symmetric monoidal (∞,1)-category of spectra
A free monad is a free object relative to a forgetful functor whose domain is a category of monads.
This general concept has many different specific incarnations, since there are potentially many different such forgetful functors. Suppose $C$ is a category, and write $Mnd(C)$ for the category whose objects are monads on $C$ and whose morphisms are natural transformations commuting with the monad structure maps; i.e. it is the category of monoids in the monoidal category of endofunctors with composition. Then we have a string of forgetful functors:
where $End(C)$ denotes the category of endofunctors of $C$, and $PtEnd(C)$ denotes the category of pointed endofunctors, i.e. endofunctors $F$ equipped with a natural transformation $Id\to F$. A free monad can then be considered as a free object relative to any one of these forgetful functors.
In general, these forgetful functors cannot be expected to have left adjoints, i.e. there will not be a "free monad functor", but individual objects can often be shown to generate free monads. One general case in which this is true is when $C$ is locally presentable and we consider monads and endofunctors which are accessible, i.e. preserve sufficiently highly filtered colimits. Suppose for the sake of argument that $C$ is locally finitely presentable (the higher-ary case is analogous). Then we can restrict the above string of forgetful functors to the finitary monads, i.e. those preserving filtered colimits, to obtain:
where the subscript $f$ denotes restriction to finitary things, and $ob(C)_f$ is the set of compact objects of $C$. In this case, all these forgetful functors do have left adjoints, and moreover at least the functors $Mnd_f(C) \to End_f(C)$ and $Mnd_f(C) \to [ob(C)_f,C]$ are monadic. (This is shown in the papers cited below.) The construction is by a convergent transfinite composition.
For example, the left adjoint to $Mnd_f(C) \to End_f(C)$, shows that there exists a “free finitary monad” on any finitary endofunctor. Note, though, that this does not automatically imply that the “free finitary monad” on a finitary endofunctor is also a “free monad” on that endofunctor, i.e. that as a free object it satisfies the requisite universal property relative to all objects of $Mnd(C)$, not merely those lying in $Mnd_f(C)$. It is, however, generally true that this is the case: free finitary monads are also free monads.
We say that a monad $T$ is algebraically-free on an endofunctor $F$ if the category $T Alg_{mnd}$ of $T$-algebras (in the sense of algebras for a monad) is equivalent to the category $F Alg_{endo}$ of $F$-algebras (in the sense of algebras for an endofunctor). N.B.: Any such equivalence must be an isomorphism $T Alg_{mnd} \cong F Alg_{endo}$, because the underlying functors from the categories of algebras in each case are amnestic isofibrations. See remarks at the article monadicity theorem on monadicity vis-à-vis strict monadicity.
A priori, being algebraically free is different from being free. However, one can show the following.
Any algebraically-free monad is free.
First observe that for a (perhaps pointed) endofunctor $F$ and a monad $T$, to give a functor $T Alg_{mnd} \to F Alg_{endo}$ over $C$ is equivalent to giving a (pointed) transformation $F\to T$, and if $F$ is a monad then such a functor takes values in $F Alg_{mnd}$ iff the transformation $F\to T$ is a monad morphism. Thus, if $F Alg_{endo} \cong T Alg_{mnd}$, then for any other monad $S$, (pointed) transformations $F\to S$ correspond to maps $S Alg_{mnd} \to F Alg_{endo} \cong T Alg_{mnd}$ and hence to monad morphisms $T\to S$, i.e. $T$ is free on $F$.
If $C$ is locally small and complete, then any free monad is algebraically-free.
For any object $x\in C$, the assumptions ensure that the codensity monad of $x$ exists. This is the right Kan extension of $x\colon 1\to C$ along itself, which we write as $\langle x,x\rangle$. The universal property of Kan extensions means that for any endofunctor $F$, to give a map $F x \to x$ (i.e. to make $x$ into an $F$-algebra) is the same as to give a natural transformation $F\to \langle x,x\rangle$. Moreover, one can check that if $F$ is a pointed endofunctor (resp. a monad), then the map $F x \to x$ is a pointed (resp. monad) algebra iff the corresponding transformation $F\to \langle x,x\rangle$ is a morphism of pointed endofunctors (resp. of monads). Therefore, if $T$ is the free monad on $F$, then applying its universal property in $Mnd(C)$ to the monad $\langle x,x\rangle$, we see that it is also algebraically-free.
Notice that this second proof relies crucially on the fact that free monads have a universal property relative to a forgetful functor whose domain is all of $Mnd(C)$, not just some subcategory of finitary or accessible monads, since $\langle x,x\rangle$ will not in general be finitary or accessible.
Perhaps the most general set-theoretically based construction of (algebraically) free monads is the transfinite construction of free algebras. (In type theory, it is natural to use instead higher inductive types.)
The monadicity of the above adjunctions can be used to give presentations of monads in terms of generators and relations. This has close connections with Lawvere theories and related ideas.
Free monads on pointed endofunctors play an important role in the construction of cofibrantly generated algebraic weak factorization systems.
Max Kelly, “A unified treatment of transfinite constructions for free algebras, free monoids, colimits, associated sheaves, and so on”
Max Kelly and John Power, “Adjunctions whose counits are coequalizers, and presentations of finitary enriched monads”
Nicola Gambino, Martin Hyland, section 6 of Wellfounded trees and dependent polynomial functors. In Types for proofs and programs, volume 3085 of Lecture Notes in Comput. Sci., pages 210–225. Springer-Verlag, Berlin, 2004 (web)
Last revised on August 8, 2018 at 11:53:01. See the history of this page for a list of all contributions to it.