How to dualize a theorem by Eilenberg and Moore about monad, comonad and adjunction?

Question

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?

Answer 1

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).

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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.

Answer 2

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.