Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 119 / 4

A monad is an endofunctor $T:\mathcal{C}\to \mathcal{C}$ with natural transformations $\eta :1_{\mathcal{C}}\to T$ and $\mu :T^2\to T$ satisfyingIts Kleisli category has the same objects as $\mathcal{C}$, with ${\mathcal{C}}_T(A,B)=\mathcal{C}(A,TB)$, identity ${\eta }_A$, and compositeThe unit laws follow from naturality of $\eta$ and the two monad unit laws. For $h:C\to TD$, the two triple composites reduce, using naturality of $\mu$, tothey agree by the monad associativity law. Thus this is a category.

Define the free functor into a Kleisli category $J$ and the right-hand functor $U$ byThe identity and composition formulas for $J$ areFor $U$, $U({\eta }_A)={\mu }_{A}T{\eta }_A=1_{TA}$, andwhere the last step is naturality of $\mu$. Hence both mappings are functors. The hom-set identity ${\mathcal{C}}_T(JA,B)=\mathcal{C}(A,UB)$ gives the adjunction $J⊣U$. Its unit is $\eta$, and its counit at $B$ is the Kleisli arrow represented by $1_{TB}:TB\to TB$. Applying $U$ to that counit gives ${\mu }_B$, so the induced monad is the specified one.

For any $L⊣R:\mathcal{D}\to \mathcal{C}$ inducing this same monad, with counit $\epsilon$, define the Kleisli comparison functorThe triangular identities give $K({\eta }_A)=1_{LA}$, and naturality of the counit together with $\mu =R\epsilon L$ gives $K(g\star f)=K(g)K(f)$. Explicitly, naturality moves ${\epsilon }_{LB}$ past $L(g)$, then moves the resulting counit past ${\epsilon }_{LC}$, reducing the composite to ${\epsilon }_{LC}L({\mu }_{C}T(g)f)$. Thus $K$ is a functor, with $KJ=L$ and $RK=U$, and it sends the Kleisli counit to ${\epsilon }_L$.

An adjunction morphism here is required to commute with the left and right adjoints and preserve their adjunction structure. Such a morphism must have the above object map; every Kleisli arrow $f$ factors as its counit after $J(f)$, so its arrow map is forced as well. This proves initiality of the Kleisli adjunction:

For $L⊣R$ inducing $T$, the Kleisli comparison functor is always full and faithful, since the adjunction givesTherefore it is part of an equivalence of categories if and only if every object of $\mathcal{D}$ is isomorphic to $LA$ for some $A$.

For an idempotent monad, multiplication is invertible. The unit laws imply $T\eta =\eta T={\mu }^{-1}$. If $(A,a)$ is an algebra for a monad, then $a{\eta }_A=1_A$, while naturality of $\eta$ givesThus $a:TA\to A$ is an isomorphism with inverse ${\eta }_A$. The algebra law $a{\mu }_A=aT(a)$ says that $a$ is a morphism of algebras for a monad from the free algebra $(TA,{\mu }_A)$ to $(A,a)$, so every object of the Eilenberg-Moore category is isomorphic to a free algebra. The criterion above yields

Use $\mathbb{N}=\{0,1,\dots \}$, as required by the formula at zero, and composition of functions as multiplication in $M$. The displayed $T$ preserves identities, and for $n>0$,both sides send zero to zero. Thus $T$ is a functor on the one-object category.

For the shift monad on order-preserving maps of natural numbers, setThese are order-preserving functions. The equations $Tf\eta =\eta f$ and $\mu T^{2}f=Tf\mu$ hold pointwise, giving the required natural transformations. For the second equation, both sides are zero at $n=0,1$, and are $f(n-2)+1$ for $n\ge 2$. The two unit laws are $\mu \eta =\mu T\eta =1$. Associativity is checked byConsequently these maps define a monad. It is not an idempotent monad, since $\mu (0)=\mu (1)$, so $\mu$ cannot be invertible.

An algebra for a monad is a map $a\in M$ with $a\eta =1$ and $aTa=a\mu$. The first equation forces $a(n+1)=n$; monotonicity then forces $a(0)=0$. Thus $a=\mu$, which satisfies the second equation by the monad associativity law. The Eilenberg-Moore category therefore has exactly one object. Its endomorphisms $h$ satisfy $h\mu =\mu Th$, and this equation holds precisely when $h(0)=0$: evaluate at zero for necessity, and at $n>0$ both sides equal $h(n-1)$.

The Kleisli comparison functor takes its only object to this only algebra and sends $f\in M$ toIt is a bijection from the Kleisli arrows to the algebra endomorphisms, with inverse $h\mapsto (n\mapsto h(n+1))$. It preserves identities and composition by the comparison construction; directly, $K(\eta )=1$ and $K(g\star f)=K(g)K(f)$. Bijectivity on objects and arrows makes it an isomorphism of categories: