# The Unapologetic Mathematician

## (Not) The Tensorator for Span 2-categories

Part of the disappointment I mentioned is that the road I was on just looked so pretty. I’ve said in various places that I agree with (what I understand to be) David Corfield’s view of mathematics as a process of telling good stories, and this was a great story, but unfortunately it just doesn’t quite ring true. Before I purge it, I want to show you the picture of the tensorator as I thought it would work. Across the top are two tensor products of one span and one object each, and across the bottom are the other two, giving the compositions in both orders. The squares (that look like triangles) at the top and bottom are pullbacks, giving the actual composite spans. Then we can put the tensor product $F\otimes G$ in the middle, and get arrows up and down from the universal properties of the pullback squares. And it even looks like a big tensor product symbol!

But ultimately there’s no way to make this span we get always be unitary, or even invertible. And all the pretty pictures in the world can’t save a deeply flawed story. Just ask Michael Bay.

## The Tensorator for Span 2-categories

I’ve just had a breakthrough today on my project to add structures to 2-categories of spans. I was hoping to generalize from the case of a monoidal structure on the base category $\mathcal{C}$ that preserved pullbacks. After some discussions with John Baez and Todd Trimble (to whom I’m much indebted), I set off on this new quest and ran into some difficulties. Finally I’ve established that in order to have a well-behaved tensorator we must assume that the monoidal structure preserves pullbacks! This is a bit of a downer, in that I was really hoping to construct a wider class of braided monoidal 2-categories with duals, but at least it covers the cases that originally drew me to the problem. With luck there will still be something interesting here. Anyhow, let’s see how this works.

First we have to consider a span $B\xleftarrow{f}A\xrightarrow{g}C$. We want this to be invertible, and further we want its inverse to be its reflection $C\xleftarrow{g}A\xrightarrow{f}B$. When we pull back $g$ over itself we get a square Here we can swap $x_l$ and $x_r$ to get a unique arrow $\beta:B\rightarrow B$ with $x_l\circ\beta=x_r$. For this to give us the identity span on $B$ we need to have $f\circ x_l=1_B=f\circ x_l\circ\beta$. This tells us that $\beta=1_B$, and we will say $f'$ for the equal arrows $x_l=x_r$, which is a right inverse for $f$: $f\circ f'=1_B$. Similarly we find a right inverse $g'$ for $g$.

Now we use the universality of the pullback in the diagram to give us a unique arrow $x$ with $f'\circ x=g'$, and similarly a unique arrow $y$ with $g'\circ y=f'$. Then we see that $f'\circ 1_B=f'=f'\circ x\circ y$, and the universality condition tells us that $1_B=x\circ y$, while $1_C=y\circ x$. And thus for the span we started with to have an inverse of the right form we must have $B\cong C$.

Now we look for a tensorator $\bigotimes_{f,g}:(f\otimes B')\circ(A\otimes g)\Rightarrow(A'\otimes g)\circ(f\otimes B)$. We start with spans $f=(A\xleftarrow{f_l}F\xrightarrow{f_r}A')$ $g=(B\xleftarrow{g_l}G\xrightarrow{g_r}B')$
and we must find a span between the pullback objects $(F\otimes B')\circ(A\otimes G)$ and $(A'\otimes G)\circ(F\otimes B)$. Further, we will want this span to have its own reflection for an inverse, as above. But as we just showed, this means that the two objects at the ends of its legs must always be isomorphic.

Now we can specialize to pick $A'=F$, $f_r=1_F$, $B=G$, and $g_l=1_G$. Then the one leg of the tensorator span will be $F\otimes G$, and so the other leg must be as well, no matter what we choose for $f_l$ and $g_r$! That is, to have any hope of finding such a well-behaved tensorator, the monoidal product on $\mathcal{C}$ must preserve pullbacks!

November 7, 2007 Posted by | Category theory | 1 Comment