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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:

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

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