# The Unapologetic Mathematician

## Representable Enriched Functors

Okay, unfortunately after a lot of work I don’t ssee an elegant way to jump right into what I thought of the other day. So, we’ll get there, but slowly.

Anyhow, we remember that categories come with representable functors, which have all sorts of nice properties. This comes over to our current context. Given a monoidal category $\mathcal{V}$, a category $\mathcal{C}$ enriched over $\mathcal{V}$, and an object $C\in\mathcal{C}$, we have the representable $\mathcal{V}$-functor $\hom_\mathcal{C}(C,\underline{\hphantom{X}}):\mathcal{C}\rightarrow\mathcal{V}$.

We should be clear about how this functor behaves on “morphisms”. First of all, remember that we don’t really have morphisms in an enriched category because we have hom-objects rather than hom-sets. Instead, what we get is a $\mathcal{V}$-natural transformation with $(A,B)$ component $\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$. Instead of a function which assigns a function between hom-sets to each morphism, we get an arrow from the hom-object (replacing the set of morphisms) to an exponential (replacing the set of functions between hom-sets).

Of course we also have the contravariant $\mathcal{V}$-functor $\hom_\mathcal{C}(\underline{\hphantom{X}},C)$, and similar comments about how it behaves on morphisms apply here.

Honestly, the first time you read this it seems pretty simple, but it’s deeper than you expect. You’re going to want to keep thinking about “elements” of a hom-object, particularly when you’re using $\mathbf{Ab}$ as your sample enriching category, but you really have to start breaking yourself of the habit of thinking in terms of “elements” of an object — even if the objects are something so familiar as abelian groups — now.

August 25, 2007 - Posted by | Category theory