My first question here.

Accordingly to M. Barr "Coequalizers and free triples" by a free triple (or free monad) generated by an endofunctor $R: X\rightarrow{X}$ we mean a triple $T=(T,\eta,\nu)$ and a natural transformation $p: R\rightarrow{T}$ such that if $T'= (T',\eta',\nu')$ is another triple and $p': R\rightarrow{T'}$ is a natural transformation, then $p'=\tau{p}$, where $\tau:T\rightarrow{T'}$ is a map of triples.

On the other hand we have always a monad in the form $(GF,\eta,G\epsilon{F})$ when we have an adjunction $F\dashv{G}$ where $\eta$ is the unit and $\epsilon$ de counit of the adjunction.

Can the endofunctor $R$ have the form of an adjunction (for example between a free and a forgetful functor) in the first definition? When does it happen and which is the relationship between the adjunction and the free monad after all?

  • $\begingroup$ What do you mean by an endofunctor "having the form of an adjunction"? $\endgroup$ – Mike Shulman May 23 '10 at 6:14
  • $\begingroup$ Being for exemple in the form $U\circ{F}$ where $F\dashv{U}$. $\endgroup$ – user6250 May 23 '10 at 8:41
  • $\begingroup$ By the way, the correct spelling is "example" not "exemple" $\endgroup$ – David White Jan 12 '13 at 19:47

There seems to be some confusion here regarding language. In your first paragraph, you define the free monad generated by an endofunctor $R$, i.e. one first fixes $R$ then gets the monad. A good toy example for such $R$ would be $U\circ F$ where $F:X\to Y$ is the free functor into some category $Y$ which is well understood and $U:Y\to X$ is the forgetful functor. So the answer to your first question is yes, it can take this form and that's a nice case to play with. But in general $R$ need not take this form. Theorem 5.5 in the paper you cite gives a sufficient condition on an endofunctor $R$ so that free monad on $R$ exists. That condition has nothing to do with $R$ coming from an adjunction.

I can interpret your second question in a couple of ways. One way is as "when is $T$ of the form $U\circ F$?'' This is trivial; it's well-known that a monad $T$ is always of the form $U\circ F$ (see wikipedia), often for more than one choice of $U,F$. A better way to interpret your question is "given a monad $T$, how do I determine if it's the free monad generated by some $R$ which takes the form $U\circ F$?" This appears to be a non-trivial problem. Theorem 5.4 in the paper gives one situation where you can determine that your monad is free and recover $R$, but this doesn't classify all such situations. Anyway, even in this nice case where you get $R$ in hand, there are plenty of times such $R$ could fail to be in the form $U\circ F$. For instance, this MO answer shows that any such $R$ would need to be a homotopy equivalence on the nerve of $X$. It should not be too hard to construct an endofunctor for which this fails, since there are plenty of self-maps of simplicial complexes which are not homotopy equivalences.


I put this idea:

a monad in a (small) category $\mathcal{C}$ is equivalent to a 2-funtor $Mnd\to Cat$ form the free monad $Mnd$ to the 2-category $Cat$. A concrete description of Mnd is: it has one object $\ast$, the hom-category Mnd(0, 0) is the category $\Delta$ of finite ordinals and order-preserving functions, and composition $\Delta \times \Delta\to \Delta$ is ordinal sum. We have the 2-subcategory $L\subset \Delta$ ($L$ for loop) with the unique object $\ast$, full as subcategory, locally discrete.THis inclusion induce the forgetful functor $U: (T, \mu, \eta)\mapsto T$. Now the Left-Kan-extention (as Cat-enriched functor i.e. 2-functor) give the left adjoint of $U$ or in poor words the free-monad of a endofunctor.


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