All Questions
Tagged with monads universal-algebra
14 questions
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, ...
- 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?
...
- 12.9k
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). ...
- 5,230
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 ...
- 66.8k
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 $M$:
...
- 9,111
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 ...
- 118k
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}{\...
- 555
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 (...
- 73.2k
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 ...
- 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 ...
CommunityBot
- 1
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 ...
- 53.3k
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 ...
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 ...
- 1,680
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 ...
- 6,162