26

Probably the reason "monadicity" gets connected with descent (and the associated terminology of descent theory) is because of its relevance to the question of descent for rings.
If you're talking about a morphism of rings $\phi:A\to B$ there is a functor $-\otimes_AB:Mod_A\to Mod_B$. Then you can ask the question: Can I recover the category $Mod_A$ from $...

answered Dec 6 '15 at 13:54

Jonathan Beardsley

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23

I think of monads in terms of algebraic theories.
Monads are substantially more general than this intuition suggests! Here is a better intuition: monads are categorified idempotents.
The point of idempotents (acting on, say, a module) is to pick out nice subobjects: subobjects that are so nice that they are simultaneously subobjects and quotient objects (...

answered Dec 6 '15 at 17:40

Qiaochu Yuan

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21

Of course, Zhen has answered the question correctly under the reasonable assumption that what is being asked is whether the usual forgetful functor $U: \mathrm{Tych} \to \mathrm{Set}$ is monadic (strictly speaking, it's functors that are or are not monadic, not categories!). One might wonder whether there is some other weird functor $G: \mathrm{Tych} \to \...

answered Jun 28 '13 at 16:49

20

As far as I know, the first paper on this was:
Ross Street, The formal theory of monads. Journal of Pure and Applied Algebra 2 (1972), 149-168.
He called them monad functors. For the same thing but with the direction of the natural transformation reversed, he called them monad opfunctors. You can also consider the case where the natural transformation ...

19

No. In fact any full subcategory of $\mathbf{Top}$ that contains all the discrete spaces cannot be monadic over $\mathbf{Set}$ unless it contains only discrete spaces. Indeed, for any such subcategory $\mathcal{C}$, the forgetful functor $\Gamma : \mathcal{C} \to \mathbf{Set}$ has a left adjoint $\Delta : \mathbf{Set} \to \mathcal{C}$ which sends a set to ...

16

The heart of the diagram chase is: All the operations and equations involved in an algebra structure on X have target X. (More generally: the target of each operation/equation could be any limit-preserving functor of the algebra-specified-so-far.)
For each of the kinds of “algebra” you describe, the structure is built up by sequentially adding operations ...

answered Feb 17 at 10:42

Peter LeFanu Lumsdaine

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15

Todd's comment provides an important limitation on what you can do here, but here's what I think is the most interesting way to answer your question. Define an endofunctor $L^+$ on the category of sets by
$$
L^+(X) = \sum_{n \geq 1} X^n
$$
for sets $X$, where $\sum$ means coproduct (disjoint union). Thus, an element of $L^+(X)$ is a nonempty finite list of ...

14

Here are the examples of comonads that I personally find most helpful. First from topology:
The universal covering is an idempotent comonad on (suitably nice) pointed topological spaces. The functor takes a pointed space $(X,x_0)$ and gives the space of homotopy classes of paths starting at $x_0$. The counit forgets the path and just keeps the endpoint, the ...

13

For $X$ an infinite set, the monad $(-)^X$ (induced from the comonoid structure on $X$ with respect to cartesian product) does not preserve reflexive coequalizers. See page 538 of this paper by Adámek, Koubek, and Velebil. Correspondingly, the forgetful functor for the category of algebras does not preserve reflexive coequalizers (hence also cannot ...

answered Jan 5 '13 at 4:00

13

I don't know where this observation was first made, but the proof is short.
Let $M$ be a monad on $Set$ such that every $M$-algebra has at most one element. For every set $A$, the set $M(A)$ has the structure of an $M$-algebra (a free one), so $M(A)$ has at most one element. On the other hand, the unit of the monad gives us a map $A \to M(A)$. Since ...

13

The bag monad is not polynomial.
Any polynomial endofunctor must preserve pullbacks: $f^*$ and $\Pi_g$ preserve all limits since they’re right adjoints, while $\Sigma_h$, being just the forgetful functor from a slice category, is well known (and easily seen) to preserve all connected limits.
However, the bag monad doesn’t preserve pullbacks. Write $B$ for ...

answered Jun 12 '18 at 17:09

Peter LeFanu Lumsdaine

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12

Note that "taking the opposite" is an operation not just on categories, but also on functors and on natural transformations. Given a functor $F: C \to D$, there is an associated functor $F^{op}: C^{op} \to D^{op}$. (Notice that $(-)^{op}$ preserves the direction -- is covariant -- on functors.) Given a natural transformation $\eta: F \to G$ for functors $F, ...

answered Jul 16 '14 at 21:57

11

I've just spent some time working out for myself the details of what an exponential monad is supposed to be, so I might as well post what I learned here. (I don't know that I'll ever have a reason to write it up more formally.)
Here is the correct definition. Given symmetric monoidal $(C,\otimes, 1)$, with finite coproducts $+$ and initial object $0$, an ...

11

I'll go ahead and turn my comment into an answer. It does form a monad, but (probably) not a very interesting one. Namely, first note that any pair of adjoint functors $L:\mathcal{C}\leftrightarrows \mathcal{D}:R$ is associated to a monad $RL$ on $\mathcal{C}$. Multiplication is given by the map $RLRL\overset{LR\to 1}{\to} RL$ and unit is $1\to RL$. Here the ...

