# The Unapologetic Mathematician

## Categories with Zero Morphisms

Todd Trimble cleared up the confusion with pointed sets and zero morphisms. The problem is that my sources were wrong to say that the monoidal structure on $\mathbf{pSet}$ is the categorical product. First, here’s the answer in Todd’s own words:

The zero morphism statement is fine as stated. It just
follows from the fact that the standard monoidal product
on pointed sets is smash product.

FWIW, here are details. Recall that the smash product
of two pointed sets A, B (with basepoints 0_A, 0_B) is

A /\ B = A x B / ({0_A} x B) \/ (A x {0_B})

where the indicated quotient means the denominator is
identified with a point (and we define the basepoint of
A /\ B to be this point).

If C is a category enriched in pointed sets, and a and b
are objects of C, we define the zero morphism 0_ab
from a to be to be to be the basepoint of hom(a, b).
Notice that composition

hom(b, c) /\ hom(a, b) –> hom(a, c)

takes a pair (g, 0_ab) [similarly, a pair (0_bc, f)] to 0_ac,
by definition of smash product. Hence, composition of
a zero morphism with another morphism, on either side,
yields a zero morphism.

Truth be told I’d been thinking about these “smash products”, but they aren’t the regular categorical product I’d been assured worked.

Okay, let’s go over this a bit. Given pointed sets $(X,x_0)$ and $(Y,y_0)$ we define their smash product $(X\wedge Y,(xy)_0)$ as follows. First we take everything in $X$ other than $x_0$ and call it $\tilde{X}$. Similarly, we say $\tilde{Y}$ is everything in $Y$ except the point $y_0$. We take the product $\tilde{X}\times\tilde{Y}$ and then throw in the new point $(xy)_0$, which we use as the new special point.

Another way to look at it is to take the regular product $X\times Y$, but to “smash” it down a bit. We say that every pair of the form $(x_0,y)$ or $(x,y_0)$ is “really the same”, and smash all of them together into one special point.

The identity object for this monoidal structure on $\mathbf{pSet}$ is the set $(\{a,b\},a)$, as you should check. Also verify that the smash product is associative (up to isomorphism, naturally).

Now a $\mathbf{pSet}$-category $\mathcal{C}$ has a special morphism in each hom-set, which we’ll (leadingly) call $0:A\rightarrow B$. If we take any other arrow $f:B\rightarrow C$, together they form the pair $(0,f)\in\hom_\mathcal{C}(A,B)\wedge\hom_\mathcal{C}(B,C)$. But since $0\in\hom_\mathcal{C}(A,B)$ is the special point of that set, this pair (and any other pair of the form $(0,g)$ or $(g,0)$ is the marked point of the smash product. Then the composition function has to send this special point to the special point $0\in\hom_\mathcal{C}(A,C)$. Voila: zero morphisms. Compose one with anything and you get another zero morphism back.

August 21, 2007 Posted by | Category theory | 1 Comment

## The Underlying Category

In the setup for an enriched category, we have a locally-small monoidal category, which we equip with an “underlying set” functor $V(C)=\hom_{\mathcal{V}_0}(\mathbf{1},C)$. This lets us turn a hom-object into a hom-set, and now we want to extend this “underlying” theme to the entire 2-category $\mathcal{V}\mathbf{-Cat}$.

Okay, we could start by finding “underlying” analogues for each piece of the whole structure, but there’s a better way. We just take the setup of the “underlying set” from our monoidal categories and port it over to our 2-categories of enriched categories.

In particular, there’s a $\mathcal{V}$-category $\mathcal{I}$ that has a single object $I$ and $\hom_\mathcal{I}(I,I)=\mathbf{1}$. This behaves sort of like a “unit $\mathcal{V}$-category”, and we define $(\underline{\hphantom{X}})_0:\hom_{\mathcal{V}\mathbf{-Cat}}(\mathcal{I},\underline{\hphantom{X}})$. This is a 2-functor from $\mathcal{V}\mathbf{-Cat}$ to $\mathbf{Cat}$, and it assigns to an enriched category the “underlying” ordinary category. Let’s look at this a bit more closely.

A $\mathcal{V}$-functor $F:\mathcal{I}\rightarrow\mathcal{C}$ picks out an object $F(I)\in\mathcal{C}$, while a $\mathcal{V}$-natural transformation $\eta:F\rightarrow G$ consists of the single component $\eta_I:\mathbf{1}\rightarrow\hom_\mathcal{C}(F(I),G(I))$ — an element of $V(\hom_\mathcal{C}(F(I),G(I)))$. Thus the underlying category $\mathcal{C}_0$ has the same objects as $\mathcal{C}$, while $\hom_{\mathcal{C}_0}(A,B)$ is the “underlying set” of $\hom_\mathcal{C}(A,B)$.

Given a $\mathcal{V}$-functor $T:\mathcal{C}\rightarrow\mathcal{D}$ we get a regular functor $T_0:\mathcal{C}_0\rightarrow\mathcal{D}_0$. It sends the object $F:\mathcal{I}\rightarrow\mathcal{C}$ of $\mathcal{C}_0$ to the object $T\circ F:\mathcal{I}\rightarrow\mathcal{D}$ of $\mathcal{D}_0$. Its action on arrows of $\mathcal{C}_0$ (natural transformations of functors from $\mathcal{I}$ to $\mathcal{C}$ shouldn’t be too hard to work out.

Given a $\mathcal{V}$-natural transformation $\eta:S\rightarrow T$ of $\mathcal{V}$-functors we get a natural transformation $\eta_0:S_0\rightarrow T_0$. Its component $\eta_{0A}:S(A)\rightarrow T(A)$ in $\mathcal{D}_0$ is an element of an “underlying hom-set” — an arrow from $\mathbf{1}$ to the appropriate hom-object. But this is just the same as the component $\eta_A$ of the $\mathcal{V}$-natural transformation we started with, so we don’t really need to distinguish them.

At this point, some of these conditions tend to diverge. The ordinary naturality condition for a transformation between functors acting on the underlying categories turns out to be weaker than the $\mathcal{V}$-naturality condition for a transformation between $\mathcal{V}$-functors, for example. In general if I start talking about $\mathcal{V}$-categories then everything associated to them will be similarly enriched. If I mean a regular functor between the underlying categories I’ll try to say so. That is, once I lay out $\mathcal{V}$-categories $\mathcal{C}$ and $\mathcal{D}$, then if I talk about a functor $F:\mathcal{C}\rightarrow\mathcal{D}$ I automatically mean a $\mathcal{V}$-functor. If I mean to talk about a regular functor $F:\mathcal{C}_0\rightarrow\mathcal{D}_0$ I’ll say as much. Similarly, if I assert a natural transformation $\eta:S\rightarrow T$ I must mean $\mathcal{V}$-natural, or I would have said $\eta:S_0\rightarrow T_0$.

## 2-functors

Of course along with 2-categories, we must have 2-functors to map from one to another.

So, what’s a 2-functor? Since we defined a 2-category as a category enriched over $\mathbf{Cat}$, a 2-functor should be a functor enriched over $\mathbf{Cat}$. That is, it consists of a function on objects and a functor for each hom-category, each of which consists of a function on 1-morphisms (the objects of the hom-category) and a function for each set of 2-morphisms. Then there are a bunch of relations.

Let’s expand this a bit. A 2-category $\mathcal{C}$ has a collection $\mathrm{Ob}(\mathcal{C})$ of objects, a collection $\mathrm{Ob}(\hom_\mathcal{C}(A,B))$ of 1-morphisms for each pair $(A,B)$ of objects, and a collection $\hom_{\hom_\mathcal{C}(A,B)}(f,g)$ of 2-morphisms for each pair $(f,g)$ of 1-morphisms between the same pair of objects. And all the same remarks go for another 2-category $\mathcal{D}$.

