# The Unapologetic Mathematician

## An awful lot of natural maps

There are a bunch of natural (in both senses) maps we can consider now. Some look all but tautological over $\mathbf{Set}$, and we may have used them in the past without comment. In the enriched context, though, we should go over them.

1. The family of arrows $\eta_C:\mathbf{1}\rightarrow\hom_\mathcal{D}(S(C),T(C))$ is extraordinarily $\mathcal{V}$-natural exactly when $\eta_C:S(C)\rightarrow T(C)$ is $\mathcal{V}$-natural.
2. For a $\mathcal{V}$-functor $T:\mathcal{C}\rightarrow\mathcal{D}$, the map $T_{A,B}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{D}(T(A),T(B))$ is natural in both $A$ and $B$.
3. Similarly, if $T:\mathcal{A}\otimes\mathcal{B}\rightarrow\mathcal{C}$ is a $\mathcal{V}$-functor, then $T(\underline{\hphantom{X}},1_B):\hom_\mathcal{A}(A,A')\rightarrow\hom_\mathcal{C}(T(A,B),T(A',B))$ is $\mathcal{V}$-natural. And it’s also natural in $B)$.
4. In particular, $\hom_\mathcal{C}(C,\underline{\hphantom{X}})_{A,B}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$ is natural in all three variables.
5. Putting together naturalities 1 and 4 tells us that for a morphism $f:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$, the transformation $\hom_\mathcal{C}(1_C,f):\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$ is natural.
6. The “evaluation” $e:Z^Y\otimes Y\rightarrow Z$ is natural in both variables.
7. Since we built compositions from evaluations and the closure adjunction, the arrow $\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$ is natural in all variables.
8. The identity functor on $\mathcal{C}$ has an identity natural transformation, so by naturality 1 we see that $i_C:\mathbf{1}\rightarrow\hom_\mathcal{C}(C,C)$ is natural.
9. All the monoidal structural isomorphisms — $\alpha_{A,B,C}:(A\otimes B)\otimes C)\rightarrow A\otimes(B\otimes C)$, $\lambda_A:\mathbf{1}\otimes A\rightarrow A$, $\rho:A\otimes\mathbf{1}\rightarrow A$, and $\gamma_{A,B}:A\otimes B\rightarrow B\otimes A$ — are natural in all variables.
10. We can start with $(Z^Y)^X$ and hit it with $\underline{\hphantom{X}}\otimes Y$ to get $(Z^Y\otimes Y)^{X\otimes Y}$. Then we can evaluate to get $Z^{X\otimes Y}$. This is an isomorphism we call $p_{X,Y,Z}^{-1}$, corresponding to the adjunction $\hom_\mathcal{C}(A\hom_\mathcal{C}(B,C))\cong\hom_\mathcal{C}(A\otimes B,C)$, and it’s natural in all variables.
11. We can compose the following arrows:
• $\lambda_X^{-1}:X\rightarrow\mathbf{1}\otimes X$
• $j_{X\otimes Y}\otimes1_X:\mathbf{1}\otimes X\rightarrow(X\otimes Y)^{X\otimes Y}\otimes X$
• $p_{X,Y,X\otimes Y}\otimes1_X:(X\otimes Y)^{X\otimes Y}\otimes X\rightarrow((X\otimes Y)^Y)^X\otimes X$
• $e_{X,(X\otimes Y)^Y}:((X\otimes Y)^Y)^X\otimes X\rightarrow(X\otimes Y)^Y$

and we get the “coevaluation” $d_{X,Y}$ — the counit of the closure adjunction. And thus the coevaluation is $\mathcal{V}$-natural in all variables.

12. We can compose the coevaluation $d_{X,\mathbf{1}}:X\rightarrow(X\otimes\mathbf{1})^\mathbf{1}$, and then use the right unit isomorphism to get a $\mathcal{V}$-natural isomorphism $X\rightarrow X^\mathbf{1}$.
13. In general, a family $f:T(D,D,A,B)\otimes S(E,E,A,C)\rightarrow R(F,F,B,C)$ is $\mathcal{V}$-natural in any of its variables if the corresponding variables are natural in $\overline{f}:T(D,D,A,B)\rightarrow R(F,F,B,C)^{S(E,E,A,C)}$

Whew. That’s a mouthful. It can be instructive to sit down and try to interpret some of these in the context of categories enriched over $\mathbf{Set}$, so when you recover I’d advise taking a look at that.

