Associated idempotent monad of a monad

Let $C$ be a complete, well-powered category, and let $M:C\to C$ be a monad with unit $u:1\to M$ and multiplication $m:MM\to M$. Then there is a universal idempotent monad, giving a right adjoint to

$$\mathrm{IdempotentMonad}(C)\hookrightarrow \mathrm{Monad}(C)$$

Given a monad $M$, define a functor $M\prime $ as the equalizer $Mu$ and $uM$:

$$M\prime \hookrightarrow M\stackrel{\stackrel{uM}{\to}}{\underset{Mu}{\to}}MM.$$

This $M\prime $ acquires a monad structure (lemma; proof given below). It might not be an idempotent monad (although it will be if $M$ is left exact). However we can apply the process again, and continue transfinitely. Define ${M}_{0}=M$, and if ${M}_{\alpha}$ has been defined, put ${M}_{\alpha +1}={M}_{\alpha}\prime $; at limit ordinals $\beta $, define ${M}_{\beta}$ to be the inverse limit of the chain

$$\dots \hookrightarrow {M}_{\alpha}\hookrightarrow \dots \hookrightarrow M$$

where $\alpha $ ranges over ordinals less than $\beta $.

Since $C$ is well-powered (i.e., since each object has only a small number of subobjects), the large limit

$$E(M)(c)=\underset{\alpha \in \mathrm{Ord}}{\mathrm{lim}}{M}_{\alpha}(c)$$

exists for each $c$. Hence the large limit $E(M)=\underset{\alpha \in \mathrm{Ord}}{\mathrm{lim}}{M}_{\alpha}$ exists as an endofunctor. The underlying functor

$$\mathrm{Monad}(C)\to \mathrm{Endo}(C)$$

reflects limits (irrespective of size), so $E=E(M)$ acquires a monad structure defined by the limit. Let $\eta :1\to E$ be the unit and $\mu :EE\to E$ the multiplication of $E$.

**Claim:**$E$ is idempotent.

For this it suffices to check that $\eta E=E\eta :E\to EE$. This may be checked objectwise. So fix an object $c$, and for that particular $c$, choose $\alpha $ so large that projections ${\pi}_{\alpha}(c):E(c)\to {M}_{\alpha}(c)$ and ${\pi}_{\alpha}E(c):EE(c)\to {M}_{\alpha}E(c)$ are isomorphisms. Clearly then ${u}_{\alpha}{M}_{\alpha}(c)={M}_{\alpha}{u}_{\alpha}c$, since ${\pi}_{\alpha}:E\to {M}_{\alpha}$ factors through the equalizer ${M}_{\alpha +1}\hookrightarrow {M}_{\alpha}$. Then, since ${\pi}_{\alpha}$ is a monad morphism, we have

$$\begin{array}{ccc}\eta E(c)& =& ({\pi}_{\alpha}{\pi}_{\alpha}(c){)}^{-1}(u{M}_{\alpha}(c)){\pi}_{c}\\ & =& ({\pi}_{\alpha}{\pi}_{\alpha}(c){)}^{-1}({M}_{\alpha}u(c)){\pi}_{c}\\ & =& E\eta (c)\end{array}$$

as required.

Finally we must check that $M\mapsto E(M)$ satisfies the appropriate universal property. Suppose $T$ is an idempotent monad with unit $v$, and let $\varphi :T\to M$ be a monad map. We define $T\to {M}_{\alpha}$ by induction: given ${\varphi}_{\alpha}:T\to {M}_{\alpha}$, we have

$$({u}_{\alpha}{M}_{\alpha}){\varphi}_{\alpha}={\varphi}_{\alpha}{\varphi}_{\alpha}(vT)={\varphi}_{\alpha}{\varphi}_{\alpha}(Tv)=({M}_{\alpha}{u}_{\alpha}){\varphi}_{\alpha}$$

so that ${\varphi}_{\alpha}$ factors uniquely through the inclusion ${M}_{\alpha +1}\hookrightarrow {M}_{\alpha}$. This defines ${\varphi}_{\alpha +1}:T\to {M}_{\alpha +1}$; this is a monad map. The definition of ${\varphi}_{\alpha}$ at limit ordinals, where ${M}_{\alpha}$ is a limit monad, is clear. Hence $T\to M$ factors (uniquely) through the inclusion $E(M)\hookrightarrow M$, 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).

By naturality of the unit $u$, we have that $u:{1}_{C}\to M$ equalizes the pair $Mu,uM:M\overrightarrow{\to}MM$. Thus $u$ factorizes uniquely through the equalizer $i:M\prime \to M$ as $u=i\circ u\prime $; this defines the unit $u\prime :{1}_{C}\to M\prime $. We define a multiplication $m\prime :M\prime M\prime \to M\prime $ as the unique map that renders the left square below commutative:

$$\begin{array}{ccccc}M\prime M\prime & \stackrel{iM\prime}{\to}& MM\prime & \stackrel{\stackrel{uMM\prime}{\to}}{\underset{MuM\prime}{\to}}& MMM\prime \\ {}_{m\prime}\downarrow & & {}_{m\circ Mi}\downarrow & & {\downarrow}_{Mm\circ MMi}\\ M\prime & \underset{i}{\to}& M& \stackrel{\stackrel{uM}{\to}}{\underset{Mu}{\to}}& MM\end{array}$$

where the existence of $m\prime $ is ensured by showing that $m\circ Mi\circ iM\prime $ equalizes the pair $uM,Mu$. This in turn follows if we show that the right square is *serially* commutative. The top square of the series (involving components of $uM$) commutes by naturality of $u$. To see that the bottom square (involving components of $Mu$) commutes, notice that both legs of the square are $M$-algebra maps, and so (by freeness of the domain $MM\prime $) it suffices to check the claim that

$$Mm\circ MMi\circ MuM\prime \circ uM\prime =Mu\circ m\circ Mi\circ uM\prime .$$

But the left-hand side is

$$\begin{array}{ccc}M(m\circ Mi\circ uM\prime )\circ uM\prime & =& uM\circ (m\circ Mi\circ uM\prime )\\ & =& uM\circ (m\circ uM\circ i)\\ & =& uM\circ i\end{array}$$

whereas the right-hand side is

$$Mu\circ m\circ uM\circ i=Mu\circ i$$

and since $uM\circ i=Mu\circ i$, the claim is verified.

From there, the monad axioms are tedious but straightforward to verify. (I may come back to this later.)

Revised on November 8, 2013 08:25:20
by
Todd Trimble