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For $F⊣G$ inducing $T=GF$, the Eilenberg-Moore comparison functor has a left adjoint exactly when every algebra $(A,a)$ has a coequalizer $q_a:FA\to L(A,a)$ of $Fa,{\epsilon }_{FA}:FTA\rightrightarrows FA$. Transposition identifies maps equalizing this pair with monad algebra morphisms $(A,a)\to K(D)$, proving both directions of the criterion. The unit is $\lambda =G{q}_a{\eta }_A$ and satisfies $G{q}_a=\lambda a$. In the base category, $a:TA\to A$ is a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$, split by ${\eta }_A$ and ${\eta }_{TA}$. Consequently $\lambda$ is invertible exactly when $G{q}_a$ is also a coequalizer. The unit is invertible everywhere precisely when $G$ preserves all these specified coequalizers; preservation of arbitrary coequalizers is not required.

If each standard counit presentation is a coequalizer, the Eilenberg-Moore comparison functor is full and faithful. An monad algebra morphism $\alpha :GB\to GC$ makes ${\epsilon }_{C}F\alpha$ equalize $FG{\epsilon }_B$ and ${\epsilon }_{FGB}$; its unique descent is the required morphism $B\to C$. The triangle identities for an adjunction make $G{\epsilon }_B$ a split epimorphism, so descent has underlying arrow exactly $\alpha$. Epimorphic cancellation at ${\epsilon }_B$ proves faithfulness.

When each comparison functor has a left adjoint, iterating the Eilenberg-Moore comparison functor produces the monadic tower. Its monadic length is the least number of comparison steps needed to reach an equivalence; an equivalence has length zero and a non-equivalence that is already monadic has length one.

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying the unit and multiplication of a monad lawsAn algebra for a monad is an object $A$ equipped with $a:TA\to A$ such thatA morphism of algebras for a monad $f:(A,a)\to (B,b)$ is a morphism $f:A\to B$ satisfying $fa=bT(f)$. Identities and composition obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^T$ and its forgetful functor $U^T(A,a)=A$.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with induced monad $T=UF$, the Eilenberg-Moore comparison functor isThe functor $U$ is monadic when it has such a left adjoint and this comparison is an equivalence of categories; a stricter convention requires an isomorphism of categories over $\mathcal{C}$. We will identify which limit-lifting conclusion each convention supports in part (iii).

Write the adjunction unit and adjunction counit of $F⊣G$ as $\eta ,\epsilon$. Its induced monad is $T=GF$, with multiplication $\mu =G{\epsilon }_F$. The Eilenberg-Moore comparison functor isThe unit law for $a_d$ is $G{\epsilon }_d{\eta }_{Gd}=1_{Gd}$, a triangle identity for an adjunction. Naturality of $\epsilon$ at ${\epsilon }_d$ gives ${\epsilon }_{d}FG{\epsilon }_d={\epsilon }_d{\epsilon }_{FGd}$; applying $G$ gives $a_{d}T{a}_d=a_d{\mu }_{Gd}$. Naturality at $h$ gives the monad algebra morphism equation for $Gh$. Thus $K$ is well-defined.

For each $d$, consider the fork in $\mathcal{D}$It commutes by the naturality equation just used. Its image under $G$ is the fork $T^2(Gd)\rightrightarrows T(Gd)\to Gd$ of the preceding question, with parallel morphisms ${\mu }_{Gd},T{a}_d$ and final map $a_d$. It is split using $s={\eta }_{Gd}$ and $t={\eta }_{T(Gd)}$. By the assumed reflection, ${\epsilon }_d$ is the coequalizer of this pair in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ for $h,k:d\to e$, naturality gives $h{\epsilon }_d={\epsilon }_{e}FGh={\epsilon }_{e}FGk=k{\epsilon }_d$. A coequalizer is an epimorphism, so $h=k$.

For fullness, let $u:K(d)\to K(e)$ be a monad algebra morphism. Thus $u:Gd\to Ge$ satisfies $u{a}_d=a_{e}Tu$. Set $v={\epsilon }_{e}Fu:FGd\to e$. We verify explicitly that $v$ equalizes the two morphisms of the displayed fork. The monad algebra morphism equation givesNaturality of $\epsilon$ first at $Fu$ and then at ${\epsilon }_e$ givesThe coequalizer therefore supplies a unique $h:d\to e$ with $h{\epsilon }_d=v$. Applying $G$ yields $Gh{a}_d=Gv=a_{e}Tu=u{a}_d$. The map $a_d$ is split epic with section ${\eta }_{Gd}$, so cancellation gives $Gh=u$. This proves fullness, and the earlier cancellation proves faithfulness. HenceNo claim that $K$ is essentially surjective is needed; that stronger conclusion needs the corresponding existence hypotheses.

For $T=GF$, define the Eilenberg-Moore comparison functorThe algebra unit law is a triangle identity. Naturality of the counit at ${\epsilon }_D$ gives its algebra associativity law; naturality at $h:D\to E$ gives $Gh{a}_D=a_{E}T(Gh)$, making $K(h)$ an algebra morphism.

Suppose first that $G$ reflects G-split coequalizers. For every $D$, consider the counit presentationIt equalizes the pair by counit naturality. After applying $G$ it becomes $T^{2}GD\rightrightarrows TGD\stackrel{a_D}{\to }GD$, with parallel maps ${\mu }_{GD},T(a_D)$. This is split: take $s={\eta }_{GD}$ and $t={\eta }_{TGD}$. The identities $a_{D}s=1$, ${\mu }_{GD}t=1$ and $T(a_D)t={\eta }_{GD}{a}_D$ follow from the algebra law, a monad unit law, and naturality of $\eta$. Reflection therefore makes ${\epsilon }_D$ a coequalizer in $\mathcal{D}$.

For faithfulness, if $Gh=Gk$ then naturality gives $h{\epsilon }_D={\epsilon }_{E}FGh={\epsilon }_{E}FGk=k{\epsilon }_D$. Coequalizers are epic, so $h=k$.

For fullness, let $\alpha :GD\to GE$ be an algebra morphism, so $\alpha a_D=a_{E}T\alpha$. Put $h_0={\epsilon }_{E}F\alpha :FGD\to E$. Its two composites with the counit-presentation pair are equal: by counit naturality they are ${\epsilon }_{E}F(a_{E}T\alpha )$ and ${\epsilon }_{E}F(\alpha a_D)$. Hence there is a unique $h:D\to E$ with $h{\epsilon }_D=h_0$. Applying $G$ gives $Gh{a}_D=a_{E}T\alpha =\alpha a_D$. Since $a_D$ is split epic, $Gh=\alpha$. This proves full faithfulness directly by descent.

Conversely suppose $K$ is full and faithful. We need the following explicit algebra lifting fact. For algebra morphisms $f,g:(A,a)\rightrightarrows (B,b)$ with a split underlying coequalizer $q:B\to Q$, the arrow $qb:TB\to Q$ equalizes $Tf,Tg$. The split diagram is preserved by $T$, so $Tq$ is their coequalizer and there is a unique $c:TQ\to Q$ with $cTq=qb$. Precomposing the unit equation with the epimorphism $q$ gives $c{\eta }_{Q}q=q$; precomposing the associativity equation with the epimorphism $T^{2}q$ gives equality by $bT(b)=b{\mu }_B$. Thus $c$ is an algebra action. If an algebra map $h:B\to (Z,z)$ equalizes $f,g$, its unique underlying factor $k:Q\to Z$ satisfies $kcTq=hb=zTh=zTkTq$. Cancelling $Tq$ shows $kc=zTk$. This proves monad algebra forgetful functor creates split coequalizers.

Now take any existing equalizing arrow $e:B\to D$ in $\mathcal{D}$ whose image is such a coequalizer. The lifting fact makes the unique compatible action on $GD$ its already given action $a_D$, since $Ke$ preserves that action. Hence $Ke$ is a coequalizer of $Kf,Kg$ in the algebra category. For any arrow $h:B\to Z$ equalizing the pair, its unique algebra factor $KD\to KZ$ lifts to an arrow $D\to Z$ by fullness of $K$; faithfulness gives its factorization equation and uniqueness. Therefore $e$ is a coequalizer in $\mathcal{D}$. We have proved the full comparison and reflection of split coequalizers equivalence

An adjunction $L⊣U$ induces the monad $T=UL$ with unit $\eta$ and multiplication $U\epsilon L$. The Eilenberg-Moore comparison functor sends $B$ to $(UB,U{\epsilon }_B)$. A monadic adjunction is one for which that comparison is an equivalence of categories. Iterating comparisons, when their left adjoints exist, gives a tower; its monadic length is the number of steps until equivalence, with length zero for an equivalence and length one for a monadic adjunction that is not already an equivalence.

The precise monadicity theorem says that a right adjoint is monadic exactly when it reflects isomorphisms and creates coequalizers of pairs whose images admit split coequalizers.

Write $U_r:{\mathcal{C}}_r\to {\mathcal{C}}_{r-1}$ for one-step forgetting. For $r=1$, the free extension of nested partial unary operations is the familiar set $A\times \mathbb{N}$, with ${\alpha }_1(a,j)=(a,j+1)$ and unit $a\mapsto (a,0)$. A map into a set with endomorphism ${\beta }_1$ extends uniquely by $(a,j)\mapsto {\beta }_1^{j}f(a)$.

For $r\ge 2$, letKeep $A$ and all its old operations, and adjoin the disjoint set $S(A)\times \mathbb{N}$. On the new points define ${\alpha }_1(a,j)=(a,j+1)$; every old operation ${\alpha }_i$ with $i\ge 2$ is undefined there. Define the new top operation only at the old eligible points, by ${\alpha }_r(a)=(a,0)$ for $a\in S(A)$.

These domains satisfy the defining rules of the nested partial unary operation category: the new points have no ${\alpha }_1$-fixed points, so none has any higher operation defined. Given $f:A\to U_{r}B$, extend it byThe expression is defined because $f$ sends eligible points to eligible points and ${\beta }_1$ is total. It preserves every defined operation; preservation of ${\alpha }_r$ at old points determines $j=0$, and preservation of ${\alpha }_1$ determines the whole chain. Thus it is the unique extension, proving $L_r⊣U_r$.

For monadicity, $U_r$ reflects isomorphisms: an isomorphism of the lower structures takes $S(A)$ bijectively to $S(B)$, so the inverse of a map preserving the new operation preserves it too.

Next prove the split-coequalizer lifting for a fixed-point-domain operation. Let $f,g:A\rightrightarrows B$ be ${\mathcal{C}}_r$-morphisms and suppose their lower images have a split coequalizer $q:U_{r}B\to Q$, with $s:Q\to U_{r}B$, $t:U_{r}B\to U_{r}A$ satisfyingEquip $Q$ with ${\beta }_Q(z)=q{\beta }_B(sz)$ on its required eligible domain. The lower-structure maps $s,t$ preserve eligibility. For eligible $x\in B$,so $q$ preserves the top operation. If $h:B\to D$ coequalizes $f,g$, its lower factor $\overline{h}:Q\to U_{r}D$ satisfiesThus it is a morphism of the full structures. Uniqueness of the lifted operation follows from the eligible section $s$ and surjectivity of $q$. This creates the specified coequalizer, so the Beck monadicity theorem proves each one-step adjunction is monadic.

Finally, every free top operation just constructed has no fixed points: for $r=1$ it shifts the free chain, while for $r\ge 2$ it sends old eligible points to new distinct points and is undefined on the new points. Freely adjoining any further operation therefore changes nothing, since its required domain is empty. For $n>m$, the monad induced on ${\mathcal{C}}_m$ by the long free-and-forgetful adjunction is consequently the same monad, including unit and multiplication, as that for ${\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$.