So a 2-functor $F$ has

• a function $F:\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{D})$
• for each pair $(A,B)$ of objects of $\mathcal{C}$, a functor $F:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{D}(F(A),F(B))$, each consisting of
• a function $F:\mathrm{Ob}(\hom_\mathcal{C}(A,B))\rightarrow\mathrm{Ob}(\hom_\mathcal{D}(F(A),F(B)))$
• for each pair $(f,g)$ of 1-morphisms from $A$ to $B$ a function $F:\hom_{\hom_\mathcal{C}(A,B)}(f,g)\rightarrow\hom_{\hom_\mathcal{D}(F(A),F(B))}(F(f),F(g))$

Now the composition functors $\circ$ give us functions for composing 1-morphisms and “horizontally” composing 2-morphisms, and the hom-categories give us functions $\cdot$ for “vertically” composing 2-morphisms. For each object we have an identity 1-morphism, and for each 1-morphism we have an identity 2-morphism. A 2-functor will preserve all these structures. First of all, since there are functors between the hom-categories the vertical composition is preserved, along with the identity 2-morphisms. The diagrams for enriched functors say that the identity 1-morphisms, the composition of 1-morphisms, and the horizontal composition of 2-morphisms are all preserved.

August 18, 2007 Posted by | Category theory | 1 Comment

## The 2-category of Enriched Categories

So we know that $\mathbf{Cat}$ — the collection of all categories — forms a 2-category with functors as 1-morphisms and with natural transformations as 2-morphisms. It shouldn’t surprise us, then, that the collection $\mathcal{V}\mathbf{-Cat}$ of all categories enriched over a monoidal category $\mathcal{V}$ also comprises a 2-category.

We need two concepts to make this go through: an “enriched” notion of a functor, and an “enriched” notion of a natural transformation. As we might expect, both of them will just be written out like the familiar notions of functor and natural transformation, but substituting the new monoidal structure for the cartesian monoidal structure of $\mathbf{Set}$.

First of all, a functor from category $\mathcal{C}$ to $\mathcal{D}$ is a function $F:\mathrm{Ob}(\mathcal{C})\rightarrow\mathrm{Ob}(\mathcal{D})$, along with functions $F:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{D}(F(A),F(B))$ for each hom-set. So to define a $\mathcal{V}$-functor we keep the function on objects and replace the functions between hom-sets with morphisms between hom-objects. Of course, these must preserve compositions and identities, as encoded in the following diagrams:

which by now should look very familiar.

A natural transformation $\eta:F\rightarrow G$ between two functors from $\mathcal{C}$ to $\mathcal{D}$ picks out a morphism $\eta_C:F(C)\rightarrow G(C)$ in $\mathcal{D}$ for each object $C$ in $\mathcal{C}$, subject to a “naturality” condition. To find an analogue of picking out a morphism from a hom-set we use the same trick we did for the identity: we pick a morphism from $\mathbf{1}$ to a hom-object. That is, a $\mathcal{V}$-natural transformation consists of an $\mathrm{Ob}(\mathcal{C})$-indexed family of arrows $\eta_C:\mathbf{1}\rightarrow\hom_\mathcal{D}(F(C),G(C))$, which make the following diagram commute:

You should try to write this diagram out in the case of $\mathbf{Set}$ to verify that it becomes the familiar naturality square in that context.

Now the exact same constructions we used to compose natural transformations “vertically” and “horizontally” apply to $\mathcal{V}$-natural transformations, and the same arguments we used in the case of $\mathbf{Cat}$ apply to give a 2-category $\mathcal{V}\mathbf{-Cat}$ of categories, functors, and natural transformations, all enriched over the monoidal category $\mathcal{V}$.

August 17, 2007 Posted by | Category theory | 3 Comments

## 2-Categories

Here’s another example of an enriched category. This one is extremely important, and to a certain extent it’s been my goal in my coverage of category theory. It’s also a very basic chunk of what they talk about over at the n-Category Café.

The monoidal category we use is $\mathbf{Cat}$. If set-theoretical questions make you nervous, read this as “small categories”. It does work out for general categories, though. We know that $\mathbf{Cat}$ is cartesian, and thus monoidal. We can take pairwise products of categories, and the terminal category is $\mathbf{1}$ — the category with one object and one (identity) morphism. The “underlying” functor gives the collection of objects in a given category.

