### Context

#### Algebra

higher algebra

universal algebra

## Theorems

#### 2-Category theory

2-category theory

## Definition

###### Definition

An idempotent monad is a monad $\left(T,\mu ,\eta \right)$ on a category $C$ such that one (hence all) of the following equivalent statements are true:

1. $\mu :TT\to T$ is a natural isomorphism.

2. All components of $\mu :TT\to T$ are monomorphisms.

3. The maps $T\eta ,\eta T:T\to TT$ are equal.

4. For every $T$-algebra ($T$-module) $\left(M,u\right)$, the corresponding $T$-action $u:TM\to M$ is an isomorphism.

5. The forgetful functor ${C}^{T}\to C$ (where ${C}^{T}$ is the Eilenberg-Moore category of $T$-algebras) is a full and faithful functor.

6. There exists a pair of adjoint functors $F⊣U$ such that the induced monad $\left(\mathrm{UF},UϵF\right)$ is isomorphic to $\left(T,\mu \right)$ and $U$ is a full and faithful functor.

###### Proof of equivalence (in more than one way).

$1⇒2$ is trivial.

$2⇒3$ Compositions $\mu \circ T\eta$ and $\mu \circ \eta T$ are always the identity (unit axioms for the monad), and in particular agree; if $\mu$ has all components monic, this implies $T\eta =\eta T$.

$3⇒4$ Compatibility of action and unit is $u\circ {\eta }_{M}={\mathrm{id}}_{M}$, hence also $T\left(u\right)\circ T\left({\eta }_{M}\right)={\mathrm{id}}_{TM}$. If $T\eta =\eta T$ then this implies ${\mathrm{id}}_{M}=T\left(u\right)\circ {\eta }_{TM}={\eta }_{M}\circ U$, where the naturality of $\eta$ is used in the second equality. Therefore we exhibited ${\eta }_{M}$ both as a left and a right inverse of $u$.

$4⇒1$ If every action is iso, then the components of multiplication ${\mu }_{M}:TTM\to TM$ are isos as a special case, namely of the free action on $TM$.

$4⇒5$ For any monad $T$, the forgetful functor from Eilenberg-Moore category ${C}^{T}$ to $C$ is faithful: a morphism of $T$-algebras is always a morphism of underlying objects in $C$. To show that it is also full, we consider any pair $\left(M,u\right)$, $\left(M\prime ,u\prime \right)$ in ${C}^{T}$ and must show that any $f:M\to M\prime$ is actually a map $f:\left(M,u\right)\to \left(M\prime ,u\prime \right)$; i.e. $u\prime \circ Tf=f\circ u$. But we know that ${\eta }_{M},{\eta }_{M\prime }$ are inverses of $u,u\prime$ respectively and the naturality for $\eta$ says ${\eta }_{M\prime }\circ f=Tf\circ {\eta }_{M}$. Compose that equation with $u$ on the right and $u\prime$ on the left with the result (notice that we used just the invertibility of $u$).

$5⇒6$ Trivial, because the Eilenberg-Moore construction induces the original monad by the standard recipe.

$6⇒3$ By $6$ the counit $ϵ$ is iso, hence $UϵF$ has a unique 2-sided inverse; by triangle identities, $T\eta$ and $\eta T$ are both right inverses of $UϵF$, hence 2-sided inverses, hence they are equal.

$6⇒1$ If $F⊣U$ is an adjunction with $U$ fully faithful, then the counit $ϵ$ is iso. since $D\left(\mathrm{FU}X,Y\right)\simeq C\left(\mathrm{UX},\mathrm{UY}\right)\simeq D\left(X,Y\right)$ where the last equivalence is since $U$ is full and faithful; hence by essential unicity of the representing object there is an iso $\mathrm{FUX}\stackrel{\sim }{\to }X$.; let $X=Y$ then the adjoint of this identity is the counit of the adjunction; since the hom objects correspond bijectively, the counit is an isomorphism. Hence the multiplication of the induced monad $\mu =UϵF$ is also an iso.

Part 5 means that in such a case ${C}^{T}$ is, up to equivalence a full reflective subcategory of $C$. Conversely, the monad induced by any reflective subcategory is idempotent, so giving an idempotent monad on $C$ is equivalent to giving a reflective subcategory of $C$.

In the language of stuff, structure, property, an idempotent monad may be said to equip its algebras with properties only (since ${C}^{T}\to C$ is fully faithful), unlike an arbitrary monad, which equips its algebras with at most structure (since ${C}^{T}\to C$ is, in general, faithful but not full).