10

The free multiset functor is not polynomial in the standard sense; it is though in a categorified sense if you somehow keep track of the different ways two expressions are the same thanks to commutativity: for that you need to pass from the category Set to the 2-category of groupoids. See Data Types with Symmetries and Polynomial Functors over Groupoids.

10

Lawvere theories can be thought of as "cartesian operads." That is, we have an analogy
$$\text{Lawvere theories} : \text{cartesian monoidal categories} :: \text{operads} : \text{symmetric monoidal categories}.$$
Consider the 2-category of symmetric monoidal cocomplete categories (where the monoidal structure distributes over colimits in both variables), ...

answered Jun 28 '18 at 13:51

Qiaochu Yuan

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10

Given a topology on a set $X$, let $2^X$ be the poset of subsets of $X$ ordered by inclusion. Then the interior operator for the topology is a comonad on $2^X$. In fact the topologies on $X$ correspond precisely to the finite-limit-preserving comonads on $2^X$. The coalgebras of the comonad are precisely the open sets.
Given a topological space $X$, define ...

9

Disclaimer: I am not an expert on model categories.
$\newcommand{\MM}{\mathcal{M}}
\newcommand{\pair}{\mathsf{P}}
\newcommand{\dom}{\operatorname{dom}}
\newcommand{\codom}{\operatorname{codom}}
\newcommand{\colim}{\operatorname*{colim}}
\newcommand{\hocolim}{\operatorname*{hocolim}}
\newcommand{\id}{\mathrm{id}}
\newcommand{\op}{\mathrm{op}}
\newcommand{\Map}...

9

What you are describing is an example of Max Kelly's notion of club, closely connected with the concept of operad. The original references date back to the 70's; one reference is
G.M.Kelly. On clubs and doctrines. In Category Seminar, Sydney 1972/1973. Springer LNM 420, pp. 181-257 (1974).
Actually, a club is defined to be a monoid in a monoidal category ...

answered Feb 20 '13 at 20:41

8

This looks like a special case of the fact that given a monad $T$ whose underlying functor has a right adjoint, that right adjoint $C$ acquires a comonad structure (mated to the structure of the monad), and the category of $M$-algebras is equivalent to the category of $C$-coalgebras. (Here $T$ is $M \times -$, with right adjoint $(-)^M$.)
This observation ...

answered Jul 12 '14 at 15:24

8

(Below by "$2$-category" I mean "bicategory.")
It's more fun to think about opposites, not in the $2$-category of categories, functors, and natural transformations, but in the $2$-category of categories, bimodules / distributors / profunctors, and natural transformations of bimodules. I prefer the bimodule terminology so let me use that. Recall that a $(C, ...

answered Jul 16 '14 at 22:20

Qiaochu Yuan

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8

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. ...

answered Jul 17 '17 at 0:16

8

I don't know what morphisms you intend for the category of finite-dimensional Hilbert spaces, but it doesn't actually matter. The answer is no, there are no interesting adjunctions between the category of groupoids and the category of finite-dimensional Hilbert spaces.
More generally,
If $C$ is a complete and cocomplete category and $D$ is a small ...

8

Over and under categories of a presentable category are presentable. This is Proposition 1.57 in Adámek, Rosický, Locally Presentable and Accessible Categories.
If $T : \mathcal{C} \to \mathcal{C}$ preserves $\lambda$-directed colimits and $\mathcal{C}$ is $\lambda$-presentable, then the category of $T$-algebras is also $\lambda$-presentable. This is proved ...

7

As for question (1): assuming cocomplete cartesian monoidal category means that cartesian products distribute over colimits (and I know from past experience that you do mean this, Martin), then it's true that the monad $T$ preserves reflexive coequalizers. The main thing you need is that finitary power functors $c \mapsto c^n$ preserve reflexive coequalizers....

answered Jan 15 '13 at 17:31

7

To make a link between Todd's answer and Qiaochu's answer: if $\mathbb{W}$ is a 2-category, then a functor:
$$(-)^* \colon \mathbb{W}^{co} \rightarrow \mathbb{W}$$
is called a "duality involution" if it is self-inverse and (pseudo) naturally satisfies:
$$\mathit{DFib}(A\times B, C) \approx \mathit{DFib}(A, B^* \times C)$$
where $\mathit{DFib}(X, Y)$ is ...

7

These questions were studied (in the dual case of comonads, née cotriples) by Applegate and Tierney in their 1970 paper Iterated cotriples.
The answer to your first question is that yes, this functor always has an adjoint as long as $\mathcal{D}$ is cocomplete. The adjoint can be constructed using a sequential colimit in $\mathcal{D}$ of the ...

answered Mar 18 '18 at 10:05

Mike Shulman

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7

It's a little weird I guess to consider "contravariant monads"; I don't think I've seen them defined before. If one considers such a notion at just the 1-categorical level, then a question is: what should an algebra map $f: A \to B$ mean? The obvious diagram one writes down would be
$$\begin{array}{l}
TA & \stackrel{a}{\to} & A \\
\uparrow \; Tf &...

answered Oct 21 '18 at 15:30

7

Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C \to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.
A fibration in a 2-category may be defined in terms of ...

7

This is covered in this English.SE question. In short, people were not all very happy about the term "triple", and tried to come up with something better. Jean Bénabou suggested "monad" during lunch at a meeting in 1966, and it was quickly adopted; for example, it appears in the titles of Anders Kock's and Eduardo Dubuc's theses from 1967 and 1969 ...

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