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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:

First prove the exact lifting statement for the forgetful functor $U^T:{\mathcal{C}}^T\to \mathcal{C}$. Take any diagram in a category of algebras for a monad $(A_j,a_j)$ whose underlying diagram has a chosen categorical limit $p_j:L\to A_j$. The arrows $a_{j}T(p_j):TL\to A_j$ form a categorical cone, since every diagram arrow is a morphism of algebras for a monad. Thus the universal property gives a unique arrow $a:TL\to L$ satisfyingThe projections are jointly monic: two arrows into $L$ with the same composites with every $p_j$ are equal. Therefore the algebra identities can be checked after these projections. For the unit,For multiplication,The algebra law for $a_j$ makes the right-hand sides equal. Thus $(L,a)$ is an algebra for a monad, and each $p_j$ is a morphism of algebras for a monad. Its action is unique with this property.

Given an algebra cone $q_j:(B,b)\to (A_j,a_j)$, let $q:B\to L$ be its unique underlying mediating map. Thenso $qb=aT(q)$ by joint monicity. Hence $q$ is an algebra map, proving the lifted cone is limiting. This establishes that the monad algebra forgetful functor creates limits, without any assumption that $T$ preserves limits.

For a monadic adjunction, the comparison satisfies $U=U^{T}K$. If $K$ is an isomorphism of categories over the base, the preceding unique lift transports back literally, so $U$ is a limit-creating functor. If monadic means that $K$ is only an equivalence of categories, the precise conclusion is creation of limits up to isomorphism: the lifted algebra is isomorphic to an object $K(D)$, and its limiting cone transports along that isomorphism. Full faithfulness supplies the arrows and their uniqueness.

This distinction concerns prescribed underlying objects, not existence of limits. Literal creation does not follow from an arbitrary equivalence alone. For example, the inclusion of the one-object category into the two-object indiscrete category, choosing one of its objects, is an equivalence of categories and is monadic under the equivalence convention. Either object of the target is terminal, but the other chosen terminal object has no literal object preimage. Thus the strict reading of the requested assertion needs the strict comparison convention; the equivalence-invariant statement is proved above.

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.

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.

A functor admitting a further left adjoint must itself be a right adjoint and preserve all existing categorical limits. A functor admitting a further right adjoint must preserve all existing colimits. Failure to preserve a single terminal object, equalizer, initial object or coproduct therefore rules out the corresponding extension. In the adjoint chain for the objects of a category, connected components fail an equalizer and the indiscrete category functor fails a binary coproduct.