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

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: