Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It was asked here whether every functor is a composition of adjoint functors. The answer is no, because all adjoint functors induce homotopy equivalences on the nerve, and we can construct functors that do not induce homotopy equivalences.

My question is the following: can all functors inducing homotopy equivalences on the nerve be expressed as compositions of adjoint functors?

share|improve this question
    
One thing to note: in order to make this question precise, one needs a notion of homotopy equivalence when the category is not small. –  Lunasaurus Rex Dec 28 '12 at 12:05
    
Side note, here is the meta: tea.mathoverflow.net/discussion/1502/… –  David Corwin Dec 29 '12 at 22:24
add comment

1 Answer

The answer is no. Let $C$ be a category such that the unique map from $C$ to the terminal category is a composition of $n$ adjoints. Then $C$ has an object $x_0$ such that every other object of $C$ can be connected to $x_0$ by a zigzag of length at most $n$; this is easy to prove by induction on $n$.

In particular, let $R$ be the "infinite zigzag", the unique poset such that $|BR|$ is homeomorphic to $\mathbb{R}$. Then $BR$ is contractible, but the unique map from $R$ to the terminal category cannot be a composition of adjoints. Note, however, that this map is a transfinite composition of adjoints (of length $\omega$). It seems plausible to me that any functor between (small) categories which is an equivalence on nerves could be a transfinite composition of adjoints.

share|improve this answer
7  
A totally different way to get a counterexample: note that any composition of adjoints must have a functor going the other way which provides a homotopy inverse on nerves (namely, the composition of all the adjoint functors going the other way). It's easy to construct examples of finite posets $X$ and $Y$ such that there is a map $X\to Y$ inducing an equivalence but no map $Y\to X$ inducing an equivalence. –  Eric Wofsey Dec 28 '12 at 13:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.