Okay, so what’s a category $\mathcal{C}$ enriched over $\mathbf{Cat}$? We’ve got a collection of objects, and for each pair of objects $(A,B)$ in $\mathcal{C}$ we have a category $\hom_\mathcal{C}(A,B)$ of morphisms. In each of these we have a collection of objects (called “1-morphisms”), and for each pair $(f,g)$ of 1-morphisms in $\hom_\mathcal{C}(A,B)$ we have a collection of morphisms $\hom_{\hom_\mathcal{C}(A,B)}(f,g)$ (called “2-morphisms”).

Wow, that looks confusing. Okay, let’s say it again a little differently. We have:

• a collection of objects (“0-morphisms”)
• collections of 1-morphisms that go from one object to another
• collections of 2-morphisms that go from one 1-morphism between a pair of objects to another 1-morphism between the same pair of objects

There’s also a “composition” functor between the categories of 1-morphisms. This takes a 1-morphism from $A$ to $B$ and one from $B$ to $C$ and gives a composite 1-morphism from $A$ to $C$. Since it’s a functor, it also acts on 2-morphisms. If $\phi:f\rightarrow g$ is a 2-morphism in $\hom_\mathcal{C}(A,B)$ (that is, both $f$ and $g$ go from $A$ to $B$) and $\xi:h\rightarrow k$ is a 2-morphism in $\hom_\mathcal{C}(B,C)$, then we get a composite 2-morphism $\xi\circ\phi:h\circ f\rightarrow k\circ g$. Of course, we also can take $\phi:f\rightarrow g$ and $\xi:g\rightarrow h$ and get a composite 2-morphism $\xi\cdot\phi:f\rightarrow h$ by using the composition in the category $\hom\mathcal{C}(A,B)$. The composition functor $\circ$ is associative.

For each object $C$ there’s an identity 1-morphism $1_C\in\hom_\mathcal{C}(C,C)$. And then it has an identity 2-morphism $1_{1_C}:1_C\rightarrow1_C$. The 1-morphism $1_C$ acts as the identity for the composition functor $\circ$, and it’s easy enough to verify that $1_{1_C}$ is not only the identity for the composition $\cdot$ in $\hom_\mathcal{C}(C,C)$, but it’s also the identity for the composition $\circ$ of 2-morphisms.

We call this structure a “2-category”, or more specifically a “strict 2-category”. We’ll get to weak ones eventually.

So do we know any good examples? Sure. The first is $\mathbf{Cat}$ itself! Here the objects are categories, the 1-morphisms are functors between categories, and the 2-morphisms are natural transformations between functors. In fact we already saw right when we defined a natural transformation that given a pair of categories we have a category of functors between them, which is halfway to having a 2-category right there! And then we know we have both compositions of 2-morphisms because those are just the “horizonatal” and “vertical” compositions we first needed when we talked about units and counits of adjunctions.

Speaking of adjunctions, they give another 2-category: $\mathbf{Adj}$. The objects here again are categories, but now the 1-morphisms are adjunctions between categories. And then we have conjugate pairs between adjunctions, with the “horizontal” and “vertical” compositions between them as our 2-morphisms.

And as a last example, what’s a 2-category $\mathcal{M}$ with one object? Well, we have our object $\mathbf{*}$, and a category $\hom_\mathcal{M}(\mathbf{*},\mathbf{*})$. Any two 1-morphisms (the objects of this category) can be composed with each other by $\circ$, and there’s an identity 1-morphism. Now let’s just shift our language and say “object” instead of “1-morphism”, “morphism” instead of “2-morphism”, $\otimes$ instead of $\circ$, and $\circ$ instead of $\cdot$. What we’re left with is exactly the definition of a strict monoidal category! That is: just as a category with one object is a monoid, so a 2-category with one object is a monoidal category!

There are a lot of 2-categories out there, and we’ll be mentioning many more as the time goes on.

August 16, 2007 Posted by | Category theory | 12 Comments

## 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

## Enriched Categories II

So we have the basic data of a category $\mathcal{C}$ enriched over a monoidal category $\mathcal{V}$. Of course, what I left out were the relations that have to hold. And they’re just the same as those from categories, but now written in terms of $\mathcal{V}$ instead of $\mathbf{Set}$: associativity and identity relations, as encoded in the following commutative diagrams:

Notice how these are very similar to the axioms for a monoidal category or a monoid object. And this shouldn’t be unexpected by now, since we know that a monoid is just a (small) category with only one object. In fact, if we only have one object in a $\mathcal{V}$-enriched category we get back exactly a monoid object in $\mathcal{V}$!

