1
$\begingroup$

Let $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{C}\to\mathcal{E}$ be two full and faithful functors from a small category $\mathcal{C}$ to categories $\mathcal{D}$ and $\mathcal{E}$ that are both both complete and cocomplete.

Clearly the left Kan extension $\text{Lan}_FG:\mathcal{D}\to\mathcal{E}$ exists. Is it immediate that it has a right adjoint? If yes, is the right adjoint $\text{Ran}_G F:\mathcal{E}\to\mathcal{D}$?

I have a couple of examples where (I think) this is the case, so I wonder if this is a general result.

  • 1
    You can say something if one among $F,G$ has an adjoint on some side. Is it the case? In fact there's no hope for a "general" statement: if this result was always true without further assumptions on $G$ it would imply that every subcategory is co/reflective, which is false (talke $F=1$, then you're saying that $Lan_1G=G\dashv Ran_G1$). – 2017-01-31
  • 0
    A counterexample: include fully and faithfully any object of a category in a category without terminal object (like, I don't know, $\{[0]\}\hookrightarrow \Delta^\text{op}$; this tautological functor has a right adjoint iff $\Delta^\text{op}$ has a terminal object, iff $\Delta$ has an initial object; there is no such object in $\Delta$) – 2017-01-31
  • 0
    @FoscoLoregian One of them is the Yoneda embedding... – 2017-02-01
  • 1
    If that's the case then you have a general theorem. If $F : {\cal A}\to {\cal C}$ goes from a small to a cocomplete category, then $\text{Lan}_yF\dashv \text{Lan}_Fy$ (so not a right Kan extension, but another left!); that's called "nerve-realization paradigm". – 2017-02-01
  • 0
    @FoscoLoregian Ah thanks! Is this still true if $G$ is not exactly the "Yoneda embedding" to $[\mathcal{C}^{\text{op}},\mathbf{Set}]$ but rather to a full subcategory of it that contains the representables, e.g. some limit-preserving presheaves? – 2017-02-01
  • 0
    If one among $F,G$ is the Yoneda embedding there are no assumptions on the other functor, the adjunction $\text{Lan}_yF\dashv \text{Lan}_F y$ is for free. You call the right adjoint the "nerve", because it coincides with $\hom(F,1)$, and if $F$ is dense, then "nerve" is a fully faithful functor. – 2017-02-01
  • 0
    @FoscoLoregian I am not talking about the ``other'' functor, but $y$ itself. I am asking what if this $y$ has the same effect as the Yoneda embedding, but it does not land to all presheaves but rather to some full subcategory that contains the representables...I don't know if there is a name for such a thing, maybe "the restricted Yoneda embedding". – 2017-02-01

0 Answers 0