Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 18 / 6

For the monad $(T,\eta ,\mu )$, an algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. A morphism of algebras for a monad $f:(A,a)\to (B,b)$ satisfies $fa=bTf$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad identities verify the algebra laws. For the forgetful functor $U$, the free adjunction iswith inverse $v\mapsto bTv$. The algebra law makes the latter an monad algebra morphism. The identities $h{\eta }_A\mapsto bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$ and $bTv{\eta }_A=b{\eta }_{B}v=v$ prove the bijection.

In the category of adjunctions inducing a fixed monad, objects are adjunctions $F⊣G$ with their induced monad identified with $T$. A morphism to $F^{\prime }⊣G^{\prime }$ is a functor $H$ between the right-hand categories satisfying $G^{\prime }H=G$, $HF=F^{\prime }$ and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of $\epsilon$ at ${\epsilon }_B$ gives the multiplication law. Naturality at $f$ makes $Gf$ an monad algebra morphism. Moreover $UK=G$, $KF=F^T$ because $G{\epsilon }_{FA}={\mu }_A$, and $K{\epsilon }_B$ is the free-adjunction counit at $K(B)$. Hence $K$ is a morphism into the Eilenberg-Moore adjunction.

For any other such $H$, its underlying object at $B$ must be $GB$. Counit compatibility forces its algebra action to be $G{\epsilon }_B$, and the forgetful functor, which is a faithful functor, forces $H(f)=Gf$. Thus $H=K$. This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.

We prove full faithfulness from counit coequalizers. Let $B,C\in \mathcal{D}$ and let $\alpha :K(B)\to K(C)$ be an monad algebra morphism. ThusSet $u={\epsilon }_{C}F\alpha :FGB\to C$. Its composites with the two arrows of the printed presentation are equal: naturality of $\epsilon$ givesIn the last equality we used naturality at $F\alpha$. The coequalizer property therefore gives a unique $v:B\to C$ satisfying $v{\epsilon }_B=u$.

Apply $G$ to that identity. The algebra-morphism equation gives $GvG{\epsilon }_B=\alpha G{\epsilon }_B$. The triangle identity makes $G{\epsilon }_B$ a split epimorphism with section ${\eta }_{GB}$, so $Gv=\alpha$. This proves fullness of $K$.

If $v,w:B\to C$ have $Gv=Gw$, naturality gives $v{\epsilon }_B={\epsilon }_{C}FGv={\epsilon }_{C}FGw=w{\epsilon }_B$. The counit is epic since it is a coequalizer, so $v=w$. Thus $K$ is a faithful functor. Both parallel arrows matter: the converted TeX loses the second one, ${\epsilon }_{FGB}$, which is visible in the original PDF.