# Categories which are both monadic and comonadic over another category

I heard a professor say that $\lambda$-rings are both monadic and comonadic over commutative rings. Remark 2.11(a) on the nlab page says the same.

1. What does it mean, intuitively, that a category is both monadic and comonadic over another category?
2. What follows from this formally?
3. What are some additional examples of such phenomena?

## migrated from math.stackexchange.comJul 16 '17 at 18:16

This question came from our site for people studying math at any level and professionals in related fields.

• Lambda-rings are coalgebras over a certain comonad (see Section 16.59 of Hazewinkel's "Witt vectors, part I"). If I understand the terminology correctly, this should yield that the forgetful functor from lambda-rings to rings is comonadic. On the other hand, the "orthodox" definition of lambda-rings (via the $\lambda^n$ maps and the axioms they satisfy) should (I am not sure here) yield that lambda-rings are algebras over a certain monad, and thus the forgetful functor is monadic. – darij grinberg Jun 26 '17 at 14:43
• @darijgrinberg forgive me if I misunderstood your comment. I'm not asking what (co)monadicity means, but rather what "bimonadicity" means morally e.g what's a nice mantra about bimonadic functors. – Arrow Jun 26 '17 at 14:48
• Ah. I actually misunderstood it differently: I thought you asked for a proof of the (co)monadicity. On the philosophical question I have no answer. – darij grinberg Jun 26 '17 at 15:05

I understand it like this: if a monad $M: C \to C$ has a right adjoint $K: C \to C$, then that right adjoint carries a comonad structure which is mated to the monad structure, and the category of $M$-algebras is canonically equivalent to the category of $K$-coalgebras. All examples of functors that are simultaneously monadic and comonadic are of this type. This observation goes back to the original 1965 Eilenberg-Moore paper, the one where the term "Eilenberg-Moore category" comes from; see section 3 of that paper.

So: intuitively you can just think of it as the case where a monad $M$ is cocontinuous, or the underlying functor $U: Alg_M \to C$ is cocontinuous. This intuition is literally correct if $M$ is e.g. a total category, e.g., a topological category or a locally presentable category -- a situation where cocontinuous functors are the same as left adjoints.

An important example is where $M$ is a monoid and one considers the forgetful functor $Set^M \to Set$. This has a left adjoint $M \times -: Set \to Set^M$, and a right adjoint $(-)^M: Set \to Set^M$. Rather more generally, one has analogously for any category $C$ a monadic functor $Set^C \to Set/C_0$ (with $C_0$ the set of objects of $C$). The left adjoint takes a $C_0$-fibered set $X \to C_0$ to $X \times_{C_0} C_1$ with the evident $C$-action, and I leave it to the reader to construct the right adjoint.

This last example plays an important role in topos theory: if $E$ is a topos and $C$ is an internal category in $E$, then the category $E^C$ of internal copresheaves is monadic and comonadic over the slice topos $E/C_0$. Being the category of coalgebras of a comonad that is left exact (because it has a left adjoint, the associated monad), $E^C$ is also a topos: this is one of the classical theorems of topos theory.

• There might be cases when the category is both monadic and comonadic but the forgetful functors are different. For example, there are monads which have left adjoint comonads. Say, $X\times-$ and $-^X$ in a cartesian closed category (but algebras are not always equivalent to coalgebras). – მამუკა ჯიბლაძე Jul 17 '17 at 4:05
• Yes, definitely it's better to refer to a functor as monadic rather than a category, since the latter leaves which forgetful functor unspecified. – Todd Trimble Jul 17 '17 at 10:09
• @ToddTrimble do you have any references for the important monoid example? I'm studying it currently and want to know where to look. – Morgan Rogers Mar 14 at 8:35
• @MorganRogers It's mentioned for instance in Mac Lane-Moerdijk's book on topos theory. But there are a number of interesting examples beyond the group case which is the best studied case. For example, the category of reflexive graphs is equivalent to $Set^M$ for a certain monoid $M$; see also ncatlab.org/nlab/show/graphic+category. – Todd Trimble Mar 14 at 13:13
• @ToddTrimble thanks. I presume in the former case you mean Sheaves in Geometry and Logic, specifically Chapter V. These are clearly motivating examples of monadic and comonadic functors. However, there is no explicit use of the fact that these functors are simultaneously monadic and comonadic as far as I can tell; these facts are used independently. Would you agree? – Morgan Rogers Mar 14 at 16:21

One important example of a monadic category over $C$ is one obtained by adjoining to $C$ a system of $n$-ary operations satisfying some relations. For example, the category of rings with involution is monadic over sets: it can be given by two binary operations ($+,\times$), two $0$-ary operations ($0,1$), and two $1$-ary operations (negation and the involution). Perhaps if you adopt a suitably enlightened definition of operations with relations then all monadic functors would arise in this way.
Now a $1$-ary operation is just a morphism, so you can also view it as a $1$-ary operation in the opposite category. So if you're adding only $1$-ary operations, then you should get a category which is both monadic and comonadic over the original category.
Indeed, the usual way of defining lambda-rings is by adjoining a bunch of $1$-ary operations to the category of commutative rings. The monad is the free lambda-ring functor, and the comonad is the big Witt vector functor (also called the co-free lambda-ring functor). Another example is the category formed from objects of the original category $C$ together with an action of your favorite group (or monoid) $G$. In representation theory, the monad is often called the induced representation functor, and the comonad is called the co-induced representation functor.
Some things that follow formally from this set up are that the new category has all the same kinds of limits and colimits that the original category $C$ has, and the forgetful functor preserves them. Beck's theorem says that some kind of converse is also true.