Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 10

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying, at every object $A$,These are the associativity and unit laws. An algebra for a monad is $(A,a:TA\to A)$ withA monad algebra morphism $h:(A,a)\to (B,b)$ is a morphism $h:A\to B$ satisfying $ha=bTh$. Identity morphisms satisfy this equation, and if $h,k$ satisfy it, then $kha=kbTh=cTkTh=cT(kh)$. Hence these objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$, with the forgetful functor $U(A,a)=A$ and $Uh=h$.

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus

For $(A,a)$, consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: ${\mu }_A$ satisfies the monad associativity law, while $Ta$ is the image under $F^T$ of $a$. The map $a$ is a monad algebra morphism because $a{\mu }_A=aTa$, and the same equation says it equalizes the pair.

Suppose $h:F^T(A)\to (B,b)$ is a monad algebra morphism with $h{\mu }_A=hTa$. Put $k=h{\eta }_A:A\to B$. Naturality of $\eta$ and the unit law giveMoreover, since $h$ is a monad algebra morphism,Thus $k$ is a monad algebra morphism and factors $h$ through $a$. If $k^{\prime }a=h$, then $k^{\prime }=k^{\prime }a{\eta }_A=h{\eta }_A=k$, proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.

The underlying fork is a split coequalizer: take $q=a$, $s={\eta }_A$ and $t={\eta }_{TA}$. Then $qs=1_A$, ${\mu }_{A}t=1_{TA}$, and $(Ta)t=sa$. Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in ${\mathcal{C}}^T$.