The first comparison category is thus equivalent to ${\mathcal{C}}_{m+1}$. Its algebra action evaluates the new formal value by ${\alpha }_{m+1}$, so the comparison from ${\mathcal{C}}_n$ retains precisely that next operation and is the remaining forgetful functor. Iteration recovers one additional operation at each step. These remaining comparisons are not equivalences while an operation is still forgotten: take a two-element set with all retained operations equal to the identity, and choose the next operation to be the identity in one extension and the transposition in another. Higher operations can be identities in the first case and undefined in the second. The identity on the retained structure is not a morphism between these extensions, so the forgetful functor is not full. This includes ${\mathcal{C}}_1\to \mathbf{Set}$.

The monadic tower for nested partial unary operations therefore has exactly the claimed length:

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.

For $L⊣U:\mathcal{D}\to \mathcal{C}$, let $T=UL$ be the induced monad. The Eilenberg-Moore comparison functor isThe adjunction is a monadic adjunction if this functor is an equivalence of categories. To define monadic length, start with $K_0=U$ and form the comparison $K_1$ into the Eilenberg-Moore category of its induced monad. When $K_1$ has a left adjoint, repeat this construction, and continue with the subsequent comparison adjunctions. The length is the least number of comparison steps after which the comparison is an equivalence; an equivalence has length zero and a monadic non-equivalence has length one. The successive left adjoints will all be explicit in this example.

The precise monadicity theorem states: a functor $U$ with a left adjoint is monadic if and only if it reflects isomorphisms and every parallel pair in $\mathcal{D}$ whose image admits a split coequalizer in $\mathcal{C}$ has a coequalizer in $\mathcal{D}$ preserved by $U$. Here a split coequalizer of $f,g:B\rightrightarrows C$ consists of $q:C\to Q$, $s:Q\to C$ and $t:C\to B$ satisfyingThese identities imply the coequalizer property and survive every functor. We shall identify the relevant algebra categories directly, which proves monadicity without having to check quotient constructions separately.

Write $U_m^{m+1}:{\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$ for one step in the nested partial unary operation category tower. For $m=0$, its left adjoint isA function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\omega }_1^{k}f(a)$, proving the adjunction.

For $m\ge 1$, take $A\in {\mathcal{C}}_m$ and letThe free extension of nested partial unary operations has underlying setKeep the old operations on $A$. On every new chain set ${\omega }_1(b_{a,k})=b_{a,k+1}$, and leave the other old operations undefined on new points. Define the new operation by ${\omega }_{m+1}(a)=b_{a,0}$ for $a\in S_m(A)$, and nowhere else. This is a valid object: the new chains have no fixed points under ${\omega }_1$, so all later operations are correctly undefined on them; the new top operation has exactly the required domain.

If $f:A\to U_m^{m+1}B$ preserves the old operations, it carries $S_m(A)$ into $S_m(B)$. Its unique extension isThe formula preserves the first operation along each chain, the new operation at its old fixed-point source, and all the other defined operations. Conversely these requirements force the formula. This proves the one-step adjunction in every case.

Let $T_m=U_m^{m+1}{F}_m^{m+1}$. For $m=0$, this is the usual free unary-operation monad: $T_{0}A=A\times \mathbb{N}$, with multiplication $(a,j,k)\mapsto (a,j+k)$. An algebra action $h:A\times \mathbb{N}\to A$ obeys $h(a,0)=a$ and $h(h(a,j),k)=h(a,j+k)$. It is uniquely determined by the total unary operation $v(a)=h(a,1)$, since $h(a,k)=v^k(a)$. Algebra morphisms are exactly functions commuting with $v$. Thus ${\mathcal{C}}_0^{T_0}\cong {\mathcal{C}}_1$.

For $m\ge 1$, no new chain point belongs to $S_m(T_{m}A)$, so $S_m(T_{m}A)=S_m(A)$. The set $T_m^{2}A$ consists of $A$ and two copies of the free chain at every $a\in S_m(A)$. Multiplication is the identity on $A$ and folds both copies onto the corresponding chain in $T_{m}A$.