If $T$ is idempotent, then it follows in particular that an object of $C$ admits at most one structure of $T$-algebra, that this happens precisely when the unit ${\eta }_{X}:X\to TX$ is an isomorphism, and in this case the $T$-algebra structure map is ${\eta }_{X}^{-1}:TX\to X$. However, it is possible to have a non-idempotent monad for which any object of $C$ admits at most one structure of $T$-algebra, in which case $T$ can be said to equip objects of $C$ with property-like structure; an easy example is the monad on semigroups whose algebras are monoids.

###### Remark

Let us be in a $2$-category $K$. Part of the structure of an idempotent monad $\left(C,T,\eta ,\mu \right)$ in $K$ is of course an idempotent morphism $T:C\to C$. More precisely (Definition 1.1.9) considers $\mu$ as part of the structure such that an idempotent 1-cell has a 2-isomorphism $\mu :\mathrm{TT}\to T$ such that $\mu T=T\mu$. Equivalently an idempotent morphism is a normalized pseudofunctor from the two object monoid $\left\{*,e\right\}$ with ${e}^{2}=e$ to $K$.

Recall that a splitting of an idempotent $\left(T,\mu \right)$ consists of a pair of 1-cells $I:D\to C$ and $R:C\to D$ and a pair of 2-isomorphisms $a:\mathrm{RI}\to {\mathrm{id}}_{D}$ and $b:T\to \mathrm{IR}$ such that $\mu ={b}^{-1}\left(I\circ A\circ R\right)\left(b\circ b\right)$ where $\circ$ denotes horizontal composition of 2-cells. Equivalently an splitting of an idempotent is a limit or a colimit of the defining pseudofunctor. If $K$ has equalizers or coequalizers, then all its idempotents split.

Now if $\left(I,R,a,b\right)$ is a splitting of an idempotemt monad, then $R⊣I$ are adjoint. And in this case the splitting of an idempotent is equivalently an Eilenberg-Moore object for the monad $\left(C,T,\eta ,\mu \right)$. In this case $D$ is called an adjoint retract of $C$.

(Peter Johnstone, sketches of an elephant, B 1.1.9, p.248-249)

###### Remark

Equivalences (resp. cores) in an allegory are precisely those symmetric idempotents which are idempotent monads (resp. comonads). In an allegory the following statements are equivalent: all symmetric idempotents split, idempotent monads split, idempotent comonads split. A similar statement holds at least for some 2-categories.

(Peter Johnstone, sketches of an elephant, B 1.1.9, p.248-249)

Peter Johnstone, sketches of an elephant, B 1.1.9, p.248-249

## Algebras of an idempotent monad

###### Proposition

Let $\left(M,\eta ,\mu \right)$ be an idempotent monad on a category $E$. The following conditions on an object $e$ of $E$ are equivalent:

1. The object $e$ carries an $M$-algebra structure.

2. The unit $\eta e:e\to Me$ is a split monomorphism.

3. The unit $\eta e$ is an isomorphism.

(It follows from 3. that there is at most one algebra structure on $e$, given by $\xi =\left(\eta e{\right)}^{-1}:Me\to e$.)

###### Proof

The implication 1. $⇒$ 2. is immediate. Next, if $\xi :Me\to e$ is any retraction of $\eta e$, we have both $\xi \circ \eta e={1}_{e}$ and

$\begin{array}{ccccc}\eta e\circ \xi & =& \left(M\xi \right)\left(\eta Me\right)& & \text{naturality of}\phantom{\rule{thinmathspace}{0ex}}\eta \\ & =& \left(M\xi \right)\left(M\eta e\right)& & \text{see definitions above}\\ & =& M\left(\xi \circ \eta e\right)& & \text{functoriality}\\ & =& {1}_{Me}& & \end{array}$\array{ \eta e \circ \xi & = & (M \xi)(\eta M e) & & \text{naturality of}\, \eta \\ & = & (M \xi)(M \eta e) & & \text{see definitions above} \\ & = & M(\xi \circ \eta e) & & \text{functoriality} \\ & = & 1_{M e} & & }

so 2. implies 3. Finally, if $\eta e$ is an isomorphism, put $\xi =\left(\eta e{\right)}^{-1}$. Then $\xi \circ \eta e={1}_{e}$ (unit condition), and the associativity condition for $\xi$,

$\xi \circ \mu e=\xi \circ M\xi ,$\xi \circ \mu e = \xi \circ M \xi,

follows by inverting the naturality equation $\eta Me\circ \eta e=M\eta e\circ \eta e$. Thus 3. implies 1.