## Extraordinary Naturality

Now that we’ve gone back and rewritten the definition of naturality, let’s push it a bit.

First, notice that if we’re enriching over $\mathbf{Set}$ (in “ordinary” categories) then $\hom_\mathcal{D}(1_{S(A)},\eta_{B})$ means “take a morphism from $S(A)$ to $S(B)$ and follow it with $\eta_{B}$“. On the other hand, $\hom_\mathcal{D}(\eta_A,1_{T(B)})$ means “first do $\eta_A$, then follow it with a morphism from $T(A)$ to $T(B)$“. This recipe gives us back exactly the old naturality square, so $\mathbf{Set}$-natural transformations are exactly the ordinary natural transformations we’re familiar with!

So let’s take this reformulation of the naturality condition and tweak it. Instead of considering a family of arrows (in $\mathcal{D}_0$) $\eta_C:S(C)\rightarrow T(C)$, let’s move the variable over from the left to the right and consider a family $\beta_C:K\rightarrow T(C,C)$. Here, $K$ is an object of $\mathcal{D}$, and $T:\mathcal{C}^\mathrm{op}\otimes\mathcal{C}\rightarrow\mathcal{D}$ is a bifunctor. Now we say that the $\beta_C$ are the components of an “extraordinary $\mathcal{V}$-natural transformation” if the following diagram commutes:

This looks bizarre at first, though clearly it’s related to our revision of the enriched naturality diagram. It turns out that we’ve seen this sort of naturality before, though. If we read the diagram in $\mathbf{Set}$, consider a monoidal category $\mathcal{M}$ with duals, and use the functor $T(A,B)=A^*\otimes B$, then this is exactly the sort of naturality we find in the duality arrows $\eta_M:\mathbf{1}\rightarrow M^*\otimes M$!

Dually, we can define extraordinary $\mathcal{V}$-naturality for a family of morphisms $\gamma_C:T(C,C)\rightarrow K$. Write out this diagram, and show that the duality arrows $\eta_M:M\otimes M^*\rightarrow\mathbf{1}$ provide an example.

As another exercise, take these extraordinary naturality diagrams and work out the interpretation in $\mathbf{Set}$ explicitly. That is, actually start with some morphism $f:A\rightarrow B$ in the upper left-hand corner, and evaluate it all around. When we did this for our new $\mathcal{V}$-naturality diagram above we got our old naturality squares back. What do we get for extraordinary $\mathbf{Set}$-naturality?

August 30, 2007 Posted by | Category theory | 1 Comment

## Who’s Working For Journals and Against Mathematicians?

Over at The n-Category Café, they’ve unmasked him as none other than Dr. Evil!

August 29, 2007 Posted by | Uncategorized | 1 Comment

## Enriched Naturality Revisited

Let’s look back at the enriched versions of representable functors. If we fix an object $C\in\mathcal{C}$ we have a $\mathcal{V}$-natural transformation $\hom_\mathcal{C}(C,\underline{\hphantom{X}})_{B,C}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$. This corresponds under the closure adjunction to $\circ:\hom_\mathcal{C}(A,B)\otimes\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$. There’s a similar transformation for composition on the other side.

Now remember that the closest thing we have to a “morphism” in an enriched category is an element of the underlying set of a hom-object. That is, we can talk about an arrow $f:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$. We often abuse the language and say that this is a morphism from $A$ to $B$, which in fact it’s a morphism in the underlying category.

Now, even though this $f$ isn’t really a morphism in our enriched category, we can still come up with a morphism sensibly called $\hom_\mathcal{C}(C,f):\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$. Here’s how it goes:

• We start with $\hom_\mathcal{C}(C,A)$ and use the left unit isomotphism to move to $\mathbf{1}\otimes\hom_\mathcal{C}(C,A)$.
• We now hit $\mathbf{1}$ with our morphism $f$ to land in $\hom_\mathcal{C}(A,B)\otimes\hom_\mathcal{C}(C,A)$.
• Finally, we compose to end up in $\hom_\mathcal{C}(C,B)$.

We can similarly take $g:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$ and construct a morphism $\hom_\mathcal{C}(g,C):\hom_\mathcal{C}(B,C)\rightarrow\hom_\mathcal{C}(A,C)$.

These composites should look familiar from the definition of enriched naturality for a transformation $\eta:S\rightarrow T$. In fact, we have a more compact diagram to replace that big hexagon:

Notice here that the right and bottom arrows in this square expand out to become the top and bottom of a hexagon, and we slip the functors $S$ and $T$ into place.

August 29, 2007 Posted by | Category theory | 2 Comments

## The Internal Monoidal Product

As we’re talking about enriched categories, we’re always coming back to the monoidal category $\mathcal{V}$. This has an underlying category $\mathcal{V}_0$, which is then equipped with a monoidal product — an ordinary functor $\otimes:\mathcal{V}_0\times\mathcal{V}_0\rightarrow\mathcal{V}_0$. But as usual we don’t want to work with ordinary categories and functors unless we have to.

Luckily, we can turn this monoidal product into a $\mathcal{V}$-functor between $\mathcal{V}$-categories: $\mathrm{Ten}:\mathcal{V}\otimes\mathcal{V}\rightarrow\mathcal{V}$. Here, $\mathrm{Ten}$ refers to “tensor product”. On objects we do the same thing as before — $\mathrm{Ten}(X,Y)=X\otimes Y$ — because the objects of the $\mathcal{V}$-category $\mathcal{V}$ are the same as those of the ordinary category $\mathcal{V}_0$. But now we have to consider how this functor should act on the hom-objects. So we recall that we define the internal hom-functor as $\hom_\mathcal{V}(X,Y)=Y^X$, using the closed structure on $\mathcal{V}$.

So to have a $\mathcal{V}$-functor we need morphisms $\mathrm{Ten}:\hom_{\mathcal{V}\otimes\mathcal{V}}((X_1,Y_1),(X_2,Y_2))\rightarrow\hom_\mathcal{V}(X_1\otimes Y_1,X_2\otimes Y_2)$. On the left we defined the hom-object for the product $\mathcal{V}$-category as $\hom_\mathcal{V}(X_1,X_2)\otimes\hom_\mathcal{V}(Y_1,Y_2)$, which is then defined as $X_2^{X_1}\otimes Y_2^{Y_1}$. On the right we have the exponential $(X_2\otimes Y_2)^{X_1\otimes Y_1}$.

But by the closure adjunction an arrow $X_2^{X_1}\otimes Y_2^{Y_1}\rightarrow(X_2\otimes Y_2)^{X_1\otimes Y_1}$ is equivalent to an arrow $(X_2^{X_1}\otimes Y_2^{Y_1})\otimes(X_1\otimes Y_1)\rightarrow(X_2\otimes Y_2)$. Now we can just swap around the factors on the left to get $(X_2^{X_1}\otimes X_1)\otimes(Y_2^{Y_1}\otimes Y_1)$, and then inside each set of parentheses we can use the evaluation morphism we get from the closure adjunction, leaving us with $X_2\otimes Y_2$. Putting together the swap and the evaluations we get the arrow we want. And then the closure adjunction flips this to the morphism we needed to define the monoidal product on hom-objects.

