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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: