Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 18 / 6

A presheaf on $0\to 1$ is an arrow $r:B=X(1)\to A=X(0)$ in sets. A morphism of presheaves is a commuting square. DefineA commuting square from $r$ to $\Delta S$ is determined by its map $A\to S$, while a square from $\Delta S$ to $r$ is determined by its map $S\to B$. A square from $r$ to $\nabla S$ is also determined by its map $B\to S$, because the other component is the unique map $A\to 1$. Therefore these Hom-set bijections establish the adjoint chain for evaluation in an arrow category:There is a further left adjoint to $\Pi$: $\Lambda (S)=(\varnothing \to S)$. A square $\Lambda S\to r$ is exactly a map $S\to A$, proving $\Lambda ⊣\Pi$.

There is no right adjoint to $\nabla$. If there were, $\nabla$ would be a left adjoint and preserve the initial object. But it sends the empty set to $(\varnothing \to 1)$, whereas the initial object of this arrow category is $(\varnothing \to \varnothing )$. They are not isomorphic. Hence $\Pi$ does have a left adjoint, while $\nabla$ has no right adjoint.

Let $D(S)$ be the discrete category on $S$, and $I(S)$ the indiscrete category, with exactly one morphism between each ordered pair of objects. A functor $D(S)\to \mathcal{A}$ is precisely a map $S\to O(\mathcal{A})$; a functor $\mathcal{A}\to I(S)$ is precisely a map $O(\mathcal{A})\to S$. Thus $D⊣O⊣I$.

Let $C(\mathcal{A})={\pi }_0\mathcal{A}$ be the set of connected components, where objects are related by finite zigzags of morphisms in either direction. A functor $\mathcal{A}\to D(S)$ must take the endpoints of every morphism to the same point, so it is exactly a map ${\pi }_0\mathcal{A}\to S$. This gives the adjoint chain for the objects of a categoryIt cannot extend to the left. The two functors from the terminal category to $I(\{0,1\})$ selecting distinct objects have empty equalizer in $\mathbf{Cat}$. Applying $C$ gives two identical maps between singleton sets, whose equalizer is a singleton. Thus $C$ does not preserve equalizers and cannot be a right adjoint.

It cannot extend to the right either. The coproduct of two copies of $I(1)$ in $\mathbf{Cat}$ is the discrete two-object category. But $I(1⨿1)$ is the indiscrete two-object category, with morphisms between its distinct objects. They are not isomorphic, or even equivalent. Hence $I$ does not preserve binary coproducts and cannot be a left adjoint. Neither end of this chain has a further adjoint.