###### Theorem (Fakir)

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

$\mathrm{IdempotentMonad}\left(C\right)↪\mathrm{Monad}\left(C\right)$IdempotentMonad(C) \hookrightarrow Monad(C)
###### Proof

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

$M\prime ↪M\stackrel{\stackrel{uM}{\to }}{\underset{Mu}{\to }}MM.$M' \hookrightarrow M \stackrel{\overset{u M}{\to}}{\underset{M u}{\to}} M M.

This $M\prime$ acquires a monad structure. 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

$\dots ↪{M}_{\alpha }↪\dots ↪M$\ldots \hookrightarrow M_{\alpha} \hookrightarrow \ldots \hookrightarrow M

where $\alpha$ ranges over ordinals less than $\beta$. This defines the monad ${M}_{\alpha }$ inductively; below, we let ${u}_{\alpha }$ denote the unit of this monad.

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

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

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

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

reflects limits (irrespective of size), so $E=E\left(M\right)$ acquires a monad structure defined by the limit. Let $\eta :1\to E$ be the unit and $\mu :EE\to E$ the multiplication of $E$. For each $\alpha$, there is a monad map ${\pi }_{\alpha }:E\to {M}_{\alpha }$ defined by the limit projection.

###### Lemma

$E$ 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 ${\pi }_{\alpha }\left(c\right):E\left(c\right)\to {M}_{\alpha }\left(c\right)$ and ${\pi }_{\alpha }E\left(c\right):EE\left(c\right)\to {M}_{\alpha }E\left(c\right)$ are isomorphisms. In particular, ${\pi }_{\alpha }{\pi }_{\alpha }\left(c\right):EE\left(c\right)\to {M}_{\alpha }{M}_{\alpha }\left(c\right)$ is invertible.

Now ${u}_{\alpha }{M}_{\alpha }\left(c\right)={M}_{\alpha }{u}_{\alpha }c$, since ${\pi }_{\alpha }:E\to {M}_{\alpha }$ factors through the equalizer ${M}_{\alpha +1}↪{M}_{\alpha }$. Because ${\pi }_{\alpha }$ is a monad morphism, we have

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

as required.

Finally we must check that $M↦E\left(M\right)$ 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

$\left({u}_{\alpha }{M}_{\alpha }\right){\varphi }_{\alpha }={\varphi }_{\alpha }{\varphi }_{\alpha }\left(vT\right)={\varphi }_{\alpha }{\varphi }_{\alpha }\left(Tv\right)=\left({M}_{\alpha }{u}_{\alpha }\right){\varphi }_{\alpha }$(u_\alpha M_\alpha)\phi_\alpha = \phi_\alpha \phi_\alpha (v T) = \phi_\alpha \phi_\alpha (T v) = (M_\alpha u_{\alpha})\phi_\alpha

so that ${\varphi }_{\alpha }$ factors uniquely through the inclusion ${M}_{\alpha +1}↪{M}_{\alpha }$. 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 $E\left(M\right)↪M$, as was to be shown.

### Examples

Let $A$ be a commutative ring, and let $f:A\to B$ be a flat (commutative) $A$-algebra. Then the forgetful functor

${f}^{*}={\mathrm{Ab}}^{f}:{\mathrm{Ab}}^{B}\to {\mathrm{Ab}}^{A}$f^\ast = Ab^f\colon Ab^B \to Ab^A

from $B$-modules to $A$-modules has a left exact left adjoint ${f}_{!}=B{\otimes }_{A}-$. The induced monad ${f}^{*}{f}_{!}$ on the category of $B$-modules preserves equalizers, and so its associated idempotent monad $T$ may be formed by taking the equalizer

$T\left(M\right)\to B{\otimes }_{A}M\stackrel{\stackrel{{f}^{*}{f}_{!}\eta M}{\to }}{\underset{\eta {f}^{*}{f}_{!}M}{\to }}B{\otimes }_{A}B{\otimes }_{A}M$T(M) \to B \otimes_A M \stackrel{\overset{f^\ast f_! \eta M}{\to}}{\underset{\eta f^\ast f_! M}{\to}} B \otimes_A B \otimes_A M

(To be continued. This example is based on how Joyal and Tierney introduce effective descent for commutative ring homomorphisms, in An Extension of the Galois Theory of Grothendieck. I would like to consult that before going further – Todd.)