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For small categories, the object functor $O$ has left adjoint the discrete category functor and right adjoint the indiscrete category functor. Connected components, defined by zigzags of morphisms, give a left adjoint ${\pi }_0$ to the discrete functor. There is no further left adjoint: ${\pi }_0$ fails the equalizer of two functors selecting distinct objects of the indiscrete two-object category. There is no further right adjoint: the indiscrete functor fails the coproduct of two singleton sets.

Use the preservation obstructions to extending an adjoint chain: a functor having a further left adjoint must preserve categorical limits, while one having a further right adjoint must preserve colimits.

At the left end of the presheaf category chain, $\Lambda$ does not preserve the terminal object. The terminal categorical presheaf has singleton value everywhere, whereas $(\Lambda 1)(S)=\varnothing$ because $S\ne \varnothing$. Thus $\Lambda$ cannot itself be a right adjoint. At the right end, $\nabla$ does not preserve the initial object: $(\nabla \varnothing )(\varnothing )=\{\ast \}$ since the empty open set is proper. Thus $\nabla$ cannot itself be a left adjoint.

For the category of small categories chain, take two functors from the terminal category $1$ to the two-object indiscrete category $I\{0,1\}$, selecting its distinct objects. Their equalizer is the empty category. But their images under $C={\pi }_0$ are the same function between singleton sets, whose equalizer is a singleton. Hence $C$ does not preserve equalizers and has no further left adjoint. Finally, $I1⨿I1$ is the two-object discrete category, while $I(1⨿1)$ has arrows between its distinct objects. The canonical comparison is not an isomorphism of categories, so $I$ fails to preserve this coproduct and has no further right adjoint.

Let $C\mathcal{A}={\pi }_0\mathcal{A}$ be the set of connected components of a category: two objects of a category are identified when joined by a finite zigzag of morphisms. Let $DA$ be the discrete category on $A$, and let $IA$ be the indiscrete category on $A$, with exactly one morphism for every ordered pair of objects.

A functor $\mathcal{A}\to DA$ must give the same object value along every arrow, hence along every zigzag. Conversely a function ${\pi }_0\mathcal{A}\to A$ uniquely defines such a functor. A functor $DA\to \mathcal{A}$ is just a choice of an object of a category for each element of $A$. A functor $\mathcal{A}\to IA$ is likewise determined by an arbitrary object function, since its arrow images are forced and always compose correctly. ThereforeThis is the adjoint chain for the objects of a category:

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.