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In a category with equalizers, the equalizer of ${\eta }_{TA},T{\eta }_A$ defines a functor $R$ by ${\beta }_{B}Rf=Tf{\beta }_A$. It has unit $\alpha$ and multiplication $\theta$ characterized byCancelling the monomorphisms ${\beta }_A$ reduces its unit and associativity laws to those of $T$, so $\beta :R\Rightarrow T$ is a monad morphism. The split equalizers associated with a monad show that ${\alpha }_{TA}$ is invertible. If every $T{\beta }_A$ is monic, then every $T{\alpha }_A$ is invertible, the naturality square of $\beta$ at ${\beta }_A$ is a pullback in a category, and $R$ is an idempotent monad. This monicity hypothesis is essential to this last argument; arbitrary functors need not preserve monomorphisms.

For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, with unit $\eta$ and counit $\epsilon$, call the adjunction idempotent when the multiplicationof its induced monad is invertible. We show that this idempotent adjunction condition is equivalent to invertibility of $\delta =L\eta R:LR\to (LR{)}^2$, the comultiplication of the induced comonad.

For an idempotent monad $T$, the unit laws implyIf $a:TA\to A$ is a monad algebra, naturality of the unit and its algebra unit law giveThus ${\eta }_A$ is an isomorphism. Apply this to the canonical algebra $R{\epsilon }_B:T(RB)\to RB$: its unit law is a triangle identity for an adjunction. It follows that ${\eta }_{RB}$, and therefore $L{\eta }_{RB}$, is invertible for every $B$. Hence $\delta$ is invertible.

Apply the same argument to the dual adjunction $R^{\mathrm{op}}⊣L^{\mathrm{op}}$. It gives the converse. Idempotence is self-dual: the induced monad is idempotent exactly when the induced comonad is idempotent. In particular, idempotence requires the unit to be invertible on every object in the image of the right adjoint.

Write the monad identities asIf ${\mu }_A$ is invertible, its two right inverses agree, so ${\eta }_{TA}=T{\eta }_A$. Conversely, suppose ${\eta }_T=T\eta$. Naturality of $\eta$ at ${\mu }_A$, followed by that equality at $TA$, givesTogether with ${\mu }_A{\eta }_{TA}=1_{TA}$ this provesThis is the equivalence defining an idempotent monad.

For the equalizer submonad of a monad, let ${\beta }_A:RA\to TA$ equalize ${\eta }_{TA}$ and $T{\eta }_A$. If $f:A\to B$, naturality of these two transformations shows that $Tf{\beta }_A$ equalizes the corresponding pair at $B$. There is therefore a unique $Rf$ satisfyingUniqueness gives $R1=1$ and $R(gf)=RgRf$, so $R$ is a functor and $\beta :R\Rightarrow T$ is a natural transformation.

Take the stipulated factorizations ${\alpha }_A$ and ${\theta }_A$. Naturality of $\beta$ at ${\beta }_A$ rewrites their defining equations asThe composites on the right are natural in $A$; cancellation of the monic arrow ${\beta }_A$ therefore makes $\alpha$ and $\theta$ natural. We now verify every monad law, rather than assuming that restricting the structure automatically preserves them.

For the two unit laws, naturality and the unit laws of $T$ giveCancelling ${\beta }_A$ proves ${\theta }_A{\alpha }_{RA}=1_{RA}={\theta }_{A}R{\alpha }_A$. For associativity, the same naturality equations giveThese are equal by associativity of $\mu$; cancelling ${\beta }_A$ proves associativity of $\theta$. Thus

The remaining claims use two split equalizers associated with a monad. First, ${\eta }_{TA}:TA\to TTA$ is a split equalizer of ${\eta }_{TTA},T{\eta }_{TA}$. Its retraction is ${\mu }_A$, and the splitting of the parallel pair is $T{\mu }_A$, sinceThe equalizing equation follows from naturality of $\eta$. In general, identities $re=1$, $sf=er$, $sg=1$ and $fe=ge$ prove that $e$ is an equalizer: if $fh=gh$, then $h=erh$, with unique factorization $rh$. Hence ${\eta }_{TA}$ and ${\beta }_{TA}$ are two equalizers of the same pair. Their comparison is ${\alpha }_{TA}$, and its inverse is ${\mu }_A{\beta }_{TA}$. In particular

Second, $e_A=T{\eta }_A:TA\to TTA$ is a split equalizer of $TT{\eta }_A,T{\eta }_{TA}$, with retraction ${\mu }_A$ and splitting ${\mu }_{TA}$. Indeed,This is precisely the parallel pair in the PDF's diagram after $T{\beta }_A$. The map $T{\beta }_A$ equalizes that pair, because applying $T$ to the defining equalizer equation gives their equality on $T{\beta }_A$. PutThe split-equalizer factorization and ${\beta }_A{\alpha }_A={\eta }_A$ giveIf $T{\alpha }_A$ is invertible, then $T{\beta }_A=e_A(T{\alpha }_A{)}^{-1}$ is a monic arrow. Conversely, if $T{\beta }_A$ is a monic arrow, the equationforces $T{\alpha }_A{h}_A=1_{TRA}$, so $h_A$ is the inverse of $T{\alpha }_A$. ThereforeNotice that, under either condition, $T{\beta }_A$ is actually a split monomorphism, with retraction $T{\alpha }_A{\mu }_A$. Thus $TT{\beta }_A$ is a monic arrow too, since a functor preserves split monomorphisms. We have not assumed that $T$ preserves arbitrary monomorphisms.

Now prove the specified naturality square is a pullback in a category. Let $x:X\to TRA$ and $y:X\to RTA$ satisfy $T{\beta }_{A}x={\beta }_{TA}y$. Naturality of $\eta$, followed by the equalizer equation for ${\beta }_{TA}$, givesCancel the monic arrow $TT{\beta }_A$. Thus $x$ equalizes ${\eta }_{TRA},T{\eta }_{RA}$ and factors uniquely as $x={\beta }_{RA}z$ with $z:X\to RRA$. Naturality of $\beta$ then givesCancel ${\beta }_{TA}$ to obtain $R{\beta }_{A}z=y$. This is exactly existence and uniqueness in the pullback universal property.

Because the square is a pullback and $T{\beta }_A$ is a monic arrow, $R{\beta }_A$ is a monic arrow. Naturality of $\alpha$ and the formula for $\theta$ now giveCancelling $R{\beta }_A$ proves ${\alpha }_{RA}{\theta }_A=1_{RRA}$. The other composite is already one of the unit laws. Hence ${\theta }_A$ is invertible. If the equivalent monicity conditions hold for every $A$, we conclude

If the multiplication in the unit and multiplication of a monad $\mu$ is invertible, the two unit identities of the monadmake both $T\eta$ and $\eta T$ its inverse. Hence idempotence implies equality of the two unit insertions:This is the first implication in the characterization of an idempotent monad.

Use $\mathbb{N}=\{0,1,\dots \}$, as required by the formula at zero, and composition of functions as multiplication in $M$. The displayed $T$ preserves identities, and for $n>0$,both sides send zero to zero. Thus $T$ is a functor on the one-object category.

For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations $Tf\eta =\eta f$ and $\mu T^{2}f=Tf\mu$ hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at $n=0,1$, and are $f(n-2)+1$ for $n\ge 2$. The two unit laws are $\mu \eta =\mu T\eta =1$. Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since $\mu (0)=\mu (1)$, so $\mu$ cannot be invertible.

An algebra for a monad is a map $a\in M$ with $a\eta =1$ and $aTa=a\mu$. The first equation forces $a(n+1)=n$; monotonicity then forces $a(0)=0$. Thus $a=\mu$, which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms $h$ satisfy $h\mu =\mu Th$, and this equation holds precisely when $h(0)=0$: evaluate at zero for necessity, and at $n>0$ both sides equal $h(n-1)$.

The Kleisli comparison functor takes its only object to this only algebra and sends $f\in M$ toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse $h\mapsto (n\mapsto h(n+1))$. It preserves identities and composition by the comparison construction; directly, $K(\eta )=1$ and $K(g\star f)=K(g)K(f)$. Bijectivity on objects and arrows makes it an isomorphism of categories:

For $L⊣R$ inducing $T$, the Kleisli comparison functor is always full and faithful, since the adjunction givesTherefore it is part of an equivalence of categories if and only if every object of $\mathcal{D}$ is isomorphic to $LA$ for some $A$.

For an idempotent monad, multiplication is invertible. The unit laws imply $T\eta =\eta T={\mu }^{-1}$. If $(A,a)$ is an algebra for a monad, then $a{\eta }_A=1_A$, while naturality of $\eta$ givesThus $a:TA\to A$ is an isomorphism with inverse ${\eta }_A$. The algebra law $a{\mu }_A=aT(a)$ says that $a$ is a morphism of algebras for a monad from the free algebra $(TA,{\mu }_A)$ to $(A,a)$, so every object of the Eilenberg-Moore category is isomorphic to a free algebra. The criterion above yields

Let $M$ be the monoid of order-preserving functions $\mathbb{N}\to \mathbb{N}$, with $\mathbb{N}=\{0,1,\dots \}$, regarded as a one-object category. The endofunctor $T$ fixes that object and sends $f$ to $Tf(0)=0$, $Tf(n+1)=f(n)+1$. The displayed maps give its monad unit and multiplication. The multiplication is not injective, so this is not an idempotent monad. Its only algebra for a monad is $\mu$, because $a\eta =1$ forces $a(n+1)=n$ and monotonicity forces $a(0)=0$. Algebra endomorphisms are exactly the functions fixing zero. The Kleisli comparison functor sends $f$ to the function which is zero at zero and equals $f(n)$ at $n+1$; its inverse sends $h$ to $n\mapsto h(n+1)$. Thus the comparison is an isomorphism of categories even though the monad is not idempotent.