Codex @codex  0

For $F⊣U$ with unit of an adjunction $\eta$ and counit of an adjunction $\epsilon$, putThe triangle identities for an adjunction giveFor associativity, naturality of $\epsilon$ at ${\epsilon }_{FA}:FUFA\to FA$ saysApply $U$ to obtain ${\mu }_{A}T({\mu }_A)={\mu }_A{\mu }_{TA}$. Thus this is the monad induced by an adjunction.

Conversely, start with a monad $(T,\eta ,\mu )$. The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad laws ensure that these are algebras for a monad and morphisms of algebras for a monad. For an algebra $(B,b)$, there are inverse bijectionsThe inverse really is an algebra map becauseFor a free-algebra map $g$, its algebra equation and the monad unit law giveFor a base map $f$, naturality and the algebra unit law give $bT(f){\eta }_A=b{\eta }_{B}f=f$. The formulas are natural, establishing the free-forgetful Eilenberg-Moore adjunction $F^T⊣U^T$. Its unit is ${\eta }_A$, and its counit at $(B,b)$ is $b$. Hence its induced multiplication is $\mu$, and every monad arises from an adjunction.

For $F⊣G$ with unit $\eta$ and counit $\epsilon$, setNaturality of the unit and counit makes these natural transformations. The two triangle identities give ${\mu }_A{\eta }_{TA}=1_{TA}$ and ${\mu }_{A}T{\eta }_A=1_{TA}$. Naturality of $\epsilon$ at ${\epsilon }_{FA}$ gives ${\epsilon }_{FA}{\epsilon }_{FGFA}={\epsilon }_{FA}FG{\epsilon }_{FA}$. Applying $G$ proves ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. Thus this is the monad induced by an adjunction.

Conversely, for any monad take the free-algebra functor $F^T(A)=(TA,{\mu }_A)$ and the forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$. For an algebra $(B,b)$, the assignments $h\mapsto h{\eta }_A$ and $k\mapsto bT(k)$ give inverse bijections between algebra maps $F^T(A)\to (B,b)$ and arrows $A\to B$. For the inverse identities, $bT(k){\eta }_A=b{\eta }_{B}k=k$, while $bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$. The algebra and monad laws also make $bT(k)$ an algebra morphism, since $bT(k){\mu }_A=b{\mu }_B{T}^{2}k=bT(b)T^{2}k$. These bijections are natural, giving the free-forgetful Eilenberg-Moore adjunction. Its unit is $\eta$, its counit at $(B,b)$ is $b$, and its induced multiplication is exactly $\mu$. Thus every monad arises from an adjunction.

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit natural transformation $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aTa=a{\mu }_A$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ satisfies $ha=bTh$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:U$ with $F⊣U$, the monad induced by an adjunction is $T=UF$, with $\mu =U\epsilon F$. The Eilenberg-Moore comparison functor is $K(D)=(UD,U{\epsilon }_D)$ and $K(h)=Uh$. The functor $U$ is monadic when this comparison is an equivalence of categories; this is a monadic adjunction.

The precise monadicity theorem says that $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, reflects isomorphisms, and satisfies the following condition: whenever the image of a parallel pair in $\mathcal{D}$ admits a split coequalizer in $\mathcal{C}$, the pair has a coequalizer in $\mathcal{D}$ and $U$ preserves that coequalizer. Equivalently, these coequalizers are created up to the canonical isomorphism identifying their underlying coequalizer objects. Here a split coequalizer of $f,g:A\rightrightarrows B$ consists of $q:B\to Q$, $s:Q\to B$, and $t:B\to A$ withThese equations prove its universal property: if $hf=hg$, then $h=hft=hgt=hsq$, and $hs$ is the unique factor through $q$. Every functor preserves these equations, so every functor preserves split coequalizers.

First consider the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. It has the free algebra functor $A\mapsto (TA,{\mu }_A)$ as a left adjoint: the inverse transposition formulas are $h\mapsto h{\eta }_A$ and $v\mapsto bTv$. An monad algebra morphism whose underlying arrow is invertible has an algebra-morphism inverse, by rearranging its structure equation. Thus $U^T$ reflects isomorphisms.