The underlying ordinary functor $\mathrm{Ten}_0$ of the $\mathcal{V}$-functor $\mathrm{Ten}$ is the old monoidal product $\otimes$ again. On objects we already have the same action, so we need to check that the underlying function of the morphism $\mathrm{Ten}:X_2^{X_1}\otimes Y_2^{Y_1}\rightarrow(X_2\otimes Y_2)^{X_1\otimes Y_1}$ is the same as the function $\otimes:\hom_{\mathcal{V}_0}(X_1,X_2)\times\hom_{\mathcal{V}_0}(Y_1,Y_2)\rightarrow\hom_{\mathcal{V}_0}(X_1\otimes Y_1,X_2\otimes Y_2)$. We already know that the underlying set of $B^A$ is $\hom_{\mathcal{V}_0}(A,B)$ and the cartesian product of hom-sets underlies the monoidal product of hom-objects, so we at least know that the underlying source and target objects are correct.

So what’s the underlying function? We have the arrow $\mathrm{Ten}:X_2^{X_1}\otimes Y_2^{Y_1}\rightarrow(X_2\otimes Y_2)^{X_1\otimes Y_1}$ and we need to produce a function $\mathrm{Ten}_0:\hom_{\mathcal{V}_0}(\mathbf{1},X_2^{X_1})\times\hom_{\mathcal{V}_0}(\mathbf{1},Y_2^{Y_1})\rightarrow\hom_{\mathcal{V}_0}(\mathbf{1},(X_2\otimes Y_2)^{X_1\otimes Y_1})$. In each of these hom-sets we can use the closure adjunction to get a function $\hom_{\mathcal{V}_0}(X_1,X_2)\times\hom_{\mathcal{V}_0}(Y_1,Y_2)\rightarrow\hom_{\mathcal{V}_0}(X_1\otimes Y_1,X_2\otimes Y_2)$. But this is clearly function $(f,g)\mapsto f\otimes g$ for the ordinary tensor product.

In light of this tight relationship between $\mathrm{Ten}$ and $\otimes$, I’ll usually just write $\otimes$ for each. Again, when I don’t specify whether I’m talking about the ordinary or the enriched functor I’ll default to the enriched version.

## Enriched Categorical Constructions

We’re going to need to talk about enriched functors with more than one variable, so we’re going to need an enriched analogue of the product of two categories.

Remember that the product $\mathcal{C}\times\mathcal{D}$ of two categories has the product of the object-classes as its objects, and it has pairs of morphisms for its morphisms. That is, the hom-set $\hom_{\mathcal{C}\times\mathcal{D}}((C_1,D_1),(C_2,D_2))$ is the product $\hom_\mathcal{C}(C_1,C_2)\times\hom_\mathcal{D}(D_1,D_2)$. Of course, in the enriched setting we no longer have hom-sets to work with.

So we’ll keep the same definition for the objects of our product category, but we’ll replace the definition of the hom-objects:
$\hom_{\mathcal{C}\times\mathcal{D}}((C_1,D_1),(C_2,D_2))=\hom_\mathcal{C}(C_1,C_2)\otimes\hom_\mathcal{D}(D_1,D_2)$
Now we can use the associativity and commutativity of our monoidal category $\mathcal{V}$ (remember we’re assuming it’s symmetric now) to move around factors like this:
$(\hom_\mathcal{C}(C_2,C_3)\otimes\hom_\mathcal{D}(D_2,D_3))\otimes(\hom_\mathcal{C}(C_1,C_2)\otimes\hom_\mathcal{D}(D_1,D_2))\rightarrow$
$(\hom_\mathcal{C}(C_2,C_3)\otimes\hom_\mathcal{C}(C_1,C_2))\otimes(\hom_\mathcal{D}(D_2,D_3)\otimes\hom_\mathcal{D}(D_1,D_2))$
at which point we can use the composition in each category to give a composition of the original pairs.

To get an identity, we use $\mathbf{1}\cong\mathbf{1}\otimes\mathbf{1}$ and then hit the left copy of $\mathbf{1}$ with the identity morphism for the object $C\in\mathcal{C}$ and the right copy with the identity morphism for $D\in\mathcal{D}$.

What about the opposite category? Well, it works pretty much the same as before. We just define $\hom_{\mathcal{C}^\mathrm{op}}(A,B)=\hom_\mathcal{C}(B,A)$. For an identity, we just use the same $\mathbf{1}\rightarrow\hom_\mathcal{C}(C,C)=\hom_{\mathcal{C}^\mathrm{op}}(C,C)$ as before.