Now, often we’re thinking of our hom-objects as “hom-sets with additional structure”. There should be a nice way to forget that extra structure and recover just a regular category again. To an extent this is true, but for some monoidal categories $\mathcal{V}$ the “underlying set” functor isn’t really an underlying set at all. For now, though, let’s look at a familiar category of “sets with extra structure” and see how we get the underlying set out of the category itself.

Again, the good example to always refer back to for enriched categories is $\mathbf{Ab}$, the category of abelian groups with tensor product as the monoidal structure. We recall that the functor giving the free abelian group on a set is left adjoint to the forgetful functor from abelian groups to sets. That is, $\hom_\mathbf{Ab}(F(S),A)\cong\hom_\mathbf{Set}(S,U(A))$. We also know that we can consider an element of the underlying set $U(A)$ of an abelian group as a function from a one-point set into $U(A)$. That is, $\hom_\mathbf{Set}(\{*\},U(A))\cong U(A)$. Putting these together, we see that $U(A)\cong\hom_\mathbf{Ab}(\mathbb{Z},A)$, since $\mathbb{Z}$ is the free abelian group on one generator.

But $\mathbb{Z}$ is also the identity object for the tensor product! The same sort of argument goes through for all our usual sets-with-structure, telling us that in all these cases the “underlying set” functor is represented by the monoidal identity $\mathbf{1}$, which is the free object on one generator. We take this as our general rule, giving the representable functor $V(\underline{\hphantom{X}})=\hom_{\mathcal{V}_0}(\mathbf{1},\underline{\hphantom{X}}):\mathcal{V}_0\rightarrow\mathbf{Set}$. In many cases (but not all!) this is the usual “underlying set” functor, but now we’ve written it entirely in terms of the monoidal category $\mathcal{V}$!

As time goes by, we’ll use this construction to recover the “underlying category” of an enriched category. The basic idea should be apparent, but before we can really write it down properly we need to enrich the notions of functors and natural transformations.

## Enriched Categories

I’d like to move on now to another way of blending various structures. We’ve seen that in certain situations the set of morphisms between two objects in a category naturally has deeper structure itself. For example, the set of homomorphisms between two abelian groups is itself an abelian group, because abelian groups are modules over the commutative ring $\mathbb{Z}$. More generally, the set of homomorphisms between two $R$-modules naturally has the structure of a $\mathbb{Z}$-module, and sometimes more.

We need a good way of talking about this sort of thing, where we replace hom sets by “hom objects” in some other category $\mathcal{V}$. When this happens we say that our category is “enriched” over $\mathcal{V}$. So to rephrase what I said above, the category of $R$-modules is enriched over $\mathbf{Ab}$. Similarly, locally small categories are enriched over $\mathbf{Set}$.

When we talk about categories — which usually for us means locally small categories — we are implicitly using a number of properties of $\mathbf{Set}$. In particular, to set up compositions we need to be able to take pairs of morphisms, which the cartesian product handles for us nicely: $\hom_\mathcal{C}(B,C)\times\hom_\mathcal{C}(A,B)$. We also need to be able to pick out a special morphism in each set of endomorphisms to be the identity, which we can take to be the image of a function from a one-point set to the set of endomorphisms $\hom_\mathcal{C}(C,C)$ sort of like we did for monoid objects.

