $\mathbb{Z}$ is the free abelian group on one generator.

But $\mathbb{Z}$ is also the identity object for the tensor product! The same sort of argument goes through for all our usual sets-with-structure, telling us that in all these cases the “underlying set” functor is represented by the monoidal identity $\mathbf{1}$, which is the free object on one generator. We take this as our general rule, giving the representable functor $V(\underline{\hphantom{X}})=\hom_{\mathcal{V}_0}(\mathbf{1},\underline{\hphantom{X}}):\mathcal{V}_0\rightarrow\mathbf{Set}$. In many cases (but not all!) this is the usual “underlying set” functor, but now we’ve written it entirely in terms of the monoidal category $\mathcal{V}$!

As time goes by, we’ll use this construction to recover the “underlying category” of an enriched category. The basic idea should be apparent, but before we can really write it down properly we need to enrich the notions of functors and natural transformations.

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August 14, 2007 - Posted by | Category theory

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