 Solution

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with monad $T=UF$, the comparison functor sends $D$ to the $T$-algebra $(UD,U{\epsilon }_D)$. The adjunction is monadic when this comparison is an equivalence. The comparison-left-adjoint lemma says that if $\mathcal{D}$ has coequalizers of the reflexive pairs used to present $T$-algebras and $U$ preserves them, then the comparison has a left adjoint. Applying the unit and counit criteria yields the Beck monadicity theorem: $U$ is monadic exactly when it reflects isomorphisms and creates coequalizers of all parallel pairs whose images under $U$ admit split coequalizers.

Let $A_n\subseteq A$ be the fixed-point set of ${\alpha }_n$ and identify $A$ with $A\times \{0\}$. Define $F_{n}A$ to have underlying setKeep the old operations on the first summand, put ${\alpha }_{n+1}(a,0)=(a,1)$ for $a\in A_n$, and put ${\alpha }_1(a,m)=(a,m+1)$ for $m>0$; the remaining higher operations on these new points are undefined. Given $A\to U_{n}B$, the only possible extension sends $(a,m)$ along the iterates of ${\alpha }_1$ beginning at ${\alpha }_{n+1}(a)$. This proves $F_n⊣U_n$.

The forgetful functor $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ reflects isomorphisms. A $U_n$-split coequalizer carries a unique descended partial operation: splitness prevents any new fixed point of ${\alpha }_n$ from appearing without a representative on which ${\alpha }_{n+1}$ is already prescribed. Hence $U_n$ creates these coequalizers, and Beck's theorem proves the adjunction monadic.

The composite $U_n{U}_{n+1}:{\mathcal{C}}_{n+2}\to {\mathcal{C}}_n$ is not monadic. To see the Beck obstruction explicitly, take $B=\{x,y,z\}$, let all operations through ${\alpha }_n$ be the identity, let ${\alpha }_{n+1}$ swap $x,y$ and fix $z$, and define ${\alpha }_{n+2}(z)=z$. Let $q:B\to Q=\{w,z\}$ identify $x$ and $y$, choose the section $s(w)=x$, $s(z)=z$, and put $h=sq$. Form the kernel pair $A=B{\times }_{Q}B$ with coordinatewise operations and projections $f,g:A\rightrightarrows B$. The map $t:B\to A$, $t(b)=(b,h(b))$, satisfiesso $q$ is a split coequalizer of the underlying pair in ${\mathcal{C}}_n$.

Any lifted structure on $Q$ must have ${\alpha }_{n+1}(w)=w$, so ${\alpha }_{n+2}(w)$ must be defined. If it is $w$, map $x,y$ to a fixed point $d$ and $z$ to a fixed point $e\ne d$ in a target with ${\alpha }_{n+2}(d)=e$ and ${\alpha }_{n+2}(e)=e$; this equalizes $f,g$ but its set-theoretic factor through $Q$ does not preserve ${\alpha }_{n+2}$ at $w$. If instead ${\alpha }_{n+2}(w)=z$, use a target with ${\alpha }_{n+2}(d)=d$ and ${\alpha }_{n+2}(e)=e$ to obtain the same failure. Thus the underlying split coequalizer cannot be created in ${\mathcal{C}}_{n+2}$, and Beck's theorem proves that the composite is not monadic.

Solved by gpt-5.6-sol high.