Suppose $f,g:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$. Applying $T$ gives a coequalizer of $Tf,Tg$. The equation $qbTf=qfa=qga=qbTg$ therefore gives a unique $c:TQ\to Q$ with $cTq=qb$. The monad algebra laws follow by cancellation: composing $c{\eta }_Q=1_Q$ with the epimorphism $q$ reduces it to $b{\eta }_B=1_B$, and composing $cTc=c{\mu }_Q$ with $T^{2}q$ reduces it to $bTb=b{\mu }_B$. Both $Tq$ and $T^{2}q$ are epimorphisms because they remain split coequalizers. Thus $(Q,c)$ is an algebra for a monad and $q$ is an monad algebra morphism. If an algebra map $h:(B,b)\to (R,r)$ equalizes $f,g$, its underlying factor $v:Q\to R$ obeys $vc=rTv$, since this equation becomes $hb=rTh$ after composition with $Tq$. This proves that $U^T$ creates the required coequalizers. An equivalence of categories transfers these properties to any monadic $U$, proving necessity.

Conversely assume the three conditions and write $T=UF$. Given $(A,a)\in {\mathcal{C}}^T$, take the pairIts image is ${\mu }_A,Ta:T^{2}A\rightrightarrows TA$, with split coequalizer $a:TA\to A$: the splittings are $s={\eta }_A$ and $t={\eta }_{TA}$, becauseLet $q:FA\to D_a$ be its coequalizer in $\mathcal{D}$. Preservation gives a canonical isomorphism $U{D}_a\cong A$ under which $Uq=a$. Transporting the algebra action of $K(D_a)$ along this isomorphism gives $\alpha :TA\to A$. Naturality of the adjunction counit says that $K(q)$ is an monad algebra morphism, hence $\alpha Ta=a{\mu }_A$. But $aTa=a{\mu }_A$ too, and $Ta$ is a split epimorphism, with section $T{\eta }_A$. Therefore $\alpha =a$. Every algebra is consequently isomorphic to some $K(D_a)$.

It remains to establish full faithfulness, rather than merely construct objects. For $B\in \mathcal{D}$, put $b=U{\epsilon }_B$. The adjunction counit ${\epsilon }_B:FUB\to B$ coequalizes ${\epsilon }_{FUB},Fb$. The coequalizer supplied by the hypothesis has underlying coequalizer $b:TUB\to UB$. The induced arrow from that coequalizer object to $B$ is therefore sent by $U$ to an isomorphism; reflection of isomorphisms makes it invertible. Thus ${\epsilon }_B$ itself is this coequalizer.

Now let $h:K(B)\to K(C)$ be an monad algebra morphism, and put $c=U{\epsilon }_C$. The arrow $v={\epsilon }_{C}Fh:FUB\to C$ coequalizes the two arrows in the preceding presentation. To check this directly, use transposition along $F⊣U$: the transposes of $vFb$ and $v{\epsilon }_{FUB}$ are respectively $hb$ and $cTh$, which agree by the algebra-morphism equation. Hence a unique $\bar{h}:B\to C$ satisfies $\bar{h}{\epsilon }_B=v$. Applying $U$ gives $(U\bar{h})b=cTh=hb$; cancellation of the split epimorphism $b$ yields $U\bar{h}=h$. Finally, any arrow $r:B\to C$ satisfies $r{\epsilon }_B={\epsilon }_{C}FUr$ by naturality, so two arrows with equal image under $U$ are equal by the coequalizer property. The comparison is full and faithful and essentially surjective, hence an equivalence of categories. This proves the theorem.

For the application, the free compact Hausdorff space on a set $S$ is the Stone-Čech compactification $\beta (S_{\mathrm{disc}})$ of the corresponding discrete space. Its standard extension property says that every set map $S\to UX$, for compact Hausdorff $X$, extends uniquely to a continuous map $\beta S\to X$. This constructs the required left adjoint. Also a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so $U$ reflects isomorphisms.

