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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 quest …
2
votes
What is the universal property of algebras for the codensity monad?
There is something in $Cat/B$ : $Alg(T) \to B$ is the universal monadic right adjoint through which $F$ factors.
There's probably a cleaner way to see this, but here is how I think about it :
It is ea …
3
votes
It's there a way to take a composite monad and a monad map to create a map of the composite?
So, having such a morphism doesn't help to make $QT$ or $SQ'$ into monads. To convince you of this, take $S$ to be the identity monad. … any monad $T$ and $Q$, you always have a (unique) distributive law $ST \to TS$ and a unique morphism of monad $S \to Q$, but when it comes to make $QT$ into a monad this is just two completely general monads …
17
votes
1
answer
563
views
A new (?) way of composing monads
By composition of monads, I mean given two monads $S$ and $T$, making their composite $S T$ into a monad. … I work in the cartesian monoidal category of Sets, so "monads" are just monoids (you can promote them to actual monads by looking at the endofunctor $X \times \_$ if you prefer). …
6
votes
1
answer
211
views
References requestion : Pretopos are algebras for a composed monad?
Unless I'm mistaken the "Free completion under finite limits monad" $C \mapsto C^{lex}$ and the "free co-completion monad" $C \mapsto \widehat{C}$ (the categories of small presheaves) satisfies a dist …
10
votes
0
answers
239
views
Colimits of algebras for $\infty$-Monad
Lurie is proving the existence of colimits in the special case of algebras for operads, but I don't think its proof can be extended to more general monads, but there might some things to be done... …
6
votes
Accepted
Is the category of computads for a finitary monad on $n$-globular sets cocomplete?
I would consider this as a folklore results in the area, and I wouldn't be surprised if it is written somewhere, at least for the special case of the strict $\infty$-category monads, but I couldn't find …
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 con …
6
votes
Accepted
Are infinitary monads monadic?
which is also in $S$, then the category of monads that are in $S$ is monadic over $S$. … $\square$
So the all point of the question is to find nice class of endofunctor for which free monads construction are available. …
22
votes
Accepted
Why are operads sometimes better than algebraic theories?
Morphisms of operads correspond to weakly cartesian morphisms of monads. … The details of this have been worked out by Gepner, Haugseng and Kock in $\infty$-Operads as Analytic Monads. …
7
votes
1
answer
206
views
Algebras for products or limits of monads
the category of monads on $C$ to the category of endofunctor on $C$). … Is there some assumptions under which we can describe algebra for some limits of monads in terms of algebras for the individual monads ? …