most recent 30 from http://mathoverflow.net 2013-05-26T01:06:46Z http://mathoverflow.net/feeds/question/335 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/335/is-every-functor-a-composition-of-adjoint-functors Anton Geraschenko 2009-10-12T03:51:02Z 2012-08-16T14:56:43Z

<p>My understanding of Ben's answer to <a href="http://mathoverflow.net/questions/263/what-is-the-universal-property-of-associated-graded" rel="nofollow">this question</a> 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.</p> <p>But are such functors really "not that bad"? In particular, is it true that <strong>any</strong> functor be written as the composition of a right adjoint and a left adjoint?</p>

http://mathoverflow.net/questions/335/is-every-functor-a-composition-of-adjoint-functors/336#336 Eric Wofsey 2009-10-12T04:07:55Z 2009-10-12T04:07:55Z

<p>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.</p>

http://mathoverflow.net/questions/335/is-every-functor-a-composition-of-adjoint-functors/104843#104843 Tom Leinster 2012-08-16T14:55:25Z 2012-08-16T14:55:25Z

<p>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). </p>