Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 23 / 8

With $T=GF$, unit $\eta$ and multiplication $G\epsilon F$, the Eilenberg-Moore comparison functor isHere $G{\epsilon }_d:T(Gd)=GFGd\to Gd$ is the monad algebra action, and the underlying functor satisfies $UK=G$. The triangle identities and counit naturality provide the monad algebra laws and the compatibility of $Gh$, as required for this definition.

For each algebra for a monad $(A,a)$, consider the specific pair in $\mathcal{D}$Then the exact condition isEach pair is reflexive, with common section $F{\eta }_A$: the two composite identities follow from $a{\eta }_A=1_A$ and ${\epsilon }_{FA}F{\eta }_A=1_{FA}$. Thus existence of all reflexive coequalizers is sufficient, but the displayed specified pairs give the necessary-and-sufficient condition.

To prove it, take $h:FA\to d$ and its adjoint transpose $x=Gh{\eta }_A:A\to Gd$. The transpose of $hFa$ is $xa$, by naturality of $\eta$. The transpose of $h{\epsilon }_{FA}$ isMoreover counit naturality and a triangle identity giveBecause transposition is bijective, the coequalizing equation is therefore equivalent toThe right side says exactly that $x:(A,a)\to Kd$ is a monad algebra morphism.

For sufficiency, choose a coequalizer $q_a:FA\to L(A,a)$ for every monad algebra. Its universal property and the equivalence just proved give natural bijectionsTo see explicitly that the chosen objects form a functor, an monad algebra morphism $u:(A,a)\to (B,b)$ makes $q_{b}Fu$ coequalize the pair for $a$: use $ua=bTu$ and naturality of the counit. Hence there is a unique $Lu$ with $Lu{q}_a=q_{b}Fu$. Uniqueness proves its identity and composition laws and the naturality of the displayed bijections. Thus $L⊣K$.

Conversely, suppose $L⊣K$ with unit ${\lambda }_{(A,a)}:(A,a)\to KL(A,a)$. Transpose its underlying arrow $A\to GL(A,a)$ through $F⊣G$ to obtain $q_a:FA\to L(A,a)$. Since the unit is an monad algebra morphism, the equivalence above shows that $q_a$ coequalizes the specified pair. Any coequalizing $h:FA\to d$ transposes to an monad algebra map $(A,a)\to Kd$, which factors uniquely through the unit by $L⊣K$. Transposing back gives a unique factorization of $h$ through $q_a$. Thus $q_a$ is the required coequalizer. This proves both directions of the Left adjoint to the Eilenberg-Moore comparison functor criterion.

Assume the coequalizers from part (b) exist and write them $q_a:FA\to L(A,a)$. The underlying arrow of the comparison-adjunction unit isWe first relate it to the monad algebra action. Naturality of $\eta$, the coequalizer equation and a monad unit law giveIn $\mathcal{C}$, $a:TA\to A$ is itself a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$. Indeed,For an explicit universal-property proof, if $r:TA\to X$ satisfies $rTa=r{\mu }_A$, then $s=r{\eta }_A$ obeysIts uniqueness follows by composing any equation $sa=r$ with ${\eta }_A$.

Now $G{q}_a$ equalizes $Ta$ and ${\mu }_A$. If it is a coequalizer, then $a$ and $G{q}_a$ are coequalizers of the same pair. Their unique comparison map is ${\lambda }_a$, since ${\lambda }_{a}a=G{q}_a$, and is therefore invertible. Conversely, if ${\lambda }_a$ is invertible, $G{q}_a={\lambda }_{a}a$ is an isomorphic copy of the coequalizer $a$, so it is a coequalizer too.

The forgetful functor ${\mathcal{C}}^T\to \mathcal{C}$ reflects isomorphisms: if an monad algebra morphism $u$ has an inverse as an underlying arrow, the equation $ua=bTu$ rearranges to $u^{-1}b=aT(u^{-1})$, making that inverse an monad algebra morphism. Thus underlying invertibility here is exactly invertibility of the unit component. For all monad algebras together, the comparison-adjunction unit criterion isEquivalently, the necessary-and-sufficient conditions for a left adjoint with invertible unit are existence of the coequalizers in part (b) and their preservation by $G$. No preservation of arbitrary coequalizers is required.