# What is the motivation for maps of adjunctions?

In Mac Lane, there is a definition of an arrow between adjunctions called a map of adjunctions. In detail, if a functor $F:X\to A$ is left adjoint to $G:A\to X$ and similarly $F':X'\to A'$ is left adjoint to $G':A'\to X'$, then a map from the first adjunction to the second is a pair of functors $K:A\to A'$ and $L:X\to X'$ such that $KF=F'L$, $LG=G'K$, and $L\eta=\eta'L$, where $\eta$ and $\eta'$ are the units of the first and second adjunction. (The last condition makes sense because of the first two conditions; also, there are equivalent conditions in terms of the co-units, or in terms of the natural bijections of hom-sets).

As far as I can see, after the definition, maps of adjunctions do not appear anywhere in Mac Lane. Googling, I found this definition also in the unapologetic mathematician, again with the motivation of being an arrow between adjunctions.

But what is the motivation for defining arrows between adjunctions in the first place? I find it hard to believe that the only motivation to define such arrows is, well, to define such arrows...

So my question is: What is the motivation for defining a map of adjunctions? Where are such maps used?

Besides the unapologetic mathematician, the only places on the web where I found the term ''map of adjunctions'' were sporadic papers, from which I was not able to get an answer to my question (perhaps ''map of adjunctions'' is non-standard terminology and I should have searched with a different name?).

I came to think about this when reading Emerton's first answer to a question about completions of metric spaces. In that question, $X$ is metric spaces with isometric embeddings, $A$ is complete metric spaces with isometric embeddings, $X'$ is metric spaces with uniformly continuous maps, $A'$ is complete metric spaces with uniformly continuous maps, and $G$ and $G'$ are the inclusions. Now, if I understand the implications of Emerton's answer correctly, then it
is possible to choose left adjoints $F$ and $F'$ to $G$ and $G'$ such that the (non-full) inclusions $A\to A'$ and $X\to X'$ form a map of adjunctions. This made me think whether the fact that we have a map of adjunctions has any added value. Then I realized that I do not even know what was the motivation for those maps in the first place.

[EDIT: Corrected a typo pointed out by Theo Johnson-Freyd (thanks!)]

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Excellent question! I had the same thought when I read through CFTWM but was not able to come up with any really natural motivating examples. A question for the experts: Continuous maps between topological spaces induce a pair of adjoint functors (geometric morphisms) on the topos of sheaves. Do homotopies between these maps give a morphism of adjoint functors? What should the correct notion of homotopy between geometric morphisms be? If this line of thought ends up resulting in an answer, I will post it later. –  Steven Gubkin Mar 1 '10 at 13:37
Incidentally, something is a little wrong about your first paragraph. In particular, you define F:A->X and K:A->A', and then ask about KF. So I assume that one of these maps is backwards. –  Theo Johnson-Freyd Mar 1 '10 at 17:13

One of the applications of adjoint functors is to compose them to get a monad (or comonad, depending on the order in which you compose them). A map of adjoint functors gives rise to a map of monads. So one might ask: what are maps of monads good for? Many algebraic categories (such as abelian groups, rings, modules) can be described as categories of algebras over a monad, others (for example in Arakelov geometry) are most easily described in such a way. A map of monads then gives functors between the categories of algebras over these objects.

Here is a concrete example from topology: Let $E$ be a connective generalized multiplicative homology theory, and let $H = H(-;\pi_0E)$ be ordinary homology with coefficients in $\pi_0E$. There exists a map $E \to H$ inducing an isomorphism on $\pi_0$. For a spectrum $X$, the functor $\underline{E}\colon X \mapsto E \wedge X$ gives rise to a monad, and similarly for $H$, thus we get a morphism of monads $\underline{E} \to \underline{H}$. The completion $X\hat{{}_E}$ of a spectrum $X$ at $E$ is defined to be the totalization of the cosimplicial spaces obtained by iteratively applying $\underline{E}$ to $X$. The monad map gives a natural map $X\hat{{}_E} \to X\hat{{}_H}$ which turns out to be an equivalence for connective $X$.

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Thank you very much for your answer. This is a bit too high for my current knowledge, but I am sure that I will benefit from this answer after I read Ch. VI of Mac Lane. –  user2734 Mar 1 '10 at 19:43
A very good motivation! –  Peter Arndt Mar 1 '10 at 23:46
@Tilman: I couldn’t resist taking a quick look at Ch. 6 of Mac Lane, at least to get a feeling of the terminology. I now see that your answer is probably THE answer to my question. I’ve also noticed that maps of adjunctions pop up in the theorem on the comparison functor (Theorem VI.3.1). Thank you very much for your help! –  user2734 Mar 2 '10 at 7:28

Here is an example of how one might have stumbled upon the definition of a map of adjunctions. Suppose that you are working on a research project with a collaborator. Let's call her Jane for the sake of argument. On the first day you and Jane realize that your joint research project depends partly on knowing whether a certain functor F has a right adjoint. It also depends on taking that supposed right adjoint and putting it to good use. So you really need to know what that right adjoint is. You and Jane call it a day, and agree to continue working the next day.

That night both of you are independently inspired. You wake up in the middle of the night an jot down some notes. The next morning you and Jane meet to discuss what you've each figured out. Fantastic news! Both of you have found the right adjoint to F. You immediately begin planing how you are going to solve XYZ-Big-Problem with this fabulous right adjoint. After the celebratory mood wares away, you and Jane realize with some horror the truth. Your right adjoint G is not the same as Jane's right adjoint G'. They are different functors and the adjunction structure maps are different.

Whatever are you to do? Which one should you use?

Fortunately Jane has a flash of insight. We know that two functors can be isomorphic, what about adjunctions? After thinking about this some more, you and Jane figure out that a morphism of adjunctions should be a morphism of functors which preserves the adjunction structure. You try to do this is the simplest way possible and BAM! You've rediscovered the notion of morphism of adjunction. You notice that while G and G' are not the same functor (and hence not the same adjunction) they are isomorphic adjunctions. Whew!

But now you and Jane start to seriously worry. You have your functor F and you know that your right adjoint G and Jane's right adjoint G' are isomorphic. But what happens when Prof. X comes along with his right adjoint G''? Will it be isomorphic to G and G'? Given F, how unique is its right adjoint? Even if G, G', and G'' are all isomorphic adjunctions there might be some monodromy, i.e. the isomorphism, $$G \to G' \to G'' \to G$$ might theoretically fail to be the identity.

Then you read a little farther in MacLane and you find this theorem (I'm rephrasing it with some terminology which is in vogue.

Theorem: Given a functor F, the category of right adjoints to F (with their adjunction data and with morphisms of adjunctions as morphisms) is either empty or is a contractible category (i.e. it is equivalent to a terminal category i.e. any two objects are isomorphic and that isomorphism is unique).

So you can stop worrying. Any other right adjoint G'' that Prof. X brings to you will in fact be (uniquely) isomorphic to the one you discovered.

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Thank you very much for your answer. Could you please point me to this theorem in Mac Lane? –  user2734 Mar 1 '10 at 14:07
I realized after writing this that I was hasty and misread the first paragraph of the question. So I did not actually answer the question asked. I was tempted to delete this answer, but I thought it still might be of some value to some people so I've decided to let it remain. The notion of morphism you describe seems weird to me. I would think you only have nat. isoms $KF \simeq F'L$ (and similar). In that case the question I address is motivation for morphism in the case L=K = identity and F=F', but G and G' different. –  Chris Schommer-Pries Mar 1 '10 at 14:10
It is a nice story, but it really only motivates isomorphisms of adjunctions. If someone wanted you to motivate homomorphisms of groups it wouldn't be very instructive to give an example of an isomorphism of groups - your pupil would say "hey, you just relabeled you elements!". A much more instructive example would be, say, the projection of the integers onto a finite cyclic group. Do you have a natural example of a map between adjuntions which is not an isomorpism? Also, what do you think about my comment to the original question? –  Steven Gubkin Mar 1 '10 at 14:14
@ Steven: Yeah I agree. It only motivates isomorphism of adjunction (I misread the question). Maybe it is my ignorance, but the notion describe in MacLane and in the original question seems strange to me. One the one hand it is "evil" (requires an equality where it should just be nat. isom.) and on the other hand, I don't think I've encountered very many (any?) natural examples of morphisms of adjunctions where the source and target categories were allowed to change. Oh, and I don't yet have an opinion on your comment. –  Chris Schommer-Pries Mar 1 '10 at 14:30
@ Chris Schommer-Pries: Thank you for the reference! Please don’t delete this answer, it is informative anyway. –  user2734 Mar 1 '10 at 14:32