Let be a complete, well-powered category, and let be a monad with unit and multiplication . Then there is a universal idempotent monad, giving a right adjoint to
Given a monad , define a functor as the equalizer and :
This acquires a monad structure (lemma; proof given below). It might not be an idempotent monad (although it will be if is left exact). However we can apply the process again, and continue transfinitely. Define , and if has been defined, put ; at limit ordinals , define to be the inverse limit of the chain
where ranges over ordinals less than .
Since is well-powered (i.e., since each object has only a small number of subobjects), the large limit
exists for each . Hence the large limit exists as an endofunctor. The underlying functor
reflects limits (irrespective of size), so acquires a monad structure defined by the limit. Let be the unit and the multiplication of .
Claim: is idempotent.
For this it suffices to check that . This may be checked objectwise. So fix an object , and for that particular , choose so large that projections and are isomorphisms. Clearly then , since factors through the equalizer . Then, since is a monad morphism, we have
Finally we must check that satisfies the appropriate universal property. Suppose is an idempotent monad with unit , and let be a monad map. We define by induction: given , we have
so that factors uniquely through the inclusion . This defines ; this is a monad map. The definition of at limit ordinals, where is a limit monad, is clear. Hence factors (uniquely) through the inclusion , as was to be shown.
See also this MathOverflow discussion. The discussion is reprised in the proof below (using the notation of the present article).
Proof of lemma
By naturality of the unit , we have that equalizes the pair . Thus factorizes uniquely through the equalizer as ; this defines the unit . We define a multiplication as the unique map that renders the left square below commutative:
where the existence of is ensured by showing that equalizes the pair . This in turn follows if we show that the right square is serially commutative. The top square of the series (involving components of ) commutes by naturality of . To see that the bottom square (involving components of ) commutes, notice that both legs of the square are -algebra maps, and so (by freeness of the domain ) it suffices to check the claim that
But the left-hand side is
whereas the right-hand side is
and since , the claim is verified.
From there, the monad axioms are tedious but straightforward to verify. (I may come back to this later.)