## Categories with Duals

Now we’ve got monoidal categories to categorify the notion of a monoid, we should consider what the proper analogue of an inverse is. What we’ll do is define a “dual object”, which is sort of halfway a left inverse and halfway a right inverse.

So, we say a monoidal category “has duals” if for every object there is a “dual object” and natural transformations and . We do *not* ask that these be isomorphisms, since for the best examples they’re actually not.

Instead we ask that the transformations be compatible in the sense that the following equations hold:

The first one means (reading right to left) that if we create a copy of the identity on the right with , then hit it with to create , then associate, then hit the on the left with to get the identity, then remove it with , it’s just the same as doing nothing at all! And there’s a similar interpretation of the other equation. I can hear your furiously scratching your heads and saying “huh?” right now, but I’ll come back to this with another viewpoint later.

Anyhow, we also want this duality to be a functor — a contravariant functor even — which you should have guessed when I called and natural transformations. So we need for every and arrow so that and . There are also a naturality conditions for and that you should be able to write down if you think of this as a covariant functor . The functor should also be monoidal, in the sense that (why this form? write out the condition explicitly).

The motivating example here is the category of finite-dimensional vector spaces over a field . Here we know that tensor product over gives a monoidal structure, and we’ll use the dual module of linear functionals as our functor. Indeed, if we set we clearly have a contravariant functor. We can verify that it’s monoidal in the proper sense, and that the pairing gives a natural transformation . The other natural transformation required is also there, but we don’t yet have the tools needed (unless you’ve taken some linear algebra on your own). Still, you can keep this example in mind.

Now one thing that *should* be bugging you is that we have arrows and , but there are two more it seems we should have. To handle this, we’ll insist that — that duality “is its own inverse”, so that we can just use and as the other natural transformations we need. This does indeed hold in our motivating example, but we again won’t prove it completely until later.

Note that there’s no reason to assume that or , so these still aren’t isomorphisms. If they are, though, then we really do have natural isomorphisms to replace left and right inverse rules in the definition of a group. If you’re feeling up to it, try to state (prove?) a coherence theorem that says what diagrams must commute to ensure they all do.

It looks like you’ve got some misplaced dollar signs in the penultimate paragraph, after “so that we can just use “.

Comment by Blake Stacey | July 8, 2007 |

Thanks, fixed.

Comment by John Armstrong | July 8, 2007 |

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typoes in par 4: for every f:N->M an arrow f^*:M*->N*

Comment by Avery Andrews | July 19, 2008 |

Thanks.

Comment by John Armstrong | July 19, 2008 |

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