# The Unapologetic Mathematician

## An example of an enriched category

Sorry for the delay, but the cable setup took more than I’d expected (there will me more on this over at Yankee Freak-Out). Today, I’d like to run through an example of a monoidal category, and what sort of enriched categories it gives rise to.

The category I’m interested in is the ordinal $\mathbf{2}$. Remember that this consists of the objects ${0}$ and $1$, with one non-identity arrow $0\rightarrow1$. We can make this into a monoidal category by saying $a\otimes b=ab$. Then $1$ is the monoidal identity object.

So what is a category enriched over $\mathbf{2}$? Well, first it has a collection of objects. For each pair $(A,B)$ of objects we either have the hom-object $\hom_\mathcal{C}(A,B)=0$ or $\hom_\mathcal{C}(A,B)=1$.

To have “identity morphisms” means we need an arrow $1\rightarrow\hom_\mathcal{C}(C,C)$ for each object $C$. But the only such arrow in $\mathbf{2}$ is $1\rightarrow1$, so $\hom_\mathcal{C}(C,C)=1$. For composition, we need arrows $\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$. Thus if $\hom_\mathcal{C}(A,B)$ and $\hom_\mathcal{C}(B,C)$ are both $1$, then so must be $\hom_\mathcal{C}(A,C)$.

Now we can see that this is just a different way of talking about a preorder. The identity morphism corresponds to the reflexive axiom, and the composition morphism corresponds to the transitive action. In short: $A\preceq B$ if and only if $\hom_\mathcal{C}(A,B)=1$.

Another example I’ve seen bandied about uses the category $\mathbf{pSet}$ of “pointed sets”. This is just a set with an identified “point”. For example, $(\{1,2,3\},1)$ is a pointed set, and $(\{1,2,3\},2)$ is a different pointed set. The morphisms are “pointed functions”, which have to preserve the point — a morphism $f:(X,x_0)\rightarrow(Y,y_0)$ consists of a function $f:X\rightarrow Y$ with $f(x_0)=y_0$. This category has finite products, so it’s monoidal.

The usual statement is that categories enriched over $\mathbf{pSet}$ are the same as categories with “zero morphisms”. These are like categories with zero objects, but without needing an object to factor things through. Every hom-set has a special “zero” morphism, and the composition of a zero morphism with any other morphism is another zero morphism. The problem is, whenever I try to show this it doesn’t seem to work out. I think that there’s something askew or oversimplified with the statement somewhere.

Your mission, should you choose to accept it, is to figure out what the right statement is, and to prove it. Let me know by email (if you can’t find my email you aren’t trying very hard) and I’ll post it up for all to see, and for your own greater glory.

August 15, 2007 Posted by | Category theory | 6 Comments

## Peer Pressure

Psst…

Have you signed the petition yet?

All the cool mathematicians are doing it.