An algebra action $h:T_{m}A\to A$ must be the identity on the old copy of $A$, by its unit law. Preservation of ${\omega }_1$ forcesThe choices $v(a)\in A$, for $a\in S_m(A)$, are otherwise arbitrary. There are no higher old operations to preserve on new points. Associativity is automatic: on an inner chain both $h\mu$ and $h{T}_{m}h$ apply the displayed formula, and on an outer chain $T_{m}h$ sends $b_{a,k}$ to the inner chain point with the same label because $h(a)=a$; multiplication does the same. Therefore $h\mu =h{T}_{m}h$ on every point.

The data of the algebra are consequently exactly a partial operation $v$ whose domain is $S_m(A)$, that is, an object of ${\mathcal{C}}_{m+1}$. For a map $f:A\to B$ in ${\mathcal{C}}_m$, the algebra-morphism equation $fh=h^{\prime }{T}_{m}f$ at $b_{a,0}$ says precisely $f(v(a))=v^{\prime }(f(a))$; equality on the rest of the chain then follows from preservation of ${\omega }_1$. Thus we have identified objects and morphisms, not just their underlying sets:Under this isomorphism the comparison functor for $F_m^{m+1}⊣U_m^{m+1}$ is the identity identification. Hence every one-step forgetful adjunction is monadic.

Now take $n>m$. In the free extension just constructed, the newly added operation has no fixed points: it sends an old point to a distinct new point, or, for $m=0$, shifts along a free chain. Therefore all further operations can only be undefined. Adding them freely adds no more points. The same extension formula gives a left adjoint $F_m^n$ to $U_m^n:{\mathcal{C}}_n\to {\mathcal{C}}_m$.

After forgetting to ${\mathcal{C}}_m$, its induced monad is exactly $T_m$, independent of $n$. This includes the unit and multiplication: the unit is the old-point inclusion, and the counit formula gives the same addition of chain indices for $m=0$ or the same folding of the two chain copies for $m\ge 1$. Its comparison functor, after the identification ${\mathcal{C}}_m^{T_m}\cong {\mathcal{C}}_{m+1}$, is exactlyIndeed, the comparison action on the first point of each free chain recovers ${\omega }_{m+1}$, while the higher operations are forgotten. Thus each comparison step recovers exactly one additional operation. After $n$ steps from sets, the comparison is the identity of ${\mathcal{C}}_n$.

No earlier step is an equivalence. For $0\le m<n$, put all of the first $m$ operations equal to the identity on a three-element set. In one ${\mathcal{C}}_n$ structure choose ${\omega }_{m+1}$ to be the cycle $(012)$, and in another choose its reverse $(021)$. Both have no fixed points at this stage, so every higher operation is undefined. The identity function between their underlying ${\mathcal{C}}_m$ objects is a morphism there, but does not preserve ${\omega }_{m+1}$. Hence $U_m^n$ is not full, and cannot be an equivalence. This also applies for $m=0$, when there are no old operations. We conclude

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying $\mu T\eta =\mu \eta T=1_T$ and $\mu T\mu =\mu \mu T$. Its Eilenberg-Moore category has objects $(X,a)$ with $a:TX\to X$, $a{\eta }_X=1_X$ and $a{\mu }_X=aTa$. A morphism $f:(X,a)\to (Y,b)$ satisfies $fa=bTf$. An adjunction $F⊣G$, with unit $\eta$ and counit $\epsilon$, induces $T=GF$, $\mu =G\epsilon F$. It is a monadic adjunction if the Eilenberg-Moore comparison functor $K(D)=(GD,G{\epsilon }_D)$ is an equivalence with the Eilenberg-Moore category.

A fork $A\stackrel{p,q}{\rightrightarrows }B\stackrel{r}{\to }C$ is a split coequalizer if $rp=rq$ and there are $s:C\to B$, $t:B\to A$ with $rs=1$, $pt=1$ and $qt=sr$. It is indeed a coequalizer: if $hp=hq$, then $h=hpt=hqt=hsr$, so $h$ factors uniquely through $r$, which is split epic. Every functor preserves these splitting identities, making such coequalizers absolute.