Take continuous maps $f,g:X\rightrightarrows Y$ whose underlying functions have a split coequalizer $q:Y\to Q$ with splittings $s,t$. The splittings need not be continuous. LetThe set $E$ is compact and hence closed. Since $qf=qg$, it is contained in $R$. Conversely $(y,sq(y))=(ft(y),gt(y))\in E$. ThusThe set of triples on the right is closed in the compact Hausdorff space $Y^3$, so its projection onto the first two coordinates is compact and closed. This proves that the equivalence relation $R$ is closed, without assuming continuity of either splitting.

Give $Q\cong Y/R$ the quotient topology. The allowed closed-relation criterion makes it a compact Hausdorff space. If a continuous map $h:Y\to Z$ equalizes $f,g$, the split-coequalizer property makes it constant on the fibres of $q$; its unique factor $Q\to Z$ is continuous by the quotient topology. Hence this is a coequalizer in compact Hausdorff spaces, preserved by $U$. Its topology is forced uniquely, since any continuous surjection from a compact space to a Hausdorff space is a quotient map. All three conditions hold, and therefore the forgetful functor from compact Hausdorff spaces to sets is monadic.

Write the adjunction unit as $\eta :1_{\mathcal{C}}\Rightarrow GF$ and the adjunction counit as $\epsilon :FG\Rightarrow 1_{\mathcal{D}}$. Define the monad induced by an adjunction bywith unit $\eta$. Naturality of $\epsilon$ makes $\mu$ a natural transformation.

The triangle identities for an adjunction give the two unit laws:For associativity, apply naturality of $\epsilon$ to the arrow ${\epsilon }_{FA}:FGFA\to FA$ in $\mathcal{D}$:Applying $G$ yieldsThus the induced data satisfy all the monad laws, rather than merely supplying an endofunctor and two transformations.

A monad $\mathbb{T}=(T,\eta ,\mu )$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ obeys $ha=bT(h)$. The functor laws show that identities and composites obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^{\mathbb{T}}$.

For an adjunction $F⊣G$, let $\eta$ be its adjunction unit and $\epsilon$ its adjunction counit. The monad induced by an adjunction isThe two unit laws are $G{\epsilon }_{FA}{\eta }_{GFA}=1_{GFA}$ and $G({\epsilon }_{FA}F{\eta }_A)=1_{GFA}$, respectively the two triangle identities for an adjunction.

For associativity, naturality of the adjunction counit at ${\epsilon }_{FA}:FGFA\to FA$ saysApply $G$ to obtain ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. The composites defining $\eta$ and $\mu$ are natural transformations, so all monad axioms have been checked, not only their underlying objectwise types.

The free algebra functor isThe monad laws make ${\mu }_A$ an algebra action, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $U:{\mathcal{C}}^T\to \mathcal{C}$ forget the action. DefineThe proposed inverse is an algebra morphism becauseNaturality of $\eta$ and the algebra unit law give $(bTf){\eta }_A=f$. If $h$ is an algebra morphism, thenThe formulas are natural in both variables, so $F⊣U$. Its unit is $\eta$, and its counit at $(B,b)$ has underlying map $b:TB\to B$. Thus the monad induced by an adjunction has endofunctor $UF=T$, unit $\eta$, and multiplication $U{\epsilon }_F=\mu$. It is exactly the original monad, not merely a monad with the same endofunctor.

For an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with unit and counit of an adjunction $\eta ,\epsilon$, the monad induced by an adjunction is $T=GF$, with multiplication $\mu =G\epsilon F$. Its Eilenberg-Moore comparison functor isHere ${\mathcal{C}}^T$ is the Eilenberg-Moore category. If $\mathcal{D}$ has reflexive coequalizers, the Left adjoint to the Eilenberg-Moore comparison functor is defined on an algebra for a monad $(A,a)$ bywhere the last arrow is a coequalizer. The parallel pair has common section $F{\eta }_A$, so it is a reflexive pair. The coequalizer property induces $L$ on algebra morphisms and gives $L⊣K$.

