5
$\begingroup$

Given adjoint functors $F: A \to B$, $G: B \to A$, if you then take their pullback functors $F^* : Set^B \to Set^A$ and $G^* : Set^A \to Set^B$ given by pre-composition, are these two also adjoint (assuming conditions for their adjoints to exist)? According to wikipedia http://en.wikipedia.org/wiki/Functor_category, this is true for post-composition, is it true for pre-composition? I realise these functors have Kan extensions as adjoints but can find nothing in the literature about taking the Kan extension along the adjoint of a functor.

$\endgroup$
8
  • 3
    $\begingroup$ If $F \dashv G$, then $G^\ast \dashv F^\ast$. Is that what you're asking? $\endgroup$ Commented Aug 14, 2014 at 13:37
  • 3
    $\begingroup$ However, for the precompositions $F^\ast: Set^{B^{op}} \to Set^{A^{op}}$ and $G^\ast: Set^{A^{op}} \to Set^{B^{op}}$, we do have $F^\ast \dashv G^\ast$ if $F \dashv G$. $\endgroup$ Commented Aug 14, 2014 at 13:45
  • $\begingroup$ Yes, that is what I meant :) Ah ok, it's true for $Set^{A^{op}}$? Do you know of where I could find a proof? Because surely then I could just apply this to the opposite functors $F^{op}: A^{op} \to B^{op}$ etc to get the result I need $\endgroup$ Commented Aug 14, 2014 at 13:48
  • 2
    $\begingroup$ It's an easy exercise if you use the definition of adjunction with the triangle identities. $\endgroup$
    – Zhen Lin
    Commented Aug 14, 2014 at 14:23
  • $\begingroup$ Charles, in case there was misunderstanding, my last comment was to apply to functors $F: A \to B$ and $G: B \to A$, which induce functors $Set^{B^{op}} \to Set^{A^{op}}$ and $Set^{A^{op}} \to Set^{B^{op}}$ in a hopefully obvious manner. $\endgroup$ Commented Aug 14, 2014 at 14:51

2 Answers 2

6
$\begingroup$

This is best understood via the bicategorical characterization of adjunctions in terms of unit and counit 2-morphisms. It is then clear that any bifunctor preserves adjunctions in the obvious sense. And taking presheaf categories and pullback functors is clearly a bifunctor from (small categories)$^{op}$ to categories.

$\endgroup$
2
  • $\begingroup$ I've usually heard the term bifunctor used to describe a functor from a product of categories (by analogy with bilinear map). Do you mean a pseudofunctor or a strict 2-functor? I'm not too familiar with higher category theory. $\endgroup$
    – ಠ_ಠ
    Commented Feb 27, 2017 at 23:03
  • 1
    $\begingroup$ I mean a 2-functor. Many people call (weak) 2-categories bicategories. $\endgroup$ Commented Apr 12, 2017 at 11:18
3
$\begingroup$

This result is E. Riehl, Category Theory in Context, Proposition 4.4.6.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .