Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 22 / 6

For the monad $\mathbb{T}=(T,\eta ,\mu )$ define the free algebra functor byThe unit and associativity identities of the monad make $(TX,{\mu }_X)$ an algebra for a monad, and naturality of $\mu$ makes $Tf$ a morphism of algebras for a monad. Let $G^{\mathbb{T}}$ be the forgetful functor. For $(A,a)$ in the Eilenberg-Moore category, setThe inverse candidate is a monad algebra morphism, sinceThe two composites are identities:These use respectively the unit law for a monad algebra, the algebra-morphism equation, and a unit identity of the monad. The formulas commute with precomposition in $X$ and postcomposition by monad algebra morphisms, so they form a natural bijection. Therefore the free-algebra functor is left adjoint to forgetting:Its adjunction unit is ${\eta }_X$, and its adjunction counit at $(A,a)$ is $a:(TA,{\mu }_A)\to (A,a)$.

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:

If the multiplication in the unit and multiplication of a monad $\mu$ is invertible, the two unit identities of the monadmake both $T\eta$ and $\eta T$ its inverse. Hence idempotence implies equality of the two unit insertions:This is the first implication in the characterization of an idempotent monad.

Assume $T\eta =\eta T$, and let $(A,a)$ be an algebra for a monad. Its unit identity gives $a{\eta }_A=1_A$. Naturality of $\eta$ at $a:TA\to A$, followed by the assumed equality, givesThus every algebra action is invertible, with its inverse prescribed by the unit:

Suppose every algebra action is an isomorphism. Its inverse is ${\eta }_A$, by the unit law for a monad algebra. For any underlying morphism $f:A\to B$ between two algebras for a monad, naturality of $\eta$ givesTherefore every underlying morphism is automatically a monad algebra morphism. The forgetful functor is always faithful, because a monad algebra morphism has no data beyond its underlying morphism, and it is now full as well. Hence invertible algebra actions imply

If $G^{\mathbb{T}}$ is fully faithful, the criterion proved for a right adjoint says that the adjunction counit of $F^{\mathbb{T}}⊣G^{\mathbb{T}}$ is a natural isomorphism. Its component at $(A,a)$ is precisely the algebra action $a$. In particular, its component at the free algebra for a monad $(TX,{\mu }_X)$ has underlying morphism ${\mu }_X:T^{2}X\to TX$. Thus every ${\mu }_X$ is an isomorphism, and the natural transformation $\mu$ is invertible. This closes the cycle and proves all four conditions equivalent: