Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 24 / 4

For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, with unit $\eta$ and counit $\epsilon$, call the adjunction idempotent when the multiplicationof its induced monad is invertible. We show that this idempotent adjunction condition is equivalent to invertibility of $\delta =L\eta R:LR\to (LR{)}^2$, the comultiplication of the induced comonad.

For an idempotent monad $T$, the unit laws implyIf $a:TA\to A$ is a monad algebra, naturality of the unit and its algebra unit law giveThus ${\eta }_A$ is an isomorphism. Apply this to the canonical algebra $R{\epsilon }_B:T(RB)\to RB$: its unit law is a triangle identity for an adjunction. It follows that ${\eta }_{RB}$, and therefore $L{\eta }_{RB}$, is invertible for every $B$. Hence $\delta$ is invertible.

Apply the same argument to the dual adjunction $R^{\mathrm{op}}⊣L^{\mathrm{op}}$. It gives the converse. Idempotence is self-dual: the induced monad is idempotent exactly when the induced comonad is idempotent. In particular, idempotence requires the unit to be invertible on every object in the image of the right adjoint.

The left adjoint to the morphism-set functor is the free category on independent arrows:Each copy has two objects, their identities, and one nonidentity arrow from the first object to the second. A functor $L(S)\to \mathcal{C}$ is specified by one arbitrary arrow of $\mathcal{C}$ for each $s$, since that arrow specifies the images of both objects as well. This proves the natural adjunction bijection with functions $S\to \mathrm{Mor}(\mathcal{C})$.

The induced monad on sets has $T(S)=S\times \{\text{source identity},\text{generator},\text{target identity}\}$. For a singleton $S$, its multiplication is a function from the nine-element set $T^{2}S$ to the three-element set $TS$, so it cannot be invertible. This adjunction is not idempotent.

The left adjoint is discretization $D$: give the underlying set of $X$ the discrete topology. Write $J$ for indiscretization. Every function $DX\to Y$ is continuous, and every function $X\to JY$ is continuous. The common set of underlying functions supplies the natural bijectionThe discrete-indiscrete adjunction has induced monad $JD$, which assigns the indiscrete topology to the original set. Applying it twice gives the same space, and its multiplication is the identity on that set. Thus the adjunction is idempotent. Its dual comonad $DJ$, discretization, is idempotent as well.

For a Boolean algebra $B$, let $U(B)$ be its set of ultrafilters, equivalently the Boolean homomorphisms $B\to 2$. Regard $U(B)$ as an object of ${\mathbf{Set}}^{\mathrm{op}}$. A homomorphism $k:B\to B^{\prime }$ acts by inverse image of ultrafilters, a set map $U(B^{\prime })\to U(B)$, hence an arrow $U(B)\to U(B^{\prime })$ in the opposite category.

A Boolean homomorphism $h:B\to \mathcal{P}(A)$ gives the functionConversely, a function $v:A\to U(B)$ gives $h(b)=\{a:b\in v(a)\}$. The ultrafilter laws are exactly what makes this map preserve the Boolean operations, including complements and $0,1$. These constructions are inverse and natural, proving the Boolean ultrafilter-power-set adjunctionIts unit is ${\eta }_B(b)=\{u\in U(B):b\in u\}$.

Take $B=\mathcal{P}(\mathbb{N})$, an object in the image of the right adjoint. There is a nonprincipal ultrafilter on $\mathbb{N}$, by extending the cofinite filter in the usual choice setting. The subset $P\subset U(B)$ consisting of all principal ultrafilters is not ${\eta }_B(b)$ for any $b\subseteq \mathbb{N}$: containing the principal ultrafilter at every integer forces $b=\mathbb{N}$, but ${\eta }_B(\mathbb{N})=U(B)$ also contains the nonprincipal ultrafilters. Thus ${\eta }_B$ is not surjective and is not an isomorphism. By the necessary condition proved in the root solution, this adjunction is not idempotent.