Actually, these same constructions apply to functors. If we have functors $F:\mathcal{C}\rightarrow\mathcal{C}'$ and $G:\mathcal{D}\rightarrow\mathcal{D}'$, we can assemble them into a functor $F\otimes G:\mathcal{C}\otimes\mathcal{D}\rightarrow\mathcal{C}'\otimes\mathcal{D}'$. Just define $F\otimes G(C\otimes D)=F(C)\otimes G(D)$, and use a similar definition for the morphisms. Also, given $F:\mathcal{C}\rightarrow\mathcal{D}$ we get a functor $F^\mathrm{op}:\mathcal{C}^\mathrm{op}\rightarrow\mathcal{D}^\mathrm{op}$.

Now we know we have a 2-category $\mathcal{V}\mathbf{-Cat}$ of categories enriched over $\mathcal{V}$. Since a 2-category is a category enriched over categories, we can pass to the underlying category $\mathcal{V}\mathbf{-Cat}_0$ of enriched categories. It turns out that all the foregoing discussion gives this category some nice, familiar structure.

The product of two enriched categories turns out to be weakly associative. Also, remember from our discussion of the underlying category that we have a $\mathcal{V}$-category $\mathcal{I}$. This behaves like a weak identity for the product. That is, when we equip $\mathcal{V}\mathbf{-Cat}_0$ with this product and identified object, it turns out to be a monoidal category! Even better, it’s symmetric$\mathcal{C}\otimes\mathcal{D}\cong\mathcal{D}\otimes\mathcal{C}$. And what is the opposite category but a duality on this category?

So now we can define contravariant enriched functors, as well as functors of more than one variable. As usual, you should go back and try to think of these definitions in terms of ordinary categories ($\mathbf{Set}$-categories) as well as $\mathbf{Ab}$-categories.

Incidentally, if you want to run ahead a bit, try working out how natural transformations fit into the picture. It turns out that the 2-category $\mathcal{V}\mathbf{-Cat}$ is an example of an even deeper structure I haven’t defined yet: it’s a symmetric monoidal 2-category with duals.

[UPDATE]: Excuse me.. I should have said that $\mathcal{V}\mathbf{-Cat}$ is a symmetric monoidal 2-category with a duality involution rather than “with duals”, and similarly for the underlying category. I blame my inattention on being stuck around the house all day waiting for repairmen to come by to put the dishwasher they left in my living room last Friday into the dishwasher-sized hole they put in my kitchen. Basically, the “duality involution” means that the opposite of the opposite $\mathcal{V}$-category is the original $\mathcal{V}$-category back again, and that the opposite of a tensor product is the tensor product of the opposites.

## Representable Enriched Functors

Okay, unfortunately after a lot of work I don’t ssee an elegant way to jump right into what I thought of the other day. So, we’ll get there, but slowly.

Anyhow, we remember that categories come with representable functors, which have all sorts of nice properties. This comes over to our current context. Given a monoidal category $\mathcal{V}$, a category $\mathcal{C}$ enriched over $\mathcal{V}$, and an object $C\in\mathcal{C}$, we have the representable $\mathcal{V}$-functor $\hom_\mathcal{C}(C,\underline{\hphantom{X}}):\mathcal{C}\rightarrow\mathcal{V}$.

We should be clear about how this functor behaves on “morphisms”. First of all, remember that we don’t really have morphisms in an enriched category because we have hom-objects rather than hom-sets. Instead, what we get is a $\mathcal{V}$-natural transformation with $(A,B)$ component $\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$. Instead of a function which assigns a function between hom-sets to each morphism, we get an arrow from the hom-object (replacing the set of morphisms) to an exponential (replacing the set of functions between hom-sets).

Of course we also have the contravariant $\mathcal{V}$-functor $\hom_\mathcal{C}(\underline{\hphantom{X}},C)$, and similar comments about how it behaves on morphisms apply here.