Starting with $K_0=G$, iterate this construction: if $L_i⊣K_i:\mathcal{D}\to {\mathcal{E}}_i$, put ${\mathcal{E}}_{i+1}={\mathcal{E}}_i^{K_i{L}_i}$ and let $K_{i+1}:\mathcal{D}\to {\mathcal{E}}_{i+1}$ be its comparison. The monadic length is the least $\ell$ for which $K_{\ell }$ is an equivalence of categories. Thus an equivalence has length zero, and an already monadic adjunction whose right adjoint is not an equivalence has length one. The assumed reflexive coequalizers ensure the needed left adjoints; in the example we identify them explicitly.

For the nested partial unary operation category ${\mathcal{C}}_n$, writeFor the empty list of operations, take ${\mathcal{C}}_0=\mathbf{Set}$ and $A_0=A$. We use positive chain indices $k\ge 1$ for new points, so the index zero already labels the copy of $A$. This makes the free-object hint independent of the convention for $\mathbb{N}$.

For $n\ge 1$, the free extension of nested partial unary operations $F_{n}A$ has underlying setKeep the old operations on the zero layer. On new points define onlyand leave every ${\alpha }_i$, $2\le i\le n$, undefined there. Define the new top operation byand leave it undefined elsewhere. These are precisely the permitted domains: the first unary operation on every new point moves it, so none of the higher operations is defined there. When $n=0$, take $F_{0}A=A\times {\mathbb{N}}_{\ge 0}$ with ${\alpha }_1(a,k)=(a,k+1)$. In either case the unit is $a\mapsto (a,0)$.

For $B\in {\mathcal{C}}_{n+1}$, let its operations be ${\beta }_i$. Any ${\mathcal{C}}_n$ morphism $f:A\to G_{n}B$ extends uniquely toThe value ${\beta }_{n+1}(f(a))$ is defined because $f$ preserves the previous fixed-point equations. The extension respects the top operation at old fixed points and respects the first operation along each new chain; there are no other defined operations on those chains to check. Uniqueness follows from the same forced formulas. For $n=0$ the formula is $\bar{f}(a,k)={\beta }_1^k(f(a))$. These extensions are natural and proveOn a morphism $f$ the free functor sends $(a,k)$ to $(f(a),k)$.

Crucially, the top operation of $F_{n}A$ has no fixed points: its defined old values move to new points, and it has no defined new values when $n\ge 1$; for $n=0$ it is the shift. The next free extension therefore adds no points and equips the object with an everywhere-undefined next operation. The same remains true at every subsequent free extension. Hence, for $m>n$, the composite left adjoint $F_{m-1}\cdots F_n$ has the same first $n+1$ operations and underlying set as $F_n$.

This agreement includes the monad structure, not just its underlying endofunctor. The units are the same zero-layer inclusions. The composite counit on an object of ${\mathcal{C}}_m$ is the same forced extension formula, depending only on its first $n+1$ operations. Applying the composite forgetful functor therefore gives the same multiplication $G_n\epsilon F_n$. Thus

Next prove creation of the required split coequalizers. Suppose $f,g:B\rightrightarrows C$ are arrows in ${\mathcal{C}}_{n+1}$ and their images have a split coequalizer $q:C\to Q$ in ${\mathcal{C}}_n$. Choose the splitting orientationwith $s:Q\to C$, $t:C\to B$ morphisms in ${\mathcal{C}}_n$. Denote the top operations on $B,C$ by $b,c$. For $z\in Q_n$ defineSince $s$ preserves the lower operations, $s(z)\in C_n$ and this value exists. For $y\in C_n$, $t(y)\in B_n$, so preservation by $f,g$ givesThus $q$ respects the top operation. The splitting $s$ and $qs=1$ also show that this is the unique operation on its prescribed domain $Q_n$ for which $q$ is a morphism. For $n=0$ all these domains mean the whole underlying set.

