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6 votes
0 answers
86 views

Reciprocity for algebra objects in two algebraic categories

I think this question Compact Hausdorff and C^*-algebra "objects" in a category. shows that there is no reciprocity between categories of algebra-objects of two algebraic categories. So, ...
Nik Bren's user avatar
  • 519
8 votes
1 answer
295 views

How many categories $C$ are there such that $C$ and $C^{\text{op}}$ are monadic over $\mathrm{Set}$?

In this question, bimonadic category is a category $C$ such that $C$ and $C^{\text{op}}$ are monadic over $\mathrm{Set}$. How many bimonadic categories are there? Can we classify them all? ...
Arshak Aivazian's user avatar
1 vote
1 answer
91 views

Algebras for general transfors

Algebras for endofunctors bridge the gap between functors acting on a category and structures defined in it. An algebra for an endofunctor $F$ is instantiated by some morphism $Fa \to a$, and more ...
Mathemologist's user avatar
2 votes
2 answers
180 views

Finitary endofunctors: "Support" of elements of $TM$ where $T:\mathrm{Set}\to\mathrm{Set}$ is finitary

Consider the word a.k.a. list monad $W:\mathrm{Set}\to\mathrm{Set}$ assigning to a given set $M$ the set of words over $M$. Now, given a set $M$, we can assign to every word $w\in WM$ a subset of $M$: ...
Gerrit Begher's user avatar
6 votes
0 answers
354 views

Cohomology without comonad?

TL;DR. Many cohomologies can be unified using comonads. Question: which cohomologies cannot be? For each algebraic theory, there is an adjunction, and therefore a (co)monad (or called a (co)triple). ...
Student's user avatar
  • 5,230
25 votes
2 answers
2k views

Relation between monads, operads and algebraic theories (Again)

This question (as the title obviously suggests) is similar to, or a continuation of, this question that was asked years ago on MO by a different user. The present question, though, is different from ...
Qfwfq's user avatar
  • 23.3k
8 votes
1 answer
463 views

Relating three viewpoints on the semidirect product

It's known that giving a semidirect product $(X,m)\rtimes G$ of a $G$-group $(X,m)$ with $G$ (as defined in wiki) is the same as giving a split pair over $G$, i.e a pair of arrows $H\overset{s}{\...
Arrow's user avatar
  • 10.5k
3 votes
2 answers
355 views

Lawvere theory and the Maybe monad

The Maybe monad is based on the endofunctor $- + 1$ (coproduct with the singleton set). Its Lawvere theory $L$ is supposed to be generated by one nullary operation (...
Bartosz Milewski's user avatar
9 votes
0 answers
323 views

To what kind of generalized Lawvere theory does the "free cartesian closed category" 2-monad on $\mbox{Cat}_g$ correspond?

Thinking of Cat as a mere 1-category, there is a 1-monad $\Lambda$ for the free cartesian closed category on a category. To every category X it assigns the category $\Lambda(X)$ whose objects are ...
Mike Stay's user avatar
  • 1,532
5 votes
2 answers
615 views

Coequalizers in an Eilenberg-Moore category

Last month I proved that some category $\mathbf C$ that I happen to care about is isomorphic to the Eilenberg-Moore category for a monad on the category of bounded posets $\mathbf{BPos}$. I know from ...
Gejza Jenča's user avatar
3 votes
1 answer
398 views

Iterating Monad-Comonads structures

Let $(T, \mu , \eta )$ a monad on the category $\mathscr{C}$ , with the usual EM (Eilenberg-Moore) adjunction $\langle F_T, U_T, \eta_Y, \epsilon_T \rangle: \mathscr{C}^T \to \mathscr{C}$ where ...
Buschi Sergio's user avatar
21 votes
3 answers
3k views

Relation between monads, operads and algebraic theories

I've begun to interest in algebraic theories and their categorical models: in particular monads, generalized multicategories and operads, lawvere theories and their generalization. Is there any ...
Giorgio Mossa's user avatar
6 votes
3 answers
2k views

Monad arising from operad

It's known that from every operad arises a cartesian monad whose algebras are the algebras for the operad. Leinster proved that there are different operads from which arise the same monad, in this way ...
Giorgio Mossa's user avatar
10 votes
2 answers
862 views

Right actions of operads and monads

Given an operad $A$, there is an associated monad $M_A$ given by $M_A(X) = \coprod_n A(n) \otimes X^{\otimes n}$ such that being an $A$-algebra and being an algebra over the monad $M_A$ is the same ...
Tilman's user avatar
  • 6,162