nLab
idempotent monad

Definition

An idempotent monad is a monad $(T,\mu ,\eta )$ 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 monic.

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

4. For every $T$-algebra ($T$-module) $(M,u)$, 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 an adjunction $F⊣U$ such that the induced monad $(\mathrm{FU},FϵU)$ is isomorphic to $(T,\mu )$ and $U$ is fully faithful.

Proof of equivalence (in more than one way).

$1\Rightarrow 2$ is trivial.

$2\Rightarrow 3$ 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 $T\eta =\eta T$.

$3\Rightarrow 4$ Compatibility of action and unit is $u\circ {\eta }_M={\mathrm{id}}_M$, hence also $T(u)\circ T({\eta }_M)={\mathrm{id}}_{TM}$. If $T\eta =\eta T$ then this implies ${\mathrm{id}}_M=T(u)\circ {\eta }_{TM}={\eta }_M\circ U$, 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 $u$.

$4\Rightarrow 1$ 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 $TM$.

$4\Rightarrow 5$ 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 $(M,u)$, $(M\prime ,u\prime )$ in $C^T$ and must show that any $f:M\to M\prime$ is actually a map $f:(M,u)\to (M\prime ,u\prime )$; 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 $u$).

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

$6\Rightarrow 3$ 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.

$6\Rightarrow 1$ If $F⊣U$ is an adjunction with $U$ fully faithful, then the counit $ϵ$ is iso. The induced monad has multiplication $\mu =UϵF$ hence iso; every isomorphic monad is also 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.

Algebras of an idempotent monad

Proposition

Let $(M,\eta ,\mu )$ 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 $\xi =(\eta e{)}^{-1}:Me\to e$.)

Proof

The implication 1. $\Rightarrow$ 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

$$\begin{array}{ccccc}\eta e\circ \xi & =& (M\xi )(\eta Me)& & \text{naturality of}\eta \\ & =& (M\xi )(M\eta e)& & \text{see definitions above}\\ & =& M(\xi \circ \eta e)& & \text{functoriality}\\ & =& 1_{Me}& & \end{array}$$\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 $\xi =(\eta e{)}^{-1}$. 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.

The associated idempotent monad of a monad

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

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

Proof

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

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

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

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

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

$$\mathrm{Monad}(C)\to \mathrm{Endo}(C)$$Monad(C) \to 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 $E$. For each $\alpha$, there is a monad map ${\pi }_{\alpha }:E\to M_{\alpha }$ defined by the limit projection.

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

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

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

$$(u_{\alpha }{M}_{\alpha }){\varphi }_{\alpha }={\varphi }_{\alpha }{\varphi }_{\alpha }(vT)={\varphi }_{\alpha }{\varphi }_{\alpha }(Tv)=(M_{\alpha }{u}_{\alpha }){\varphi }_{\alpha }$$(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 $M_{\alpha +1}\hookrightarrow M_{\alpha }$. 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.

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

$$T(M)\to B{\otimes }_{A}M\stackrel{\stackrel{f^{*}{f}_{!}\eta M}{\to }}{\underset{\eta f^{*}{f}_{!}M}{\to }}B{\otimes }_{A}B{\otimes }_{A}M$$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.)

Mike Shulman: How about some examples of monads and their associated idempotent monads?

Do 2-monads have associated lax-, colax-, or pseudo-idempotent 2-monads?

References

• F. Borceux, Handbook of categorical algebra, vol.2, p. 196.

• P. Gabriel and M. Zisman, Calculus of Fractions and Homotopy Theory