Monads Comonads and Their Algebras

Put $T=GF$ and $\mu =G\epsilon F$. Whiskering preserves naturality, so $T$ is an endofunctor and $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ are natural transformations.

At an object $A$, the two associativity composites are ${\mu }_A\circ T({\mu }_A)=G({\epsilon }_{FA}\circ FG({\epsilon }_{FA}))$ and ${\mu }_A\circ {\mu }_{TA}=G({\epsilon }_{FA}\circ {\epsilon }_{FGFA})$; naturality of $\epsilon$ at ${\epsilon }_{FA}:FGFA\to FA$ identifies the expressions inside $G$, so $\mu \circ T\mu =\mu \circ \mu T$.

Componentwise, ${\mu }_A\circ {\eta }_{TA}=G({\epsilon }_{FA})\circ {\eta }_{GFA}=1_{GFA}$ by the second triangle identity, while ${\mu }_A\circ T({\eta }_A)=G({\epsilon }_{FA}\circ F({\eta }_A))=1_{GFA}$ by the first; hence both unit laws hold and $(T,\eta ,\mu )$ is a monad.

Apply [L1] to the opposite adjunction $G^{\mathrm{o}\mathrm{p}}⊣F^{\mathrm{o}\mathrm{p}}$. Reversing arrows translates its endofunctor $F^{\mathrm{o}\mathrm{p}}{G}^{\mathrm{o}\mathrm{p}}$, unit, and multiplication into $FG$, $\epsilon$, and $F\eta G$ on $\mathcal{D}$.

The translated associativity equation is $H\delta \circ \delta =\delta H\circ \delta$, and the translated unit equations are $H\epsilon \circ \delta =1_H=\epsilon H\circ \delta$; these are precisely coassociativity and the two counit laws, so the displayed data form a comonad.

For an algebra $(A,a)$, functoriality gives $T(1_A)=1_{TA}$, so $1_A\circ a=a=a\circ T(1_A)$ and $1_A$ is an algebra homomorphism.

If $f:(A,a)\to (B,b)$ and $g:(B,b)\to (C,c)$ are algebra homomorphisms, then $(g\circ f)\circ a=g\circ b\circ T(f)=c\circ T(g)\circ T(f)=c\circ T(g\circ f)$, so their composite is one too.

Composition of these morphisms is the composition in $\mathcal{C}$; step 1.1 supplies its identities and step 2.1 its closure, while associativity and the identity laws are inherited from $\mathcal{C}$.

Define $F^T(A)=(TA,{\mu }_A)$ and $F^T(f)=T(f)$. Naturality of $\mu$ gives $T(f)\circ {\mu }_A={\mu }_B\circ T^2(f)$, so $T(f)$ is an algebra homomorphism; the functor laws follow from those of $T$.

At an algebra $(A,a)$ define the counit component ${ϵ}_{(A,a)}:=a:(TA,{\mu }_A)\to (A,a)$; the algebra associativity law makes it an algebra homomorphism and the homomorphism equation makes these components natural. The equations $a\circ {\eta }_A=1_A$ and ${\mu }_A\circ T({\eta }_A)=1_{TA}$ are the two triangle identities, so $F^T⊣U^T$.

The composite $U^T{F}^T$ equals $T$, the adjunction unit is $\eta$, and $U^TϵF^T$ has component ${ϵ}_{(TA,{\mu }_A)}={\mu }_A$; hence the induced multiplication is $\mu$ and the induced monad is the given one on the nose.

Expanding the definitions gives $(h\star g)\star f={\mu }_D\circ T({\mu }_D)\circ T^2(h)\circ T(g)\circ f$, while $h\star (g\star f)={\mu }_D\circ T(h)\circ {\mu }_C\circ T(g)\circ f$.

Naturality of $\mu$ at $h$ rewrites $T(h)\circ {\mu }_C$ as ${\mu }_{TD}\circ T^2(h)$, and the monad associativity equation ${\mu }_D\circ T({\mu }_D)={\mu }_D\circ {\mu }_{TD}$ identifies the two expansions in step 1.1.

Naturality of $\eta$ and the two distinct unit laws give $f\star {\eta }_A={\mu }_B\circ T(f)\circ {\eta }_A={\mu }_B\circ {\eta }_{TB}\circ f=f$ and ${\eta }_B\star f={\mu }_B\circ T({\eta }_B)\circ f=f$, so the asserted identities hold.

