Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 / 6

A monad $\mathbb{T}=(T,\eta ,\mu )$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ obeys $ha=bT(h)$. The functor laws show that identities and composites obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^{\mathbb{T}}$.

For an adjunction $F⊣G$, let $\eta$ be its adjunction unit and $\epsilon$ its adjunction counit. The monad induced by an adjunction isThe two unit laws are $G{\epsilon }_{FA}{\eta }_{GFA}=1_{GFA}$ and $G({\epsilon }_{FA}F{\eta }_A)=1_{GFA}$, respectively the two triangle identities for an adjunction.

For associativity, naturality of the adjunction counit at ${\epsilon }_{FA}:FGFA\to FA$ saysApply $G$ to obtain ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. The composites defining $\eta$ and $\mu$ are natural transformations, so all monad axioms have been checked, not only their underlying objectwise types.

Define the forgetful functor $G^{\mathbb{T}}(A,a)=A$, acting as the identity on underlying morphisms. Define the free algebra functor byThe monad laws make ${\mu }_X$ an algebra for a monad structure, and naturality of $\mu$ makes $T(h)$ an monad algebra morphism.

For $(A,a)$, define the transpose mapsThe map $\Psi (k)$ is an monad algebra morphism, because naturality of $\mu$ and the algebra associativity law giveFor $k:X\to A$, naturality of $\eta$ gives $\Phi \Psi (k)=aT(k){\eta }_X=a{\eta }_{A}k=k$. Conversely, if $h$ is an monad algebra morphism, thenPrecomposing by a map into $X$ or postcomposing by an monad algebra morphism respects both formulas. Thus the bijection is natural and $F^{\mathbb{T}}⊣G^{\mathbb{T}}$. This is the free-forgetful Eilenberg-Moore adjunction.

Its adjunction unit is the given ${\eta }_X$; its adjunction counit at $(A,a)$ is the monad algebra morphism $a:(TA,{\mu }_A)\to (A,a)$. Hence $G^{\mathbb{T}}{F}^{\mathbb{T}}=T$, and the induced multiplication is the underlying counit at $(TX,{\mu }_X)$, namely ${\mu }_X$. Therefore this adjunction induces exactly the original monad, including its unit and multiplication of a monad.