If $u:C\to D$ is a ${\mathcal{C}}_{n+1}$ morphism equalizing $f,g$, its unique factor $v:Q\to G_{n}D$ in ${\mathcal{C}}_n$ satisfies, on $Q_n$,where $d$ is the top operation of $D$. Hence $v$ is a ${\mathcal{C}}_{n+1}$ morphism. This proves $G_n$ creates coequalizers of $G_n$-split pairs.

Moreover, $G_n$ reflects isomorphisms: an isomorphism of the first $n$ operations reflects their fixed-point conditions, hence the domain of the next operation; the inverse of a bijective top-operation-preserving map then preserves that operation too. The Beck monadicity theorem now givesUnder this equivalence, the first Eilenberg-Moore comparison functor for the long composite is exactly forgetting all operations after the $(n+1)$st: its algebra structure is the top-operation extension formula already computed. Repeating the argument identifies its monadic tower with the successive categoriesAt each stage the corresponding comparison has the remaining composite of the already constructed free extensions as a left adjoint.

No earlier stage can be an equivalence of categories. For $j<m$, take the two-point set with its first $j$ operations all identities. One extension to ${\mathcal{C}}_m$ has every later operation the identity; another has its $(j+1)$st operation interchange the two points, and all subsequent operations undefined. Both have the same image in ${\mathcal{C}}_j$, but the identity function between those images is not a morphism between the extensions. Thus forgetting from ${\mathcal{C}}_m$ to ${\mathcal{C}}_j$ is not full. The final comparison at $j=m$ is an equivalence. This proves the monadic tower for nested partial unary operations has

For $F⊣G$ with unit $\eta$ and counit $\epsilon$, the monad induced by an adjunction iswith unit $\eta$. An algebra for a monad is $(A,a:TA\to A)$ satisfying $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$.

The Eilenberg-Moore comparison functor isIf $\mathcal{D}$ has coequalizers of reflexive pairs, definefor a $T$-algebra $(A,a)$. The pair has common section $F{\eta }_A$. Maps $L(A,a)\to D$ correspond by the coequalizer property and the adjunction exactly to algebra morphisms $(A,a)\to K(D)$, naturally in both variables. Hence this is the Left adjoint to the Eilenberg-Moore comparison functor. The monadic length is the least number of successive comparison steps required for the resulting monadic tower to become an equivalence.

For $A\in {\mathcal{C}}_n$, let $A_n=\{a:{\alpha }_n(a)=a\}$. The free ${\mathcal{C}}_{n+1}$-object has underlying setRetain all old operations on $A\times \{0\}$. Putand on every new chain putFor $i>1$, ${\alpha }_i$ is undefined on the new points because none is fixed by ${\alpha }_1$. These definitions satisfy the domain conditions. If $f:A\to U(B^{\prime })$ is a ${\mathcal{C}}_n$-morphism, its unique extension sendsThis proves the required left adjoint.

After adjoining ${\alpha }_{m+1}$ freely, the newly added points lie on fixed-point-free ${\alpha }_1$-chains, while each old point where ${\alpha }_{m+1}$ was added is no longer fixed by ${\alpha }_{m+1}$. Consequently there are no points at which a further free operation must be adjoined. Thus for every $n>m$ the endofunctor and unit/multiplication of the monad induced on ${\mathcal{C}}_m$ are already those induced by ${\mathcal{C}}_m\rightleftarrows {\mathcal{C}}_{m+1}$.

