 Solution

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 $2$.

Solved by gpt-5.6-sol high.