The Unapologetic Mathematician

Mathematics for the interested outsider

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


  1. […] Naturality Revisited Let’s look back at the enriched versions of representable functors. If we fix an object we have a -natural transformation . This corresponds under the closure […]

    Pingback by Enriched Naturality Revisited « The Unapologetic Mathematician | August 29, 2007 | Reply

  2. […] how do we define the dual transformation? It turns out this is the contravariant functor represented by . That is, if is a linear functional, we define . In terms of the action on […]

    Pingback by Matrices IV « The Unapologetic Mathematician | May 28, 2008 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: