The Unapologetic Mathematician

Mathematics for the interested outsider

More on units and counits

Last time we took an adjunction and came up with two natural transformations, weakened versions of the natural isomorphisms defining an equivalence. Today we’ll see how to go back the other way.

So let’s say we have an adjunction F\dashv G:\mathcal{C}\rightarrow\mathcal{D} given by natural isomorphism \Phi_{C,D}:\hom_\mathcal{D}(F(C),D)\rightarrow\hom_\mathcal{C}(C,G(D)). Remember that we defined the unit and counit by \eta_C:\Phi_{C,F(C)}(1_{F(C)}) and \epsilon_D=\Phi_{G(D),D}^{-1}(1_{G(D)}). We can take either one of these and reverse-engineer it. For instance, given an arrow f:F(C)\rightarrow D in \mathcal{D} we can calculate
\Phi_{C,D}(f)=\Phi_{C,D}(f\circ1_{F(C)})=G(f)\circ\Phi_{C,F(C)}(1_{F(C)})=G(f)\circ\eta_C
so once we know the unit of the adjunction we can calculate \Phi from it. Notice how we use the naturality of \Phi in the second equality.

Dually, we can determine \Phi^{-1} in terms of the counit. Given g:C\rightarrow G(D) in \mathcal{C}, we calculate:
\Phi_{C,D}^{-1}(g)=\Phi_{C,D}^{-1}(1_{G(D)}\circ g)=\Phi_{G(D),D}^{-1}(1_{G(D)})\circ F(g)=\epsilon_D\circ F(g)
so we can also determine the natural isomorphism of hom-sets in terms of the counit.

Of course, since we can determine the same isomorphism (technically the isomorphism and its inverse) from either the unit or the counit, they must be related. So what do these equations really tell us?

For this we have to go back to the way we compose natural transformations. The obvious way is where we have natural transformations \phi:F\rightarrow G and \theta:G\rightarrow H between three functors from \mathcal{C} to \mathcal{D}. We put them together to get (\theta\cdot\phi)_C=\theta_C\circ\phi_C:F\rightarrow H.

Less obviously, we can consider functors F_1 and F_2 from \mathcal{C} to \mathcal{D}, functors G_1 and G_2 from \mathcal{D} to \mathcal{E}, and natural transformations \phi:F_1\rightarrow F_2 and \theta:G_1\rightarrow G_2. We can put these together to get \theta\circ\phi:F_1\circ G_1\rightarrow F_2\circ G_2, defined by \theta_{F_2(C)}\circ G_1(\phi_C) or G_2(\phi_C)\circ\theta_{F_1(C)} (exercise: show that these two composites are equal).

Now what we need here is this “horizontal” composite. Let’s go back to the adjunction and take the natural transformations \eta:1_\mathcal{C}\rightarrow G\circ F and 1_G:G\rightarrow G. The components of their horizontal composite \eta\circ1_G:G\rightarrow G\circ F\circ G is then given by \eta_{G(D)}. Similarly, if we take the natural transformations 1_G and \epsilon:F\circ G\rightarrow1_\mathcal{D}, their horizontal composite has components given by G(\epsilon_D). Now the “vertical” composite of these two (1_G\circ\epsilon)\cdot(\eta\circ1_G) has components G(\epsilon_D)\circ\eta_{G(D)}. And the above formula for the adjunction isomorphism in terms of the unit tells us that this is \Phi_{G(D),D}(\epsilon_D)=\Phi_{G(D),D}(\Phi_{G(D),D}^{-1}(1_{G(D)}))=1_{G(D)}.

To put it at a bit of a higher level, if we start with the functor G, use the unit to turn it into the functor G\circ F\circ G, then use the counit to move back to G, the composite natural transformation is just the identity transformation on G. Similarly, we can show that the composite (\epsilon\circ1_F)\cdot(1_F\circ\eta) taking F to F\circ G\circ F and back to F is the identity transformation on F.

Inherent in this is also the converse statement. If we have natural transformations \eta:1_\mathcal{C}\rightarrow G\circ F and \epsilon:F\circ G\rightarrow1_\mathcal{D} satisfying these two identities, then we can use the above formulae to define a natural isomorphism \phi in terms of \eta and its inverse in terms of \epsilon. Thus an adjunction is determined by a unit and a counit satisfying these “quasi-inverse” relations.

If you’re up to it, try to see where we’ve seen these quasi-inverse relations before in a completely different context. I’ll be coming back to this later.

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