The Unapologetic Mathematician

Mathematics for the interested outsider

Monoidal Structures on Span 2-Categories II

As I just stated in my update to yesterday’s post, I’ve given the data for a monoidal structure on the 2-category \mathbf{Span}(\mathcal{C}). Now we need some conditions on the data.

  1. For any object A, the maps \underline{\hphantom{X}}\otimes A and A\otimes\underline{\hphantom{X}} are 2-functors.
  2. For x any object, 1-morphism, or 2-morphism, x\otimes\mathbf{1}=\mathbf{1}\otimes x=x
  3. For x any object, 1-morphism, or 2-morphism, and any objects A and B, we have the equalities
    • A\otimes(B\otimes x)=(A\otimes B)\otimes x
    • A\otimes(x\otimes B)=(A\otimes x)\otimes B
    • x\otimes(A\otimes B)=(x\otimes A)\otimes B
  4. For any 1-morphisms f:A\rightarrow A', g:B\rightarrow B', and h:C\rightarrow C', we have the equalities
    • \bigotimes_{A\otimes g,h}=A\otimes\bigotimes_{g,h}
    • \bigotimes_{f\otimes B,h}=\bigotimes_{f,B\otimes h}
    • \bigotimes_{f,g\otimes C}=\bigotimes_{f,g}\otimes C
  5. For any objects A and B we have 1_A\otimes B=A\otimes 1_B=1_{A\otimes B}
  6. For any 1-morphisms f:A\rightarrow A' and g:B\rightarrow B', we have \bigotimes_{1_A,g}=1_{A\otimes g} and \bigotimes_{f,1_B}=1_{f\otimes B}
  7. For any 1-morphisms f:A\rightarrow A', g:B\rightarrow B', and h:B\rightarrow B' and any 2-morphism \beta:g\Rightarrow h, we have \bigotimes_{f,h}\bullet((f\otimes B')\circ(A\otimes\beta))=((A'\otimes\beta)\circ(f\otimes B))\bullet\bigotimes_{f,g}
  8. For any 1-morphisms f:A\rightarrow A', g:A\rightarrow A', and h:B\rightarrow B' and any 2-morphism \alpha:f\Rightarrow g, we have \bigotimes_{g,h}\bullet((\alpha\otimes B')\circ(A\otimes h))=((A'\otimes h)\circ(\alpha\otimes B))\bullet\bigotimes_{f,h}
  9. For any 1-morphisms f:A\rightarrow A', g:B\rightarrow B', and g':B'\rightarrow B'', we have \bigotimes_{f,g'\circ g}=((A'\otimes g')\circ\bigotimes_{f,g})\bullet(\bigotimes_{f,g'}\circ(A\otimes g))
  10. For any 1-morphisms f:A\rightarrow A', f':A'\rightarrow A'', and g:B\rightarrow B', we have \bigotimes_{f'\circ f,g}=(\bigotimes_{f',g}\circ(f\otimes B))\bullet((f'\otimes B')\circ\bigotimes_{f,g})

Okay, a bunch of conditions. Notice here that we have stated a bunch of equalities. Most of them are at the level of 2-morphisms, and everything at that level in a 2-morphism should hold on the nose. But some of them are equalities between 1-morphisms, which our philosophy says we should weaken.

Really what we’re laying out here is a semistrict monoidal 2-category, as described in Higher Dimensional Algebra I. Just like for monoidal categories, there’s a “coherence theorem” that tells us that once we specify a number of relations, all the others we want will follow. There’s also something like a “strictification theorem”, but now we can’t just wipe away all the structure morphisms. We can only “semi-strictify” an arbitrary monoidal 2-category. So that’s what we’re doing here and avoiding all the extra conditions that would be required if we had associators and other such things floating around.

Okay, enough of what everyone (who’s crazy enough to work on this stuff) already knows. Let’s get down to what (as far as I can tell) is being worked out for the first time as I’m writing it down here.

For condition 1, let’s just consider tensoring on the left with an object, since tensoring on the right is almost exactly the same. It clearly preserves identity 1-morphisms, since A\otimes(B\stackrel{1_B}{\leftarrow}B\stackrel{1_B}{\rightarrow}B) is just A\otimes B\stackrel{1_A\otimes 1_B}{\leftarrow}A\otimes B\stackrel{1_A\otimes 1_B}{\rightarrow}A\otimes B and we know that 1_A\otimes 1_B=1_{A\otimes B} in the monoidal structure on \mathcal{C}. Similarly, an identity 2-morphism in \mathbf{Span}(\mathcal{C}) is given by an identity arrow in \mathcal{C}, and tensoring it on the left with 1_A gives back the appropriate identity arrow.

Does tensoring on the left with A preserve all three compositions? Sure. It preserves vertical composition of 2-morphisms straight off. It preserves composition of 1-morphisms becase we’re assuming that the tensor product on \mathcal{C} preserves pullbacks. The horizontal composition of 2-morphisms is a little trickier, partly because I was never very explicit on how to compose 2-morphisms like this. Here’s how it looks:
Horizontal Composition of Span 2-morphisms
Start with spans A\leftarrow X\rightarrow B, A\leftarrow X'\rightarrow B, B\leftarrow Y\rightarrow C, and B\leftarrow Y'\rightarrow C. Throw in arrows \alpha:X\rightarrow X' and \beta:Y\rightarrow Y' making the appropriate triangles commute. Then pull back X\rightarrow B\leftarrow Y and X'\rightarrow B\leftarrow Y' to get Z and Z', respectively. We can follow the arrow from Z to X and then by \alpha on to X'. Similarly we can get an arrow from Z to Y'. If we compose these with the arrows from X' and Y' down to B, the square clearly commutes, so by the universal property of the pullback there is a unique arrow Z\rightarrow Z'. This is the horizontal composite of \alpha and \beta. From here it’s straightforward to see that if I tensor everything in sight on the left with some object, everything will be preserved. And so we have shown condition 1.

Conditions 2 and 3 is all but trivial, since everything we to any object, 1-morphism, or 2-morphism to tensor it with an object is to invoke the monoidal product down on \mathcal{C}, where these associativity and unit constraints hold automatically. Similarly, 5 also follows immediately from the monoidal structure on \mathcal{C}.

If we look back at what we did last time, we see that we can have set up our pullbacks as in the diagram I showed then. In fact, I’m coming to think I was being overly cautious to even bring that up. That is, we can just take the tensorator to be the identity 2-morphism on the appropriate span. This immediately satisfies conditions 4 and 6. The remaining conditions 7, 8, 9, and 10 also fall in line once you write out the compositions in terms of the spans. But I won’t kid you, they look ugly. I took a picture of the diagram for condition 7. Above the center it’s the left of the equation and below the center it’s the right. I’ll eventually TeX these up, but for now suffice to say that if you actually draw out these span diagrams and set the tensorator to be the identity that everything works out smoothly.

So now we’ve proven that the data we laid out yesterday does, in fact, constitute the structure of a monoidal 2-category.

About these ads

October 11, 2007 - Posted by | Category theory

1 Comment »

  1. [...] on Span 2-Categories Now that we can add a monoidal structure to our 2-category of spans, we want to add something like a [...]

    Pingback by Braidings on Span 2-Categories « The Unapologetic Mathematician | October 14, 2007 | Reply


Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 386 other followers

%d bloggers like this: