# nLab free monad

### Context

#### Higher algebra

higher algebra

universal algebra

## Definition

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:

$Mnd(C) \to PtEnd(C) \to End(C) \to [ob(C),C]$

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.

## Free finitary monads

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 preserved filtered colimits, to obtain:

$Mnd_f(C) \to PtEnd_f(C) \to End_f(C) \to [ob(C)_f,C]$

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.

###### Theorem

Any algebraically-free monad is free.

###### Proof

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 $F$.

###### Theorem

If $C$ is locally small and complete, then any free monad is algebraically-free.

###### Proof

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.

## Constructions

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.)

## References

• 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)

Revised on January 8, 2014 17:14:55 by Urs Schreiber (82.113.106.24)