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. since $D(\mathrm{FU}X,Y)\simeq C(\mathrm{UX},\mathrm{UY})\simeq D(X,Y)$ 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.

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

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 $\xi$,

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

Proof

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

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

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

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

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.

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

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

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.