# The Unapologetic Mathematician

## Adjoints with parameters

Now that we know how to transform adjoints, we can talk about whole families of adjoints parametrized by some other category. That is, for each object $P$ of the parametrizing category $\mathcal{P}$ we’ll have an adjunction $F_P\dashv G_P:\mathcal{C}\rightarrow\mathcal{D}$, and for each morphism of $\mathcal{P}$ we’ll have a transformation of the adjunctions.

Let’s actually approach this from a slightly different angle. Say we have a functor $F:\mathcal{C}\times\mathcal{P}\rightarrow\mathcal{D}$, and that for each $P\in\mathcal{P}$ the functor $F(\underline{\hphantom{X}},P):\mathcal{C}\rightarrow\mathcal{D}$ has a right adjoint $G(P,\underline{\hphantom{X}})$. Then I claim that there is a unique way to make $G$ into a functor from $\mathcal{P}^\mathrm{op}\times\mathcal{D}$ to $\mathcal{C}$ so that the bijection $\phi_{C,P,D}:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ is natural in all three variables. Note that $G$ must be contravariant in $P$ here to make the composite functors have the same variance in $P$.

If we hold $P$ fixed, the bijection is already natural in $C$ and $D$. Let’s hold $C$ and $D$ fixed and see how to make it natural in $P$. The components $\phi_P:\hom_\mathcal{D}(F(C,P),D)\rightarrow\hom_\mathcal{C}(C,G(P,D))$ are already given in the setup, so we can’t change them. What we need are functions $\hom_\mathcal{D}(F(1_C,p),1_D):\hom_\mathcal{D}(F(C,P'),D)\rightarrow\hom_\mathcal{D}(F(C,P),D)$ and $\hom_\mathcal{C}(1_C,G(p,1_D)):\hom_\mathcal{C}(C,G(P',D))\rightarrow\hom_\mathcal{C}(C,G(P,D))$ for each arrow $p:P\rightarrow P'$.

For naturality to hold, we need $\hom_\mathcal{C}(1_C,G(p,1_D))\circ\phi_{P'}=\phi_P\circ\hom_\mathcal{D}(F(1_C,p),1_D)$. But from what we saw last time this just means that the pair of natural transformations $(F(1_C,p),G(p,1_D))$ forms a conjugate pair from $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ to $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$. And this lets us define $G(p,1_D)$ uniquely in terms of $F(1_C,p)$, the counit $\epsilon'$ of $F(\underline{\hphantom{X}},P')\dashv G(P',\underline{\hphantom{X}})$, and the unit $\eta$ of $F(\underline{\hphantom{X}},P)\dashv G(P,\underline{\hphantom{X}})$ by using the first of the four listed equalities.

From here, it’s straightforward to show that this definition of how $G$ acts on morphisms of $\mathcal{P}$ makes it functorial in both variables, proving the claim. We can also flip back to the original viewpoint to define an adjunction between categories $\mathcal{C}$ and $\mathcal{D}$ parametrized by the category $\mathcal{P}$ as a functor from $\mathcal{P}$ to the category $\mathbf{Adj}(\mathcal{C},\mathcal{D})$ of adjunctions between those two categories.

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July 31, 2007 - Posted by | Category theory

## 4 Comments »

1. With every new post you make I’m finding out that I know almost nothing about adjoints!

Do you plan on talking about fibered categories at some point?

Comment by ulfarsson | July 31, 2007 | Reply

2. Maybe eventually I will. I’ve got a short-term goal for the category theory, though, before I can finally move on to topology.

And analysis.

And linear algebra.

And representation theory…

Comment by John Armstrong | July 31, 2007 | Reply

3. Suppose C is a monoidal category and F is the monoidal operation. Suppose further that the functor F(_,P) has a right adjoint, g(P) (contravariant) for each object P in C.

Now it seems that you are suggesting that there is a functor

G: F(C,C)–>C such that g(P) = G(P,_)

Assuming that I am understanding you correctly,

1) Can G be thought of as a right adjoint to F?
2) Is G(P,Q) necessarily an (the) exponential object Q^P? If it is, then C is, by definition, a closed category, right?
3) Can one start with a category C, a bifunctor G and define another functor F analogously? Presumably, G would have to satisfy properties similar to the monoidal operation but not quite dual properties.

dan

Comment by Dan | September 25, 2007 | Reply

4. Dan, I think your notation is a bit confused, but I can answer question 2 definitely. Yes, this is the definition of the exponential object, and in fact that’s why I gave this definition. If $\underline{\hphantom{X}}\otimes P$ has a right adjoint $E_P(\underline{\hphantom{X}}$ for all $P$, then these right adjoints fit together to make a functor $E:\mathcal{C}^\mathrm{op}\times\mathcal{C}\rightarrow\mathcal{C}$ so that $E(P,\underline{\hphantom{X}})=E_P(\underline{\hphantom{X}})$.

In this case, the functor isn’t the “right adjoint”, but the two functors together define a parametrized family of adjunctions.

And yes, you can start with something like the exponential and extract a monoidal structure given the correct left adjoints. In fact, I did just this just the other day when I talked about ordered linear spaces.

Of course, neither of these families of adjoints has to exist. But when they do they give us a lot of useful structure to work with.

Comment by John Armstrong | September 25, 2007 | Reply