# The Unapologetic Mathematician

## Mathematics for the interested outsider

And now we go back to adjoints. Like every other structure out there, we want to come up with some analogue of a homomorphism between two adjunctions. Let’s consider the adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and $F'\dashv G':\mathcal{C}'\rightarrow\mathcal{D}'$, and try to find a good notion of a transformation from the first to the second.

We’ll proceed by considering an adjunction to consist of the pair of categories $(\mathcal{C},\mathcal{D})$ with the functors giving extra structure. Down in the land of groups and rings and such, we’d consider sets with extra structure and functions that preserved that structure. So here, naturally, we want to consider functors which preserve this extra structure. That is, a map of adjunctions consists of a pair of functors $K:\mathcal{C}\rightarrow\mathcal{C}'$ and $L:\mathcal{D}\rightarrow\mathcal{D}'$. These must preserve the structure in that $K\circ G=G'\circ L$ and $L\circ F=F'\circ K$.

But hold up a second, we’ve forgotten something else that goes into an adjunction: the isomorphism $\phi:\hom_\mathcal{D}(F(C),D)\rightarrow\hom_\mathcal{C}(C,G(D))$. Here’s a diagram showing how the map of adjunctions should play nicely with them: Equivalently we can specify an adjunction by its unit and counit. In this case the compatibility in question is a pair of equations of natural transformations: $1_K\circ\eta=\eta'\circ1_K$ and $1_L\circ\epsilon=\epsilon'\circ1_L$.

What if we’re looking at two different adjunctions between the same pair of categories? Well then we may as well try to use the appropriate identity functors for $K$ and $L$. But then it’s sort of silly to insist that $G=1_\mathcal{C}\circ G=G'\circ\mathcal{D}=G'$ on the nose, and similarly for $F'$. Instead, as we do so often, let’s weaken this equality to just a natural transformation.

We’ll say that a pair of natural transformations $\sigma:F\rightarrow F'$ and $\tau:G'\rightarrow G$ are “conjugate” if $\hom_\mathcal{C}(1_C,\tau_D)\circ\phi'_{C,D}=\phi_{C,D}\circ\hom_\mathcal{D}(\sigma_C,1_D)$. This is equivalent, in terms of the unit and counit, to any one of the following four equalities:

• $\tau=(1_G\circ\epsilon')\cdot(1_G\circ\sigma\circ1_{G'})\cdot(\eta\circ1_{G'})$
• $\sigma=(\epsilon\circ1_{F'})\cdot(1_F\circ\tau\circ1_{F'})\cdot(1_F\circ\eta')$
• $\epsilon\cdot(1_F\circ\tau)=\epsilon'\cdot(\sigma\circ1_{G'})$
• $(1_G\circ\sigma)\cdot\eta=(\tau\circ1_{F'})\cdot\eta'$

Now it’s easily verified that given a pair of categories $(\mathcal{C},\mathcal{D})$ we can build a category whose objects are adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and whose morphisms are conjugate pairs of natural transformations, which we write out in full as $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$. We compose conjugate pairs in this category in the obvious way, which we write $(\sigma',\tau')\cdot(\sigma,\tau)$.

On the other hand, if we have a pair $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$ and another $(\bar{\sigma},\bar{\tau}):(\bar{F},\bar{G},\bar{\phi})\rightarrow(\bar{F}',\bar{G}',\bar{\phi}'):\mathcal{D}\rightarrow\mathcal{E}$, then we can form the composite $(\bar{\sigma}\circ\sigma,\tau\circ\bar{\tau}):(\bar{F}\circ F,G\circ\bar{G},\bar{\phi}\cdot\phi)\rightarrow(\bar{F}'\circ F',G'\circ\bar{G}',\bar{\phi}'\cdot\phi'):\mathcal{C}\rightarrow\mathcal{E}$, which we’ll write as $(\bar{\sigma},\bar{\tau})\circ(\sigma,\tau)$. Notice the similarity of this situation with the two different compositions of natural transformations between functors.

July 30, 2007 Posted by | Category theory | 4 Comments