Honestly, the first time you read this it seems pretty simple, but it’s deeper than you expect. You’re going to want to keep thinking about “elements” of a hom-object, particularly when you’re using $\mathbf{Ab}$ as your sample enriching category, but you really have to start breaking yourself of the habit of thinking in terms of “elements” of an object — even if the objects are something so familiar as abelian groups — now.

August 25, 2007 Posted by | Category theory | 2 Comments

## Nothing today

I was getting orientated for most of today, and I think some of the grad students and I are going to hit the Vieux Carré before the undergrads start showing up en masse tomorrow, so I just don’t have it in me to post the next bit today.

But I’ve been thinking about it. I’ll also have to check up on some of my references to see if they can tell me how to do what I’m pretty sure can be done, or I’ll just have to cobble it together myself. But it’s pretty cool. I’ll give you one hint: look how I started my discussions of adjoint functors.

For now, go read the Carnival. Yes, I’ll give it a fair shot while the semester begins and see if it picks up as predicted.

## Internal Hom Functors

As Todd Trimble pointed out, things get really nice when a category is enriched over itself. That is, the morphisms from one object to another in $\mathcal{V}$ themselves have the structure of an object of $\mathcal{V}$. This trivially the case for $\mathbf{Set}$, because there’s a set of functions from one set to another. We also know that in $\mathbf{Ab}$ there’s an abelian group of homomorphisms from one abelian group to another. We say that the category has an “internal hom functor”, because the hom functor lands back inside the category itself, rather than in the category of sets.

For the moment, let’s consider a category $\mathcal{V}$ that is not only monoidal (which is needed to have an enriched category), but also symmetric and closed. Remember that “closed” means we have an adjunction $\underline{\hphantom{X}}\otimes B\dashv (\underline{\hphantom{X}})^B$ for each object $B$. In $\mathbf{Set}$ the set $A^B$ is the set of functions from $B$ to $A$, while in $\mathbf{Ab}$ it’s the abelian group of homomorphisms from $B$ to $A$. We see that these are already the internal hom functors we’re looking for in these situations.

So in general let’s take our symmetric, monoidal, closed category $\mathcal{V}$, with underlying ordinary category $\mathcal{V}_0$. The adjunction between the monoidal structure and the exponential has a counit — an arrow $A^B\otimes B\rightarrow A$ — which corresponds to “evaluation” in both of our sample cases. That is, it takes a function $f:B\rightarrow A$ and an element $b\in B$ and gives an element $f(b)\in A$. We can use this to build a category.

Start with the objects of $\mathcal{V}_0$, and define the hom-object from $B$ to $A$ as $A^B$ (using the exponential functor from the closed structure). We need to find arrows $A^B\otimes B^C\rightarrow A^C$ and $\mathbf{1}\rightarrow A^A$, and we’ll use the adjunction to do it. For composition, we have the arrow

$(A^B\otimes B^C)\otimes C\rightarrow A^B\otimes(B^C\otimes C)\rightarrow A^B\otimes B\rightarrow A$

where the first step is the associator and the other two are evaluations. This is an element of $\hom_{\mathcal{V}_0}((A^B\otimes B^C)\otimes C,A)$, so the adjunction sends it to an element of $\hom_{\mathcal{V}_0}(A^B\otimes B^C,A^C)$, as we require. For identities, we can just use the left-unit arrow $\mathbf{1}\otimes A\rightarrow A$ and pull the same trick. Now properties of adjoints give us the required relations to make this a category enriched over $\mathcal{V}$.

And finally we can check that $V(A^B)=\hom_{\mathcal{V}_0}(\mathbf{1},A^B)\cong\hom_{\mathcal{V}_0}(B,A)$, so the “underlying set” of $A^B$ is actually the set of morphisms from $B$ to $A$ in the underlying category $\mathcal{V}_0$. This justifies our suspicions that the $\mathcal{V}$-category we just built is in fact $\mathcal{V}$ itself, now as a category enriched over itself.

August 23, 2007 Posted by | Category theory | 3 Comments

## Tulane’s Math Department

I noticed something interesting about the department’s roster…

Yes it’s a shameless cross-weblog post. Deal with it :P