Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 8

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\Rightarrow T$, $\mu :T^2\Rightarrow T$, satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$, $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism $(A,a)\to (B,b)$ is an arrow $h:A\to B$ with $ha=bT(h)$. These form the Eilenberg-Moore category ${\mathcal{C}}^T$.

A functor $G:\mathcal{D}\to \mathcal{C}$ is monadic when it has a left adjoint $F$ and its comparison $K:\mathcal{D}\to {\mathcal{C}}^{GF}$ is an equivalence; this is the monadic adjunction convention. Some strict formulations require $K$ to be an isomorphism over the base. The distinction affects the meaning of literal limit creation, discussed in part (iii), but not the algebra construction.

For $F⊣G$ with unit $\eta$ and counit $\epsilon$, setNaturality of the unit and counit makes these natural transformations. The two triangle identities give ${\mu }_A{\eta }_{TA}=1_{TA}$ and ${\mu }_{A}T{\eta }_A=1_{TA}$. Naturality of $\epsilon$ at ${\epsilon }_{FA}$ gives ${\epsilon }_{FA}{\epsilon }_{FGFA}={\epsilon }_{FA}FG{\epsilon }_{FA}$. Applying $G$ proves ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. Thus this is the monad induced by an adjunction.

Conversely, for any monad take the free-algebra functor $F^T(A)=(TA,{\mu }_A)$ and the forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$. For an algebra $(B,b)$, the assignments $h\mapsto h{\eta }_A$ and $k\mapsto bT(k)$ give inverse bijections between algebra maps $F^T(A)\to (B,b)$ and arrows $A\to B$. For the inverse identities, $bT(k){\eta }_A=b{\eta }_{B}k=k$, while $bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$. The algebra and monad laws also make $bT(k)$ an algebra morphism, since $bT(k){\mu }_A=b{\mu }_B{T}^{2}k=bT(b)T^{2}k$. These bijections are natural, giving the free-forgetful Eilenberg-Moore adjunction. Its unit is $\eta$, its counit at $(B,b)$ is $b$, and its induced multiplication is exactly $\mu$. Thus every monad arises from an adjunction.

First prove strict creation for the canonical forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$. Let a diagram of algebras $(A_i,a_i)$ have a chosen underlying limit cone $p_i:L\to A_i$. The maps $a_{i}T(p_i):TL\to A_i$ form a cone: if $h:A_i\to A_j$ is a diagram arrow, then $h{a}_i=a_{j}T(h)$ and $h{p}_i=p_j$. Thus the limit property gives a unique action $a:TL\to L$ withThe projections are jointly monic. By naturality of $\eta$ and the algebra unit laws, $p_{i}a{\eta }_L=a_i{\eta }_{A_i}{p}_i=p_i$, hence $a{\eta }_L=1_L$. Likewiseso $a{\mu }_L=aT(a)$. Therefore $(L,a)$ is an algebra and all $p_i$ are algebra morphisms.

For any algebra cone $h_i:(B,b)\to (A_i,a_i)$, let $h:B\to L$ be its unique underlying factor. Then $p_{i}hb=h_{i}b=a_{i}T(h_i)=p_{i}aT(h)$, so joint monicity gives $hb=aT(h)$. Thus the factor is an algebra morphism, unique as such. The action on the prescribed $L$ was itself forced by the cone equations. This proves monad algebra forgetful functor creates limits, without assuming $T$ preserves them.

For a monadic $G$ with comparison an equivalence, transport $(L,a)$ and its cone across that equivalence. This yields a limiting cone in $\mathcal{D}$ whose image is identified with the prescribed base cone by an isomorphism, uniquely up to the corresponding cone-compatible isomorphism. Thus a monadic functor creates limits in the equivalence-invariant sense. If monadicity uses a comparison isomorphism over the base, the lift is literally on the prescribed underlying cone, proving strict creation as well.

A qualification is necessary if equivalence is used to define monadicity but literal unique lifting is demanded. Let $\mathcal{C}$ be the indiscrete category on two objects $0,1$, let $\mathcal{D}$ be the terminal category, and let $G$ select $0$. The unique $F:\mathcal{C}\to \mathcal{D}$ is left adjoint to $G$, and the comparison to the induced algebra category is an equivalence: the two algebra objects have exactly one map between each pair. Yet the chosen terminal cone with vertex $1$ in $\mathcal{C}$ has no literal lift, since the only object in the image of $G$ is $0$. The creation of limits up to isomorphism conclusion holds, because $0$ and $1$ are uniquely isomorphic. This distinguishes the two conventions rather than claiming an invalid strict conclusion from equivalence alone.