28

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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22

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 ...

19

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 '20 at 10:42

Peter LeFanu Lumsdaine

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19

After reaching out to every researcher who cited the manuscript, John Kennison was kind enough to find and scan his copy of the untitled manuscript containing the crude and precise monadicity theorems. I have uploaded it to the nLab for posterity: Jon Beck's untitled manuscript. This copy was distributed at the Conference Held at the Seattle Research Center ...

16

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 ...

13

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

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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13

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 ...

12

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 ...

12

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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12

That's a good question! I think Barwick and Haine have thought much more about this, and maybe they already know the answer? What I say below is definitely known to them. Also beware that I've written the below in a stream of consciousness, not quite knowing where it will go when I started.
I'll write "anima" for what is variously called homotopy ...

11

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.

11

Let $F: C \to D$ be a left adjoint functor. I hope I'm not saying anything stupid, but I think you can just rephrase the two conditions of Beck Monadicity theorem in terms of the left adjoint:
The condition that $U$ is conservative translate as:
The $Hom(F(x),\_)$ are jointly conservatives.
It can also be replaced by the apparently stronger condition but ...

11

I checked the TAC reprint of Beck's Triples, Algebras, and Cohomology from 1967. It is evident from the discussion at pag. 8, before Thm. 1 that tripleability was not presented in writing by Beck before his thesis (1967). There, Beck promises a paper "to appear", whose intended title was The tripleableness theorems. The TAC reprint has an editor's ...

10

(1, 2, 3) Though Linton's An outline of functorial semantics does contain the essence of the results and proofs of the monad–theory correspondence (see in particular Theorems 8.1 and 9.1 – 9.3), it is true that the terminology and style make it difficult to extract the results as we would expect to see them today. I think it is appropriate to cite this paper ...

9

(This is not an answer to your question, just a long comment.)
There is something to be careful about with ambidextrous adjunctions. When we work with an ordinary adjunction we can rest assured that there's no harm in just specifying one of the functors, because it having an adjoint is a property in the sense that the data of the adjoint functor, the unit, ...

answered Nov 22 '20 at 1:26

Qiaochu Yuan

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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

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

(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 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 ...

8

The category $\mathrm{Mfd}$ of finite dimensional manifolds sits fully faithfully inside the larger category $\mathrm{LocProMfd}$ of locally pro-finite dimensional manifolds, which basically extends manifolds by spaces locally modeled on the Fréchet space $\mathbb{R}^\mathbb{N}$. Take a finite dimensional base manifold $\Sigma$ and consider the category $\...

8

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 ...

8

I am not able to give the high-tech answer you are clearly hoping for. For me this statement is just a straight forward generalization of the known fact that, say, $\mathsf{Grp} \to \mathsf{Set}$ creates limits, and the same proof can be used. The statement is so basic that you should better watch out if any high-tech answer actually already uses this ...

answered Feb 17 '20 at 9:46

Martin Brandenburg

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8

1 and 3 are proved in Appendix A of the book “Algebraic theories”
by Jiří Adámek, Jiří Rosický, Enrico M. Vitale.
1 is Theorem A.37 (and A.41 for the multisorted version).
3 is Theorem A.21 (and A.40 for the multisorted version).

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

I don't know about geometric intuition, but this is pretty straightforward as long as $C$ is complete and cocomplete, though you can get away with substantially weaker hypotheses.
Simplicial objects in $C$ form a functor category $[\Delta^\mathrm{op},C]$, while split simplicial objects also form a functor category $[\Delta_\top, C]$ where $\Delta_\top$ 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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