For setting up the relations a category must satisfy we need to be able to build triples from pairs in two ways:
$(\hom_\mathcal{C}(C,D)\times\hom_\mathcal{C}(B,C))\times\hom_\mathcal{C}(A,B)\cong$
$\hom_\mathcal{C}(C,D)\times(\hom_\mathcal{C}(B,C)\times\hom_\mathcal{C}(A,B))$
We also need to pair a morphism with a (unique) identity morphism:
$\hom_\mathcal{C}(A,B)\cong\hom_\mathcal{C}(A,B)\times\{*\}\rightarrow\hom_\mathcal{C}(A,B)\times\hom_\mathcal{C}(A,A)$

What are the important properties of the category of sets that make it useful for these purposes? It’s just the fact that $\mathbf{Set}$ equipped with finite products (including a singleton set as terminal object) is a monoidal category! So let’s take a monoidal category $\mathcal{V}$ — a useful example to have always at hand is $\mathbf{Ab}$ — and try to use it for enrichment. As we proceed, we’ll write $\mathcal{V}_0$ for the underlying regular category (that is, forget that $\mathcal{V}$ is monoidal).

So, given such a monoidal category $\mathcal{V}$ we’ll define a $\mathcal{V}$-category $\mathcal{C}$ to consist of a class of objects $\mathrm{Ob}(\mathcal{C})$, and for each pair $(A,B)$ of objects a “hom-object” $\hom_\mathcal{C}(A,B)\in\mathcal{V}_0$. For each triple of objects $(A,B,C)$ there is a composition $\circ:\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$. For each object $A$ there is an “identity”, described by an arrow $i:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,A)$.

I’ll be spending some time on this, so let’s leave it at the definition for now. Go through and unpack it for the case of an $\mathbf{Ab}$-category, and see what the definition says such a thing should look like.

August 13, 2007 Posted by | Category theory | 11 Comments

## Correction

Todd Trimble pointed out to me a mistake I made in saying that internalizations commute. I glossed over a number of important subtleties and he set me straight.

Now I’m just getting back from a visit to Bourbon Street, since the real intarwobs and television won’t be reaching my apartment until Wednesday morning. I’ve skimmed his points and they check out, but I invite him to post them here as a comment so I don’t mangle them. Basically, I think I’m “morally” right (and I’m sure I’ve read John Baez saying somthing along these lines), but I’ve messed something up in my presentation as I attempt to get back in the saddle. I’d be glad of any clarification that can be made to see where I’ve gone Pete Tong.

These last two posts are sort of the tail end of internalization anyhow, and I was planning on moving on to something new on Monday. If they’re a little messy, so be it.

August 12, 2007 Posted by | Category theory | 1 Comment

## Internal Categories

Just like we have monid objects, we can actually define something we could sensibly call a “category object”. In this case, however, it will be a little more accurate to use the term “internal category”.

This is because a (small) category isn’t just a set with extra structure. It’s two sets with extra structure. We have a set $O$ of objects, a set $M$ of morphisms, a function $i:O\rightarrow M$ assigning the identity morphism to each object, functions $s:M\rightarrow O$ and $t:M\rightarrow O$ assigning the source and target objects to each morphism, and an arrow $\gamma:M{}_s\times_tM\rightarrow M$ telling us how to compose certain pairs of morphisms. This involves a “fibered product”, which is just the pullback in $\mathbf{Set}$. We take the arrows $s$ and $t$ from $M$ to $O$ and pull back the square to get the set of all pairs of morphisms so that the source object of one is the target of the other.

Then there are a bunch of relations which hold:

• The source of the identity arrow on an object is the object itself.
• The target of the identity arrow on an object is the object itself.
• The identity arrow on an object acts as a left and right identity for the composition.
• The source of a composition is the source of the second member of the pair.
• The target of a composition is the target of the first member of the pair.
• The composition is associative.

I’ll leave you to write these out purely in terms of the functions $m$, $i$, $s$, and $t$.

Now we can take this whole setup and drop it into any other category, as long as that category has pairwise pullbacks. If $\mathcal{C}$ does have these pullbacks, then a category internal to $\mathcal{C}$ (or a “category object”) consists of a pair of objects and four morphisms of $\mathcal{C}$, which must satisfy the above relations. Then a category internal to $\mathbf{Set}$ is a small category.

When we’re talking about categorifying something like a group, we want to replace the underlying set of a group with a small category. That is, we want to have a group object in $\mathbf{Cal}$. But we know that internalizations commute, so this is the same thing as a “category object” in groups! That is, instead of looking for a category with a multiplication functor and so on, we can instead look for a pair of groups with source, target, composition, and identity homomorphisms between them.