Eilenberg and Moore have shown that given a monad $$L$$, if $$L$$ has right adjoint $$R$$, then $$R$$ is a comonad.

I see how to dualize this result to obtain the following theorem: given a comonad $$R$$, if $$R$$ has left adjoint $$L$$, then $$L$$ is a monad. Indeed it suffices to notice that an adjunction $$L \dashv R$$ dualizes to $$R^\text{op} \dashv L^\text{op}$$.

But at the bottom of page 7 of this paper, it is written that it also "easily" dualizes to the following theorem: given a comonad $$L$$, if $$L$$ has right adjoint $$R$$, then $$R$$ is a monad. How to make this easy dualization?

• Presumably what's meant by "easily" is that the proof in the next paragraph isn't too hard. I don't think it means that it's trivial (i.e. via replacing $\mathcal C$ with its opposite). Feb 29, 2020 at 21:17
• Not exactly. Parts of the statement are dualized (monad <-> comonad) but other parts aren't (right adjoint stays right adjoint). It might be possible to generalize the statements so that they are related by a duality. Feb 29, 2020 at 21:56
• I'm not completely sure. You'd need to introduce a second category so that one part can be dualized but not the other. Feb 29, 2020 at 22:07
• Alternatively, going to 2-categories would give you multiple ways to dualize (i.e. not only $^{op}$ but also $^{co}$ and $^{coop}$). Feb 29, 2020 at 22:10
• I will post an answer and try to make drawings. :) Mar 1, 2020 at 16:06

Let $$\mathcal K$$ be a $$2$$-category. A monad in $$\mathcal K$$ is an object $$C$$ together with a $$1$$-morphism $$T \colon C \to C$$ and $$2$$-morphisms $$\eta \colon 1 \to T$$ and $$\mu \colon TT \to T$$ as well as some commuting diagrams (see the nlab). Comonads can be similarly defined.

Similarly, adjunctions can be defined internally to any $$2$$-category. A right adjoint to a $$1$$-morphism $$L \colon C \to D$$ is a $$1$$-morphism $$R \colon D \to C$$ with a unit and counit making the usual diagrams commute (modulo coherence isomorphisms).

Given that a monad structure on $$L$$ yields a comonad structure on any right adjoint $$R$$, we can get all four possible dualizations by switching out $$\mathcal K$$ for its duals.

You can check that a monad structure on $$T$$ in $$\mathcal K^{co}$$ ($$2$$-morphisms are reversed) is the same as a comonad structure on $$T$$ in $$\mathcal K$$. This also dualizes adjoints: if $$L \dashv R$$ in $$\mathcal K^{co}$$ then $$R \dashv L$$ in $$\mathcal K$$.

Interestingly, switching to $$\mathcal K^{op}$$ only dualizes adjoints: if $$L \dashv R$$ in $$\mathcal K^{op}$$ ($$1$$-morphisms are reversed), then $$R \dashv L$$ in $$\mathcal K$$.

To dualize only monads, we can use $$\mathcal K^{coop}$$ (both kinds of arrows are reversed).

Some more details, as requested. Suppose we're given an ordinary comonad $$L$$ and $$L$$ has an ordinary right adjoint $$R$$. We'd like to conclude that $$R$$ is a monad by applying the theorem that given a monad $$L$$ and a right adjoint $$R$$, $$R$$ is a comonad.

More specifically, we're going to apply the general $$2$$-category version of that theorem. We need to start with a monad, but right now we have a comonad. To switch between them, we'll work with $$\mathcal {Cat}^{co}$$ instead. This means that we have a comonad now, but also switches the adjunctions, so that our comonad has a left adjoint. (Doing just this step is equivalent to switching out the category for its opposite, i.e. the trivial duality).

Thus, we need to dualize again in a way that unswaps adjoints, but leaves monads unchanged. Taking the $$^{op}$$ of the $$2$$-category accomplishes that, so now we're working in $$\mathcal {Cat}^{coop}$$.

To spell that out explicitly, if we're given an ordinary comonad $$L$$ and an ordinary right adjoint $$R$$, then this same data is equivalently a monad $$L$$ and a right adjoint $$R$$ in $$\mathcal {Cat}^{coop}$$. Then applying the general theorem, we get a comonad structure on $$R$$ in $$\mathcal {Cat}^{coop}$$, which translates back to a monad structure on $$R$$ in in $$\mathcal {Cat}$$, i.e., an ordinary monad.

• I have little experience with 2-categories, thus I have trouble applying your answer to the 2-category Cat that is of concern in my question. Can you spell it out?
– Bob
Mar 1, 2020 at 9:07
• @Bob, I've added some details. Mar 1, 2020 at 11:56
• Great! This is clear now. Thanks. However, this assumes that the "general theorem" is true. But, as far as I know, Eilenberg and Moore proved it in the $2$-category Cat, not in any $2$-category $\mathcal K$.
– Bob
Mar 1, 2020 at 12:33
• @Bob That's true, but the general proof is almost exactly the same. Just a couple extra structural isomorphisms inserted in places. Mar 1, 2020 at 12:34
• @Bob Never say never, but I don't really think so, at least not in a way that reuses the same proof exactly. Still, it's just a matter of reversing arrows (either the direction of the functors or the direction of the morphisms), so the same proof works with the right adjustments. But without the mechanism of the opposite $2$-category, you still need to check all the details (commuting diagrams, etc.). It's not an "easy" corollary. Mar 1, 2020 at 12:51

We can see things more clearly using string diagram notation. I did not find how to draw them here (no tikz allowed) so I had to resort to images scan.

A monad structure on $$T : C→C$$ is given by a pair of natural transformations as below, satisfying certain axioms. Suppose $$G$$ is a right adjoint to $$T$$. This is given by a pair of natural transformations as below satisfying the triangle identities. Using these, we can "bend" the monad structure on $$T$$ to get a comonad structure on $$G$$ as follows (rest to check the axioms). Now, if $$G$$ is a left adjoint to $$T$$, we can do the "bending" the other way around: This corresponds to what is said in the answer of SCappelia: we have a "formal" theorem working in any $$2$$-category. We have an up-down symmetry but also a left-right symmetry, making 4 versions in total.

• On the first diagram, I understand that the right part is the multiplication, but why are there two occurrences of T in the left part?
– Bob
Mar 1, 2020 at 17:07
• Aaah you are right. I mixed things up with the unit of an adjunction, sorry! I redo the drawings. Mar 1, 2020 at 17:16
• @Bob The drawings are corrected. Mar 1, 2020 at 17:26
• My answer is essentially a reformulation of SCappella's one. Here is a comparison: to prove for instance that coproducts are associative in a category, you can either use the fact that products are associative and apply it to the dual category, or you can apply the "dual reasoning" to your category directly, without ever mentioning the dual category. But these are two formulations of exactly the same thing. Mar 1, 2020 at 19:32
• @Bob Hi. There is at least a way of visualizing the statement Zhen Lin quotes as the correct dualization using string diagrams. But in order to recover your statement as a formal dual of Zhen Lin's statement, you still need to show that the Kleisli category of a monad is a Kleisli object, and I'm not sure string diagrams would help for that. Feb 17, 2022 at 19:08