31
votes

Accepted

### Why are monadicity and descent related?

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 ...

26
votes

### Why are monadicity and descent related?

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 ...

23
votes

Accepted

### Jon Beck's untitled manuscript containing the "tripleability theorem" (i.e. the monadicity theorem)

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....

22
votes

### Big list of comonads

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 ...

Community wiki

22
votes

Accepted

### Why are operads sometimes better than algebraic theories?

First - yes, for symmetric set-operads this functor is "injective", though it is not fully faithful. It is faithful on general maps and fully faithful on isomorphisms. Its image can easily ...

19
votes

### Conceptual reason that monadic functors create limits?

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-...

18
votes

Accepted

### List is a monad, but is it a comonad with these natural transformations?

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 ...

16
votes

### Relation between monads, operads and algebraic theories (Again)

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 ...

15
votes

Accepted

### Is the Cartesian product of two finitely presented objects finitely presentable?

No. For counterexamples, see Theorems 3.8, 3.9, and 3.10 of
Finiteness properties of direct products
of algebraic structures
Peter Mayr, Nik Ruškuc
Journal of Algebra 494 (2018) 167-187.
These ...

14
votes

Accepted

### What is the polynomial functor for the Bag monad

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 ...

13
votes

### Monads on Set with trivial algebras

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 ...

13
votes

### Big list of comonads

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 $X$ ...

Community wiki

12
votes

### Is forming the Albanese variety a monad?

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 \...

12
votes

### Infinity-categorical analogue of compact Hausdorff

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 ...

11
votes

### What is the polynomial functor for the Bag monad

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 ...

11
votes

Accepted

### Characterization of functors whose right adjoint is monadic?

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 ...

11
votes

### Jon Beck's untitled manuscript containing the "tripleability theorem" (i.e. the monadicity theorem)

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 ...

10
votes

### A specific property of bi-adjunction

(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 ...

10
votes

Accepted

### Reference request for Linton's theorems on equational theories

(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 ...

9
votes

### A new (?) way of composing monads

Ok, I think I've solved the mystery, and it is a little disappointing: The point is that there is actually a distributive law lurking in the background , constructed from the $l$ in the original ...

8
votes

### Categories which are both monadic and comonadic over another category

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 $M$-...

8
votes

### monadic decomposition

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 ...

8
votes

### Adjunctions between Groupoids and Hilbert spaces

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 ...

8
votes

Accepted

### Locally presentable categories

Over and under categories of a presentable category are presentable. This is Proposition 1.57 in Adámek, Rosický, Locally Presentable and Accessible Categories.
If $T : \mathcal{C} \to \mathcal{C}$ ...

8
votes

### Big list of comonads

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 ...

Community wiki

8
votes

Accepted

### When were triples called monads for the first time?

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 ...

8
votes

### Conceptual reason that monadic functors create limits?

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, $\mathsf{Grp} \to \mathsf{Set}$ ...

8
votes

Accepted

### 2-monads for categories with a class of (co)limits

Kelly and Lack's paper On the monadicity of categories with chosen colimits answers your questions (1),(2) and (3) affirmatively. The main theorems are Theorem 6.1, 6.2 and 7.1. Their main trick is ...

8
votes

### Reference request for Linton's theorems on equational theories

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 ...

8
votes

Accepted

### When is the Eilenberg-Moore category of a monad on an ind-category itself an ind-category?

Let $T$ be a monad on an accessible category (i.e. an $\mathbf{Ind}$-category). If the underlying endofunctor of $T$ is finitary (i.e. preserves filtered colimits), then the Eilenberg–Moore category ...

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