# Is every functor a composition of adjoint functors?

My understanding of Ben's answer to this question is that even though associated graded is not an adjoint functor, it's not too bad because it is a composition of a right adjoint and a left adjoint.

But are such functors really "not that bad"? In particular, is it true that any functor be written as the composition of a right adjoint and a left adjoint?

• A similar but simpler argument: any adjunction between categories $C$ and $D$ induces a bijection between their sets (or classes) $\pi_0 C$ and $\pi_0 D$ of connected-components. So categories with different numbers of connected-components are never linked by a chain of adjoint functors. Eric's example will also do here. – Tom Leinster Aug 16 '12 at 15:02
• Avoiding the nerve functor, we can also give further examples (without different numbers of connected-components). For instance, we can use the following fact (which is similar to the facts given above and implies the observation of Tom Leinster): there is a functor $L: Cat\to Grp$ which is left adjoint to the inclusion of the 2-category of groupoids into the 2-category of categories. Clearly, adjoints are taken to adjoints by $L$. Therefore, since the adjoints of $Grp$ are equivalences, we get that adjoints are taken to equivalences of groupoids. – Fernando Mar 23 '17 at 18:53
• Then, once we know how to construct $L$, it is easy to get examples. For instance, there is no chain of adjoints between the terminal category and any category with at least one nontrivial automorphism. – Fernando Mar 23 '17 at 19:03
• In short, to construct examples, one only needs to consider (or construct) a nontrivial 2-functor $\mathcal{A} : \mathsf{Cat}\to \mathsf{X}$ in which $\mathsf{X}$ is locally groupoidal and "nontrivial" means that there are at least two nonequivalent objects of $\mathsf{X}$ in the image of $\mathcal{A}$. – Fernando Mar 24 '17 at 12:36