The Unapologetic Mathematician

Spans and Cospans II

There’s something we need to note about spans that will come in extremely handy as we start trying to add structure to our categories of spans.

Remember that we’re starting with a category $\mathcal{C}$ with pullbacks, and from this we construct the weak 2-category $\mathbf{Span}(\mathcal{C})$. It turns out that we can find $\mathcal{C}$ inside $\mathbf{Span}(\mathcal{C})$. First of all, we consider $\mathcal{C}$ as a 2-category itself by our usual trick of considering a set as a category — to every morphism in $\mathcal{C}$, just add one identity 2-morphism and nothing else.

Now we’re going to need an inclusion 2-functor. We could just look for it to preserve compositions up to some natural 2-morphism, but we can actually get preservation on the nose. Just send $f:A\rightarrow B$ to the span $A\leftarrow A\rightarrow B$, where the left arrow is the identity on $A$ and the right arrow is $f$. The composite of two such spans looks like this:

Clearly the identity arrow in $\mathcal{C}$ is sent to the identity span. Since there are only identity 2-morphisms in $\mathcal{C}$, their image under the inclusion is obvious.

This inclusion has a number of nice properties. First off, it’s always faithful. Clearly two such spans are the same only if the original arrows were the same. But we can even go further to assert that if two spans in the image of the inclusion are even related by a 2-morphism, they have to be from the same arrow in $\mathcal{C}$. Indeed, here’s the diagram:

The only possible arrow in the middle that makes the left triangle commute is the identity, and then the right triangle can only commute if $f=g$.

On the other hand, the inclusion is almost never full, even if we only ask for “essential” fullness. Indeed, if there’s any non-identity arrow then we could use it on the left of a span to make something that can’t be in the image. On the other hand, if we ask that every span have a 2-morphism from something in the image we find the diagram:

Here we can always pick $h$ to make the right side commute as long as we can find an arrow in the middle that makes the left side commute. But that would mean that every arrow $f$ in $\mathcal{C}$ has an arrow $x$ with $f\circ x=1_A$. That is, every single arrow in $\mathcal{C}$ would have to be a surjection, which is far too much to ask. And at the level of 2-morphisms there’s not nearly enough in the image to be full.

So the category $\mathcal{C}$ sits inside the 2-category $\mathbf{Span}(\mathcal{C})$ as a sub-2-category. This means (roughly) that if we get some structure on $\mathbf{Span}(\mathcal{C})$ there must be a corresponding structure on $\mathcal{C}$ itself by restriction. Then we can turn around and try to extend structures on $\mathcal{C}$ to the whole of $\mathbf{Span}(\mathcal{C})$. I’ve worked out some of the basic examples, which I’ll start in on tomorrow. But some of the interesting ones I’ll be working out as I write them up, so this should prove interesting indeed.

October 9, 2007 - Posted by | Category theory