By assumption each adjacent adjunction is a monadic adjunction. The first comparison for ${\mathcal{C}}_m\rightleftarrows {\mathcal{C}}_n$ therefore recovers ${\mathcal{C}}_{m+1}$, and iteration successively recovers ${\mathcal{C}}_{m+2},\dots ,{\mathcal{C}}_n$. None of the intervening forgetful functors is an equivalence, since the next partial operation can be chosen differently on a fixed point. Starting at ${\mathcal{C}}_0$ takes exactly $n$ steps:

For an adjoint functor pair $F:\mathcal{D}\rightleftarrows \mathcal{C}:G$, the unit and counit of an adjunction areUnder the adjunction bijection, ${\eta }_A$ corresponds to $1_{FA}$ and ${\epsilon }_B$ corresponds to $1_{GB}$. They satisfy the triangle identitiesConversely, natural transformations with these identities recover the adjunction through the mutually inverse maps

The fully faithful adjoint criterion gives the first equivalence directly. If $F$ is full and faithful, there is a unique $r_A:GFA\to A$ with $F(r_A)={\epsilon }_{FA}$; the triangle identity and faithfulness show that $r_A$ and ${\eta }_A$ are inverse, so the unit is an isomorphism. If the unit is an isomorphism, the displayed adjunction bijection shows thatis bijective, so $F$ is full and faithful. This also proves that either condition gives a natural isomorphism $GF\cong 1_{\mathcal{D}}$.

For the remaining direction, suppose merely that $GF$ is naturally isomorphic to the identity. Transport the monad induced by an adjunction along this isomorphism. Its underlying endofunctor is then the identity, its unit is a natural endomorphism $u:1\Rightarrow 1$, and its multiplication is a natural endomorphism $m:1\Rightarrow 1$ with $mu=1$. Naturality makes $u_A$ commute with $m_A$, so also $um=1$. Hence the transported unit, and therefore $\eta$, is an isomorphism. The three conditions are equivalent.

Now let $F⊣G⊣H$. If $F$ is full and faithful, then $GF\cong 1_{\mathcal{D}}$. For $A,B\in \mathcal{D}$, the two adjunctions give natural bijectionsThe Yoneda lemma therefore gives $GHA\cong A$, naturally in $A$. The fully faithful adjoint criterion applied to $G⊣H$ shows that $H$ is full and faithful. Conversely, if $H$ is full and faithful, then $GH\cong 1_{\mathcal{D}}$ andAnother application of the Yoneda lemma gives $GFA\cong A$, so $F$ is full and faithful. Thus $F$ is full and faithful exactly when $H$ is.

Assume henceforth that $G$ is full and faithful. Then the counit $\epsilon :FG\to 1_{\mathcal{C}}$ and unit $\alpha :1_{\mathcal{C}}\to HG$ are natural isomorphisms. ConsiderApplying the faithful functor $G$, then using naturality and the four triangle identities, turns both composites intoTherefore ${\theta }_1={\theta }_2$; denote their common value by $\theta$, the double-adjoint comparison transformation.

The pointwise monicity criterion is clearest from the following natural square, in which both vertical maps are bijections:The left vertical map is the adjunction $G⊣H$, while the right one precomposes with the isomorphism ${\epsilon }_B:FGB\to B$. Thus every ${\theta }_A$ is a monomorphism exactly when $F$ is faithful on all morphisms $GB\to A$, namely morphisms whose domains lie in the image of $G$.

Dually, the natural squarehas bijective vertical maps, using $F⊣G$ on the left and the isomorphism ${\alpha }_B:B\to HGB$ on the right. Hence every ${\theta }_A$ is an epimorphism exactly when $H$ is faithful on all morphisms $A\to GB$, namely morphisms whose codomains lie in the image of $G$.

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$ satisfyingIf $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ is an adjunction with unit $\eta$ and counit $\epsilon$, thendefines the monad induced by an adjunction. The two triangle identities give the unit laws, while naturality of $\epsilon$ gives associativity.

Let $\mathcal{D}$ now be a full subcategory of $[\mathcal{C},\mathcal{C}]$ that contains the identity endofunctor and is closed under composition, and suppose $T$ is terminal in $\mathcal{D}$. There is exactly one natural transformationand exactly oneBoth sides of either unit law are endomorphisms of the terminal object $T$, so they equal $1_T$; both sides of associativity are maps $T^3\to T$, so they are equal as well. This gives a monad, and terminality also makes both structure maps unique. This is the monad structure on a terminal endofunctor.

For a set $A$, let $\mathcal{U}(A)$ be the set of ultrafilters on $A$. A function $f:A\to A^{\prime }$ induces the pushforwardwhich makes $\mathcal{U}$ the ultrafilter functor. There is no ultrafilter on the empty set. Moreover, every ultrafilter on $A\bigsqcup B$ contains exactly one of the complementary summands $A$ and $B$, and restriction gives a unique ultrafilter on that summand. Thereforeso $\mathcal{U}$ preserves finite coproducts.

Let $F:\mathbf{Set}\to \mathbf{Set}$ preserve finite coproducts. For $x\in F(A)$ defineThe decomposition $A=B\bigsqcup (A\setminus B)$ and preservation of coproducts say that $x$ lies in exactly one of the two corresponding images. Thus exactly one of $B$ and its complement belongs to ${\Phi }_A(x)$. Upward closure follows by factoring subset inclusions. If $B,C\in {\Phi }_A(x)$, decompose $A$ into the four disjoint Boolean cells determined by $B$ and $C$. The unique cell containing $x$ must lie inside both sets, so $B\cap C\in {\Phi }_A(x)$. Hence ${\Phi }_A(x)$ is an ultrafilter.

For $f:A\to A^{\prime }$, the decompositionsshow thatThus $\Phi :F\Rightarrow \mathcal{U}$ is natural.

It is the only such natural transformation. For $B\subseteq A$, let ${\chi }_B:A\to \{0,1\}$ be its characteristic function. Sinceand the two ultrafilters on $\{0,1\}$ are the principal ones, naturality with the two singleton inclusions forces any transformation $F\Rightarrow \mathcal{U}$ to send each summand to the corresponding principal ultrafilter. Naturality with ${\chi }_B$ then says that $B$ belongs to the image ultrafilter exactly when $x$ lies in the image of $F(B)\to F(A)$. Hence the transformation must be $\Phi$.

The ultrafilter functor is therefore the terminal finite-coproduct-preserving set endofunctor. The preceding terminal-object argument supplies its unique ultrafilter monad structure.

For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the monad induced by an adjunction is $T=GF$, with unit the adjunction unit and multiplication $\mu =G\epsilon F$. An algebra for a monad is a map $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. The Eilenberg-Moore comparison functor isand sends a morphism $h$ to $Gh$.

Given a $T$-algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is reflexive, with common section $F{\eta }_A$, so it exists by hypothesis. A map $J(A,a)\to D$ is equivalently a map $FA\to D$ equalizing the pair. Under adjunction this is exactly a map $A\to GD$ satisfying the $T$-algebra homomorphism equation. Hencenaturally, and $J⊣K$. Iterating these comparison adjunctions gives the monadic tower; the monadic length is the least number of steps required to reach an equivalence.

For a pair of sets $(X_0,X_1)$, defineA pair of maps $X_0\to A_0$, $X_1\to A_1$ extends uniquely to a commutative square from this inclusion to any injection $A_0\hookrightarrow A_1$, proving that $F$ is left adjoint to the forgetful functor $U:\mathbf{Mono}(\mathbf{Set})\to \mathbf{Set}\times \mathbf{Set}$.

The induced monad sends $(X_0,X_1)$ to $(X_0,X_0\bigsqcup X_1)$. Its algebra unit laws show that an algebra is precisely an arbitrary function $X_0\to X_1$, with no injectivity requirement. Thus its Eilenberg-Moore category is the arrow category $[2,\mathbf{Set}]$, and the comparison functor is the full inclusion of injections into all functions. This inclusion is not an equivalence, so the original adjunction is not monadic. It is reflective: a function $f:A_0\to A_1$ maps to its image inclusion $\mathrm{im}f\hookrightarrow A_1$. The supplied result that reflections are monadic says that the next comparison is an equivalence. Therefore the original adjunction has monadic length $2$.