The precise monadicity theorem states: for an adjunction $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$, the comparison is an equivalence if and only if $G$ reflects isomorphisms and every pair in $\mathcal{D}$ whose $G$-image admits a split coequalizer has a coequalizer which $G$ preserves. Equivalently, the right adjoint creates these split-underlying coequalizers, with creation understood up to the canonical isomorphism of the prescribed underlying coequalizer. Here is a proof.

First consider the forgetful functor from monad algebras. It reflects isomorphisms: the inverse of an invertible algebra morphism also obeys the algebra-morphism identity, by multiplying that identity by the inverses. For an algebra pair with a split underlying coequalizer $r:B\to C$, all powers of $T$ preserve its coequalizer. If $b:TB\to B$ is the algebra structure on $B$, the map $rb$ coequalizes $Tp,Tq$, so there is a unique $c:TC\to C$ withThe unit law follows by composing with the epimorphism $r$: $c{\eta }_{C}r=cTr{\eta }_B=rb{\eta }_B=r$. For associativity, naturality and the algebra law giveCancel the epimorphism $T^{2}r$ to obtain $c{\mu }_C=cTc$. Thus $(C,c)$ is an algebra and $r$ is an algebra morphism. Any algebra morphism coequalizing the pair factors through $r$ on underlying objects, and its factor is an algebra morphism by cancellation of the epimorphism $Tr$. This creates the coequalizer. Transporting these facts through an equivalence proves necessity.

For sufficiency, let $(X,a)$ be a $T$-algebra. In $\mathcal{D}$ consider the free pairIts underlying pair is ${\mu }_X,Ta:T^{2}X\rightrightarrows TX$, with split coequalizer $a:TX\to X$: the splittings are $s={\eta }_X$ and $t={\eta }_{TX}$, using the algebra unit law, the monad unit law and naturality of $\eta$. By hypothesis its coequalizer $q:FX\to D$ exists and is preserved. Identify $GD$ with $X$ by the unique coequalizer isomorphism, so $Gq=a$. Naturality of the counit makes $Kq$ an algebra morphism, hence the comparison structure $c$ on $GD$ satisfies $cTa=a{\mu }_X=aTa$. The map $Ta$ is split epic because $a{\eta }_X=1$, so $c=a$. Every algebra is therefore isomorphic to a comparison algebra.

For an actual object $D$, its counit ${\epsilon }_D:FGD\to D$ is a coequalizer of the analogous free pair. It coequalizes that pair by counit naturality. Compare it with the coequalizer furnished by the hypothesis; applying $G$ gives an isomorphism, since both underlying arrows coequalize the same split pair. Reflection of isomorphisms makes the comparison map an isomorphism. This proves the counit's asserted coequalizer property.

Let $f:KD\to KE$ be an algebra morphism. Its transpose is $h_0={\epsilon }_{E}Ff:FGD\to E$. The two composites of $h_0$ with the free pair have adjunct transposes $G{\epsilon }_{E}Tf$ and $fG{\epsilon }_D$, equal because $f$ is an algebra morphism. Consequently $h_0$ factors uniquely through ${\epsilon }_D$ as $h:D\to E$. Applying $G$ gives $(Gh)G{\epsilon }_D=fG{\epsilon }_D$; the last factor is split epic, so $Gh=f$. Conversely any lift of $f$ must satisfy $h{\epsilon }_D={\epsilon }_{E}Ff$ by naturality and hence is unique. Thus $K$ is full and faithful, and the earlier construction makes it essentially surjective. The comparison is an equivalence, completing the theorem's proof.

Now take $G:\mathbf{CompHaus}\to \mathbf{Set}$. Its left adjoint exists by the permitted topological result; it can be realized as the Stone-Čech compactification of the discrete input set. A bijective continuous map from a compact space to a Hausdorff space is a homeomorphism, so $G$ reflects isomorphisms.

