Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 21 / 4

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

Let $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ forget the last partial unary operation. For an object $A$ of the nested partial unary operation category ${\mathcal{C}}_n$, setThe nesting rule implies that every earlier operation is defined and fixes such an $x$. To construct a free extension of nested partial unary operations, adjoin a disjoint infinite chain for each $x\in S_n(A)$:Keep the old operations on $A$. On each new chain put ${\alpha }_1(x,k)=(x,k+1)$; every old operation ${\alpha }_i$ with $2\le i\le n$ is undefined there, because the first operation has no fixed points there. Define the new operation only on $S_n(A)$, by ${\alpha }_{n+1}(x)=(x,0)$. These are exactly the fixed points of the old top operation, so the required domain rule is satisfied. The inclusion $A\hookrightarrow U_n{L}_{n}A$ preserves all old defined operations.

For a morphism $f:A\to U_{n}B$, preservation of the old operations makes $f(x)\in S_n(B)$ whenever $x\in S_n(A)$. Any extension to a morphism $\hat{f}:L_{n}A\to B$ must haveThe operation ${\beta }_{n+1}$ is defined at $f(x)$, and ${\beta }_1$ is total, so this formula exists. It preserves the new operation and the chain operation; there are no other defined operations on new points to check. It therefore gives the unique extension. Naturality and functoriality follow from uniqueness. ThusNotice the useful extra fact: the new top operation has no fixed points, since every value $(x,0)$ is new and every input $x$ is old. Hence freely adding any further operations adds no more points; those operations have empty domains.

Put ${\mathcal{C}}_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to {\mathcal{C}}_1$ sends $A$ to $A\times \mathbb{N}$ with ${\alpha }_1(a,k)=(a,k+1)$. A function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\beta }_1^k(f(a))$. Its induced monad is $T_{0}A=A\times \mathbb{N}$, with unit $a\mapsto (a,0)$ and multiplication $(a,k,l)\mapsto (a,k+l)$. An algebra for a monad is determined by the action of $1\in \mathbb{N}$, so its Eilenberg-Moore category is ${\mathcal{C}}_1$. Thus this first free–forgetful adjunction is a monadic adjunction without needing the given assumption for $n=0$.

The monadic length counts successive Eilenberg-Moore comparison functor steps until the comparison becomes an equivalence. Fix $0\le m<n$. For $m\ge 1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the left adjoint $F_{n,m}$ to the composite forgetful functor $U_{n,m}:{\mathcal{C}}_n\to {\mathcal{C}}_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of ${\mathcal{C}}_m$, unit and multiplication are exactly those of the one-step adjunction:as monads, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.

By the assumed one-step monadic adjunction property, or the explicit argument above when $m=0$, the Eilenberg-Moore category ${\mathcal{C}}_m^{T_m}$ is equivalent to ${\mathcal{C}}_{m+1}$. Under this equivalence the comparison ${\mathcal{C}}_n\to {\mathcal{C}}_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using ${\alpha }_1$, and evaluates the first new value using ${\alpha }_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole monadic tower for nested partial unary operations iswith the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.

It does not terminate earlier. If $m<n$, take the same underlying two-element set in two objects of ${\mathcal{C}}_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make ${\alpha }_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to ${\mathcal{C}}_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a full functor, and cannot be an equivalence. Therefore