# The Unapologetic Mathematician

## 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 $\mathcal{M}$ “has duals” if for every object $M$ there is a “dual object” $M^*$ and natural transformations $\eta_M:\mathbf{1}\rightarrow M^*\otimes M$ and $\epsilon_M:M\otimes M^*\rightarrow\mathbf{1}$. 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: $\lambda_M\circ(\epsilon_M\otimes1_M)\circ\alpha_{M,M^*,M}^{-1}\circ(1_M\otimes\eta_M)\circ\rho_M^{-1}=1_M$ $\rho_{M^*}\circ(1_{M^*}\otimes\epsilon_M)\circ\alpha_{M^*,M,M^*}\circ(\eta_M\otimes1_{M^*})\circ\lambda_M^{-1}=1_{M^*}$
The first one means (reading right to left) that if we create a copy of the identity on the right with $\rho^{-1}$, then hit it with $\eta$ to create $M^*\otimes M$, then associate, then hit the $M\otimes M^*$ on the left with $\epsilon$ to get the identity, then remove it with $\lambda$, 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 $\eta$ and $\epsilon$ natural transformations. So we need for every $f:M\rightarrow N$ and arrow $f^*:N^*\rightarrow M^*$ so that $(f\circ g)^*=g^*\circ f^*$ and $1_M^*=1_{M^*}$. There are also a naturality conditions for $\eta$ and $\epsilon$ that you should be able to write down if you think of this as a covariant functor $(\underline{\hphantom{X}})^*:\mathcal{M}^\mathrm{op}\rightarrow\mathcal{M}$. The functor should also be monoidal, in the sense that $(M\otimes N)^*\cong N^*\otimes M^*$ (why this form? write out the condition explicitly).

The motivating example here is the category $\mathbf{FinVect}_k$ of finite-dimensional vector spaces over a field $k$. Here we know that tensor product over $k$ gives a monoidal structure, and we’ll use the dual module of linear functionals as our functor. Indeed, if we set $M^*=\hom_{\mathbf{Vect}_k}(M,k)$ 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 $M\otimes M^*\rightarrow\mathbf{1}=k$. 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 $M\otimes M^*\rightarrow\mathbf{1}$ and $\mathbf{1}\rightarrow M^*\otimes M\rightarrow$, but there are two more it seems we should have. To handle this, we’ll insist that $(M^*)^*\cong M$ — that duality “is its own inverse”, so that we can just use $\eta_{M^*}$ and $\epsilon_{M^*}$ 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 $\epsilon_{M^*}\circ\eta_M=1_\mathbf{1}$ or $\epsilon_{M}\circ\eta_{M_*}=1_\mathbf{1}$, 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.

July 7, 2007 Posted by | Category theory | 10 Comments