For continuous $f,g:X\rightrightarrows Y$ with a split underlying fork $q,s,t$, let $R$ be the relation of having the same $q$-image. We claimFor the forward inclusion choose $x=t(y)$, $x^{\prime }=t(y^{\prime })$: $ft=1$ and $gt=sq$ supply the equality. For the reverse inclusion use $qf=qg$. The subset $\{(x,x^{\prime }):g(x)=g(x^{\prime })\}\subset X^2$ is closed, because $Y$ is Hausdorff, and compact. Its continuous image $R\subset Y^2$ is compact and hence closed. A closed equivalence relation on a compact Hausdorff space has a compact Hausdorff quotient; its quotient map is a coequalizer here, since it identifies exactly the generated relation. Its underlying set coequalizer is the specified one up to the unique isomorphism. Thus $G$ preserves the needed coequalizers, andThe closedness argument is the closed relation generated by a set-split pair of compact Hausdorff maps lemma; the splitting maps themselves need not be continuous.

With $T=GF$, unit $\eta$ and multiplication $G\epsilon F$, the Eilenberg-Moore comparison functor isHere $G{\epsilon }_d:T(Gd)=GFGd\to Gd$ is the monad algebra action, and the underlying functor satisfies $UK=G$. The triangle identities and counit naturality provide the monad algebra laws and the compatibility of $Gh$, as required for this definition.

Put ${\mathcal{C}}_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to {\mathcal{C}}_1$ sends $A$ to $A\times \mathbb{N}$ with ${\alpha }_1(a,k)=(a,k+1)$. A function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\beta }_1^k(f(a))$. Its induced monad is $T_{0}A=A\times \mathbb{N}$, with unit $a\mapsto (a,0)$ and multiplication $(a,k,l)\mapsto (a,k+l)$. An algebra for a monad is determined by the action of $1\in \mathbb{N}$, so its Eilenberg-Moore category is ${\mathcal{C}}_1$. Thus this first free–forgetful adjunction is a monadic adjunction without needing the given assumption for $n=0$.

The monadic length counts successive Eilenberg-Moore comparison functor steps until the comparison becomes an equivalence. Fix $0\le m<n$. For $m\ge 1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the left adjoint $F_{n,m}$ to the composite forgetful functor $U_{n,m}:{\mathcal{C}}_n\to {\mathcal{C}}_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of ${\mathcal{C}}_m$, unit and multiplication are exactly those of the one-step adjunction:as monads, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.

By the assumed one-step monadic adjunction property, or the explicit argument above when $m=0$, the Eilenberg-Moore category ${\mathcal{C}}_m^{T_m}$ is equivalent to ${\mathcal{C}}_{m+1}$. Under this equivalence the comparison ${\mathcal{C}}_n\to {\mathcal{C}}_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using ${\alpha }_1$, and evaluates the first new value using ${\alpha }_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole monadic tower for nested partial unary operations iswith the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.

It does not terminate earlier. If $m<n$, take the same underlying two-element set in two objects of ${\mathcal{C}}_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make ${\alpha }_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to ${\mathcal{C}}_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a full functor, and cannot be an equivalence. Therefore

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.

Write $F:\mathcal{X}\to \mathcal{A}$ and $G:\mathcal{A}\to \mathcal{X}$, with adjunction counit $\epsilon$. The induced monad has $T=GF$ and $\mu =G\epsilon F$. Define the Eilenberg-Moore comparison functor byThe action $a_A=G{\epsilon }_A:TGA\to GA$ satisfies $a_A{\eta }_{GA}=1_{GA}$ by a triangular equation. Naturality of $\epsilon$ at ${\epsilon }_A:FGA\to A$ givesApplying $G$ yields $a_{A}T(a_A)=a_A{\mu }_{GA}$, the other algebra identity. For $f:A\to A^{\prime }$, naturality givesso $Gf$ is a monad algebra morphism. Identities and compositions are respected because $G$ is a functor. Finally,and the same equalities hold on morphisms. Thus both comparison identities hold, strictly when $\mathbb{T}$ is the specified induced monad: