Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 8 / ii / Solution

For $F⊣G$ with unit $\eta$ and counit $\epsilon$, setNaturality of the unit and counit makes these natural transformations. The two triangle identities give ${\mu }_A{\eta }_{TA}=1_{TA}$ and ${\mu }_{A}T{\eta }_A=1_{TA}$. Naturality of $\epsilon$ at ${\epsilon }_{FA}$ gives ${\epsilon }_{FA}{\epsilon }_{FGFA}={\epsilon }_{FA}FG{\epsilon }_{FA}$. Applying $G$ proves ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. Thus this is the monad induced by an adjunction.

Conversely, for any monad take the free-algebra functor $F^T(A)=(TA,{\mu }_A)$ and the forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$. For an algebra $(B,b)$, the assignments $h\mapsto h{\eta }_A$ and $k\mapsto bT(k)$ give inverse bijections between algebra maps $F^T(A)\to (B,b)$ and arrows $A\to B$. For the inverse identities, $bT(k){\eta }_A=b{\eta }_{B}k=k$, while $bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$. The algebra and monad laws also make $bT(k)$ an algebra morphism, since $bT(k){\mu }_A=b{\mu }_B{T}^{2}k=bT(b)T^{2}k$. These bijections are natural, giving the free-forgetful Eilenberg-Moore adjunction. Its unit is $\eta$, its counit at $(B,b)$ is $b$, and its induced multiplication is exactly $\mu$. Thus every monad arises from an adjunction.