Define $F_{T}A=A$ and $F_T(f)={\eta }_B\circ f:A\to TB$. Define $U_{T}A=TA$, and for a Kleisli arrow $g:A\to TB$ put $U_T(g)={\mu }_B\circ T(g):TA\to TB$.

The Kleisli identity and associativity laws show that both assignments preserve identities and composition. Moreover ${\mathcal{C}}_T(F_{T}A,B)=\mathcal{C}(A,TB)=\mathcal{C}(A,U_{T}B)$ naturally in $A$ and $B$, so they form an adjunction.

The unit is $\eta$. The counit at a Kleisli object $B$ is represented by $1_{TB}:TB\to TB$, and applying $U_T$ gives ${\mu }_B$; the Kleisli unit laws are the triangle identities. Thus $U_T{F}_T=T$ and the induced unit and multiplication are $\eta$ and $\mu$.

A Kleisli arrow $f:A\to TB=UFB$ has adjoint transpose ${\epsilon }_{FB}\circ F(f):FA\to FB$; define $J$ by this formula and by $J(A)=FA$.

For an identity ${\eta }_A$, the first triangle identity gives $J({\eta }_A)={\epsilon }_{FA}\circ F({\eta }_A)=1_{FA}$. For $f:A\to TB$ and $g:B\to TC$, substitution of $\mu =U\epsilon F$, naturality of $\epsilon$, and the triangle identities gives $J(g\star f)=J(g)\circ J(f)$, so $J$ is a functor.

The formulas give $J{F}_T=F$ and $UJ=U_T$ on objects and arrows, and $J$ sends the Kleisli counit to $\epsilon$. Conversely, those equalities force the image of $f:A\to UFB$ to be its adjoint transpose, so they force the formula of step 1.1 on every arrow and prove uniqueness.

For $d\in \mathcal{D}$ set $K(d)=(Ud,U{\epsilon }_d)$, and for $h:d\to d^{\mathrm{\prime }}$ set $K(h)=Uh$.

The triangle identity gives $U{\epsilon }_d\circ {\eta }_{Ud}=1_{Ud}$. Naturality of $\epsilon$ at ${\epsilon }_d$ gives the algebra associativity equation for $U{\epsilon }_d$, and naturality at $h$ gives $Uh\circ U{\epsilon }_d=U{\epsilon }_{d^{\mathrm{\prime }}}\circ T(Uh)$; hence the objects and arrows in step 1.1 are algebras and algebra homomorphisms.

Directly $U^{T}K=U$, and $KFA=(UFA,U{\epsilon }_{FA})=(TA,{\mu }_A)=F^{T}A$; the Eilenberg–Moore counit at $K(d)$ is $U{\epsilon }_d$, so the counit data also agree. These strict equalities force the underlying arrow action and every structure map, proving uniqueness.

Apply [L1] to the supplied adjunction; its unique factorisation is the required morphism from the Kleisli resolution.

Apply [L2] to the same adjunction; its unique comparison is the required morphism to the Eilenberg–Moore resolution.

Steps 1.1 and 1.2 give the two asserted existence-and-uniqueness properties for each supplied adjunction. This objectwise assertion does not form a category of all resolutions.

The composite induced endofunctor is $G{G}^{\mathrm{\prime }}{F}^{\mathrm{\prime }}F=GF$, and [L1] with ${\eta }^{\mathrm{\prime }}=1$ gives $\bar{\eta }=\eta$.

Its multiplication is $G{G}^{\mathrm{\prime }}\bar{\epsilon }{F}^{\mathrm{\prime }}F$. Expanding $\bar{\epsilon }$ by [L1], then using $G^{\mathrm{\prime }}{F}^{\mathrm{\prime }}=1_{\mathcal{D}}$ and $G^{\mathrm{\prime }}{\epsilon }^{\mathrm{\prime }}{F}^{\mathrm{\prime }}=1$, reduces this transformation to $G\epsilon F$.

The endofunctor, unit, and multiplication obtained in steps 1.1 and 2.1 are exactly $GF$, $\eta$, and $G\epsilon F$, so the two induced monads agree on the nose.

The comparison sends $A$ to $(TA,{\mu }_A)$ and a Kleisli arrow $f:A\to TB$ to the algebra homomorphism ${\mu }_B\circ T(f):TA\to TB$.

For every $A,B$, the map $f\mapsto {\mu }_B\circ T(f)$ is a bijection from $\mathcal{C}(A,TB)={\mathcal{C}}_T(A,B)$ to the algebra homomorphisms $(TA,{\mu }_A)\to (TB,{\mu }_B)$; its inverse sends $h$ to $h\circ {\eta }_A$. The monad unit laws show the two composites are identities, using the algebra-homomorphism equation for $h$ in one direction.

Thus every induced hom-set map is bijective, so $M$ is fully faithful (Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors); step 1.1 also shows that its objects are precisely the free algebras, giving the claimed strict image.

The maps $a_j\circ T(p_j):TL\to U^{T}Dj$ form a cone because every arrow of $D$ is an algebra homomorphism. By [L2] there is a unique $\mathrm{\ell }:TL\to L$ with $p_j\circ \mathrm{\ell }=a_j\circ T(p_j)$ for every $j$.

After composition with every $p_j$, the equations $\mathrm{\ell }\circ {\eta }_L=1_L$ and $\mathrm{\ell }\circ T(\mathrm{\ell })=\mathrm{\ell }\circ {\mu }_L$ become the corresponding algebra laws for $a_j$ by naturality; the limit legs are jointly monic by uniqueness in [L2], so both equations hold and $(L,\mathrm{\ell })$ is a $T$-algebra.

If a cone from an algebra $(X,x)$ has underlying mediating arrow $u:X\to L$, then composing $u\circ x$ and $\mathrm{\ell }\circ T(u)$ with every $p_j$ gives the same map because each cone leg is an algebra homomorphism; joint monicity makes $u$ an algebra homomorphism. Its uniqueness and the uniqueness of $\mathrm{\ell }$ follow from the base limit, giving exactly the strict lift required by [L1], including when $\mathcal{J}$ is empty.

For any small diagram in the Eilenberg–Moore category of the induced monad on $\mathbf{S}\mathbf{e}\mathbf{t}$, [L1] supplies its underlying limit and [L2] lifts that limit; hence this Eilenberg–Moore category is complete.

Monadicity says that the comparison $K:\mathcal{D}\to {\mathbf{S}\mathbf{e}\mathbf{t}}^T$ is an equivalence. By [L3], each small limit from step 1.1 transports across $K$, so $\mathcal{D}$ is complete.

Since $T$ preserves the colimit, the maps $i_j\circ a_j:T{U}^{T}Dj\to Q$ induce a unique $a:TQ\to Q$ satisfying $a\circ T(i_j)=i_j\circ a_j$ for every $j$.

Precomposition with every $i_j$ proves $a\circ {\eta }_Q=1_Q$. Precomposition with every $T^2(i_j)$ proves $a\circ T(a)=a\circ {\mu }_Q$; these families are jointly epic because $T$ and $T^2$ preserve the colimit, so $(Q,a)$ is an algebra and every $i_j$ is an algebra homomorphism.

Any algebra cocone has a unique underlying mediating arrow $u:Q\to X$; precomposition with every $T(i_j)$ shows $u\circ a=x\circ T(u)$, so $u$ is automatically an algebra homomorphism. This proves the lifted universal property and the conditional creation claim.

Let $f:(A,a)\to (B,b)$ be an algebra homomorphism whose underlying morphism has inverse $g$. From $f\circ a=b\circ T(f)$, the inverse equations, and functoriality, one obtains $g\circ b=a\circ T(g)$ by composing with $f$ and cancelling it; hence $g$ is an algebra homomorphism.

Therefore the Eilenberg–Moore forgetful functor reflects isomorphisms: an underlying inverse is automatically an inverse inside the algebra category by step 1.1.

For a monadic $U$, its comparison $K$ is an equivalence and hence fully faithful. If $U(f)=U^{T}K(f)$ is an isomorphism, step 2.1 makes $K(f)$ an isomorphism, and full faithfulness reflects that isomorphism back to $f$; thus $U$ is conservative.

If $\mu$ is a natural isomorphism, each ${\mu }_A$ is an isomorphism and therefore monic, proving $1\Rightarrow 2$.

If every ${\mu }_A$ is monic, the monad unit laws give ${\mu }_A\circ {\eta }_{TA}=1_{TA}={\mu }_A\circ T({\eta }_A)$; cancellation by [L1] yields ${\eta }_{TA}=T({\eta }_A)$ for every $A$, proving $2\Rightarrow 3$.

Assume $\eta T=T\eta$. The unit law makes ${\eta }_{TA}$ a right inverse to ${\mu }_A$. Naturality of $\eta$ at ${\mu }_A$, followed by ${\eta }_{T^{2}A}=T({\eta }_{TA})$, gives ${\eta }_{TA}\circ {\mu }_A=T({\mu }_A)\circ T({\eta }_{TA})=1_{T^{2}A}$; thus ${\eta }_{TA}$ is also a left inverse, every ${\mu }_A$ is invertible, and $3\Rightarrow 1$.

If $a:TA\to A$ is an algebra structure, then $a\circ {\eta }_A=1_A$. Naturality of $\eta$ at $a$ and idempotence, which gives ${\eta }_{TA}=T({\eta }_A)$, imply ${\eta }_A\circ a=T(a)\circ {\eta }_{TA}=T(a\circ {\eta }_A)=1_{TA}$; hence $a={\eta }_A^{-1}$.

Conversely, if ${\eta }_A$ is invertible, put $a={\eta }_A^{-1}$. The unit axiom is immediate, and naturality of $\eta$ together with the monad unit laws gives $a\circ T(a)=a\circ {\mu }_A$; step 1.1 shows this structure is unique.

Naturality of $\eta$ shows every base morphism between two such fixed objects commutes with their inverse-unit algebra structures, so the subcategory is full. The unit ${\eta }_A:A\to TA$ is universal from $A$ to this subcategory, since $TA$ is fixed and every map to a fixed object extends uniquely through ${\eta }_A$; hence $T$ is its reflector in the sense of Reflective full subcategory and reflector.

The canonical comparison is fully faithful, and its strict image consists of the free algebras.

For each algebra $(A,a)$, the idempotent-algebra theorem gives $a={\eta }_A^{-1}$. The map ${\eta }_A:(A,a)\to (TA,{\mu }_A)$ is therefore a specified algebra isomorphism from $(A,a)$ to the free algebra on $A$, with inverse $a$.

The specified isomorphisms of step 1.2 make the fully faithful comparison of step 1.1 split essentially surjective; [L1] therefore makes it an equivalence.

By [L1] and [L3], the reflection induces the monad $T=IR$ on $\mathcal{C}$ with multiplication $I\epsilon R$. By [L4] this multiplication is a natural isomorphism, so $T$ is idempotent (Idempotent monad).

By [L2], the Eilenberg–Moore category of $T$ is the full subcategory of objects $C$ for which the reflection unit $C\to IRC$ is invertible. The comparison sends $A\in \mathcal{A}$ to $(IA,I{\epsilon }_A)$, and its essential image is exactly this fixed-object subcategory.

Fullness of $I$ makes the comparison fully faithful, and every algebra $(C,{\eta }_C^{-1})$ has the specified isomorphism ${\eta }_C:C\to IRC$ to the comparison image of $RC$. Thus the comparison is an equivalence, so $I$ is monadic by Monadic and strictly monadic functors.

The identity $1_T:T\Rightarrow T$ satisfies $1_T{\eta }^T={\eta }^T$ and $1_T{\mu }^T={\mu }^{T}T{1}_T(1_T)T$, so it is a monad morphism.

If $\alpha :T\Rightarrow S$ and $\beta :S\Rightarrow R$ are monad morphisms, then $(\beta \alpha ){\eta }^T=\beta {\eta }^S={\eta }^R$, so their vertical composite preserves the unit.

Naturality of $\beta$ gives $\beta S\circ S\alpha =R\alpha \circ \beta T$; substituting the multiplication equations for $\alpha$ and $\beta$ yields $(\beta \alpha ){\mu }^T={\mu }^{R}R(\beta \alpha )(\beta \alpha )T$, so the composite preserves multiplication.

By [L2], vertical composition is associative and the transformations in step 1.1 are identities. Steps 2.1 and 3.1 give closure under composition under the stated endofunctor-category size condition. Hence these objects and arrows form a category.

For an $S$-algebra $(A,a)$, put $a_T=a{\alpha }_A:TA\to A$. By [L1]–[L2], $a_T{\eta }_A^T=a{\eta }_A^S=1_A$, while naturality of $\alpha$ and the two multiplication equations give $a_T{\mu }_A^T=a_{T}T(a_T)$. Thus $(A,a_T)$ is a $T$-algebra.