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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?

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The answer is no, because the nerve functor turns an adjoint pair of functors between categories into inverse homotopy equivalences between spaces (this is because of the existence of the unit and counit and the fact that nerve turns natural transformations into homotopies). In particular, this means that any functor whose nerve is not a homotopy equivalence cannot be a composite of adjoints. For a very simple example, you could take the functor from the 2-object discrete category to the terminal category.

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    $\begingroup$ 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. $\endgroup$
    – Tom Leinster
    Commented Aug 16, 2012 at 15:02
  • $\begingroup$ 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. $\endgroup$
    – Fernando
    Commented Mar 23, 2017 at 18:53
  • $\begingroup$ 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. $\endgroup$
    – Fernando
    Commented Mar 23, 2017 at 19:03
  • $\begingroup$ 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} $. $\endgroup$
    – Fernando
    Commented Mar 24, 2017 at 12:36
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Here's a really trivial way to see that the answer is "no": a functor from the empty category to a nonempty category is never a composite of adjoints (since a functor from the empty category to a nonempty category is never an adjoint).

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