The Unapologetic Mathematician

Mathematics for the interested outsider

A New Definition of Spans

I was trying to add in duals to the theory, and I ran into some trouble. The fix seems to be something I’d considered a little earlier for a possible generalization, but it seems that duals force the issue. We need to tweak our definition of spans.

Now, most of the definition is fine as it stands. Our objects are just the same as in \mathcal{C}, and our 1-morphisms are spans A\stackrel{f_l}{\leftarrow}X\stackrel{f_r}{\rightarrow}B. When I first (re)invented these things, I stopped at this level. I handled the associativity not with 2-morphisms, but with defining two spans A\stackrel{f_l}{\leftarrow}X\stackrel{f_r}{\rightarrow}B and A\stackrel{g_l}{\leftarrow}Y\stackrel{g_r}{\rightarrow}B to be equivalent if there was an isomorphism \alpha:X\rightarrow Y so that f_l=g_l\circ\alpha and f_r=g_r\circ\alpha. Then I said that the 1-morphisms were equivalence classes of spans, giving me a 1-category.

Now \mathcal{C} still sits inside this category I’ll call \mathbf{Span}_1(\mathcal{C}). Indeed, if we use the same inclusion as before, the only way two arrows of \mathcal{C} can be isomorphic in \mathbf{Span}(\mathcal{C}) (and thus equal in \mathbf{Span}_1(\mathcal{C})) is for them to be equal already in \mathcal{C}.

But we want to pay attention to those 2-morphisms, and that’s where things start to get interesting. See, those arrows \alpha:X\rightarrow Y are nice as 2-morphisms, but they’re not very.. “spanny”. Instead, let’s define a 2-morphism from A\stackrel{f_l}{\leftarrow}X\stackrel{f_r}{\rightarrow}B to A\stackrel{g_l}{\leftarrow}Y\stackrel{g_r}{\rightarrow}B to be itself a span X\stackrel{\alpha_l}{\leftarrow}M\stackrel{\alpha_r}{\rightarrow}Y satisfying f_l\circ\alpha_l=g_l\circ\alpha_r and f_r\circ\alpha_l=g_r\circ\alpha_r. Here’s the picture:
The New Definition of a 2-morphism

Now, we handle the associativity at the 2-morphism level by cutting off the same way we did in \mathbf{Span}_1(\mathcal{C}) and say that a 2-morphism is really an equivalence class of spans. This makes the vertical composition of 2-morphisms just the same pullback construction as for the composition of 1-morphisms.

The horizontal composition of these 2-morphisms gets tricky. Here’s another picture:
Composition of 2-Morphisms
Here we have our two 2-morphisms and we’ve already pulled back to get the 1-morphisms for the source and target of the composition. We write \alpha:M\rightarrow A' for the composite f_r\circ\alpha_l=g_r\circ\alpha_r, and similarly \alpha' on the other side.

Now we can pull back the diagram M\stackrel{\alpha}{\rightarrow}A'\stackrel{\alpha'}{\leftarrow}M' to get an object M'' with arrows to M and M'. If we follow these arrows, then up to X and X' (respectively) the universal property of X'' gives us a unique arrow from M'' to X''. Similarly, we have a unique arrow from M'' to Y''. These arrows make the required squares commute, and so define a span from X'' to Y'' which is our composite 2-morphism.

When we compose two spans, again we only have associativity up to isomorphism. In \mathbf{Span}_1(\mathcal{C}), this becomes an equality, so we’re fine. In \mathbf{Span}(\mathcal{C}) we made this isomorphism a 2-morphism between the two composite spans. Now in \mathbf{Span}_2(\mathcal{C}) we can make this isomorphism into one leg of a span 2-morphism, and everything works out as before. The exchange identity for the two compositions of 2-morphisms also works out, but it’s even more complicated than the definition of horizontal composition.

Seriously, does anyone know of a tool that will render commutative diagrams in 3-D, like with Java or something? This is getting ridiculous.

Anyhow, I think now I can throw away the request that the monoidal structures on \mathcal{C} play nice with the pullbacks. Unfortunately, it’s getting a lot more complicated now and I have other real-world obligations I’ve got to attend to. So I think I’ll back-burner this discussion and move back to something old rather than spend too much time working this stuff out live as I have been doing.

October 15, 2007 - Posted by | Category theory


  1. It might be worth mentioning that a number of people appear to have been playing with this idea. Joshua Nichols-Barrer mentioned to me that he and Mike Schulman thought about this way of dealing with spans and spans-between-spans, and so on, to give an infinity-category of spans. I use something similar based on a cubical arrangement involving diagrams that look like products of two span diagrams (i.e. the source and target of the spans needn’t be the same, so you have spans at both ends as well as in the middle) – which itself is related to a similar construction in n dimensions which Marco Grandis has some papers about (actually cospans in that case). It seemed necessary to do this to get these things to represent cobordisms adequately. There are probably other examples I’m not aware of.

    The kind of arrangement you have here, with a common source and target for the two spans here, seems related to localization, which is used in derived categories (when we’re internal to a category of chain complexes). In that case, there is a special class of “quasi-isomorphisms” which have to appear on the left leg of the span – introducing spans as morphisms is like a way of introducing formal inverses for them, a-la localization in a ring.

    This business is on my back-burner also. Anyway, it’s interesting stuff.

    Comment by Jeffrey Morton | October 17, 2007 | Reply

  2. I suppose I didn’t explicitly mention that in my earlier installments. As I said, \mathcal{C} sits in \mathbf{Span}(\mathcal{C}), as does \mathcal{C}^\mathrm{op}, and the two inclusions give each other 1-sided inverses. If things are nice, you get actual inverses.

    And even better, the dual for a 1- or 2-morphism that comes from one of these inclusions is exactly the adjoined inverse. I think I’ll be writing this up as soon as I can, because it would be nice to have it out before my paper using things like spans in studying tangles.

    One more thing: it seems an n-tangle is (in a sense) a cospan of (n-1)-tangles, and this is where it gets all its nice structure. Specifically, if you take the braided monoidal category with duals \mathcal{T}ang and hit it with this construction, you get the category from HDA4.

    Comment by John Armstrong | October 17, 2007 | Reply

  3. The infinity-category of spans-between-spans-between-spans is closely analogous to the infinity-category of cobordisms-between-cobordisms-between-cobordisms. Cutting it off at some level and using isomorphism classes of spans is analogous to using diffeomorphism classes of cobordisms.

    This infinity-category always reminds Jim and me of the Monty Python routine: “Span, span, span, span…”

    Comment by John Baez | October 19, 2007 | Reply

  4. That’s exactly what I’m thinking of. Cobordisms are cospans, and so cobordisms between cobordisms between… are spans between spans between…

    Basically, I think the reason that everything about 2-tangles listed in HDA4 works out is that it’s secretly a special case of (the dual of) this construction.

    So here’s a far-reaching conjecture, since a man’s reach should exceed his grasp: If we start at the point (n,k) with both n and k at least 1 in your famous table, then “spanning” moves us to either (n+1,k) or (n+1,k+1), and duals come along for the ride.

    Comment by John Armstrong | October 19, 2007 | Reply

  5. Sounds like a great conjecture! I hope you prove it for some nice medium-sized n and k.

    Comment by John Baez | October 21, 2007 | Reply

  6. […] 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 that […]

    Pingback by The Tensorator for Span 2-categories « The Unapologetic Mathematician | November 7, 2007 | 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 )

Facebook photo

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

Connecting to %s

%d bloggers like this: