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Let $F:\mathcal{C}\to\mathcal{D}$ be a functor of small categories, and let $\mathcal{A}$ be a (co)complete category. The pre-composition with $F$ induces a functor $$F^\ast:\mathbf{Fun}(\mathcal{D},\mathcal{A})\to \mathbf{Fun}(\mathcal{C},\mathcal{A}).$$ A left (resp. right) Kan extension along $F$ is a left (resp. right) adjoint to $F^\ast$.

  1. Is there a sufficient and necessary conditions on $F$, so that the left and right Kan extensions coincide?
  2. What can one deduce when the left and right Kan extensions coincide? Particularly, does it imply that $F^\ast$ is an equivalence of categories?
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  • $\begingroup$ The left and right Kan extensions may coincide when $F^\ast$ is not an equivalence of categories. $\endgroup$ – user337830 Oct 27 '16 at 13:11
  • $\begingroup$ Can you tell us an example? $\endgroup$ – Berci Oct 27 '16 at 21:37
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    $\begingroup$ @Berci: let $\mathcal{D}$ be the preordered category corresponding to the preordered set $(\mathbb{N},\leq)$, let $\mathcal{C}=\mathcal{D}\times \mathcal{D}$, and let $F:\mathcal{C}\to \mathcal{D}$ be given on objects by $F(n,m)=n+m$. For an additive category $\mathcal{A}$, and for a functor $G:\mathcal{C}\to \mathcal{A}$, one has $ (\mathop{\mathbf{Lan}_F}G)_n=\bigoplus_{p+q=n}G(p,q)=(\mathop{\mathbf{Ran}_F}G)_n.$ And, it is easy to see that $F^\ast$ is not an equivalence of categories. $\endgroup$ – user337830 Oct 27 '16 at 22:36

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