# Left Kan extension and extension of functors

Consider a functor $F:\mathcal{A}\to \mathcal{B}$ between two small categories. Let $\mathcal{K}$ be a locally presentable category. Consider a functor $G:\mathcal{A}\to \mathcal{K}$, there is a natural transformation $G\Rightarrow (\mathrm{Lan}_F G)\circ F$ coming from the adjunction.

Is there a known necessary and sufficient condition for the natural transformation $G\Rightarrow (\mathrm{Lan}_F G)\circ F$ to be an isomorphism ?

By Proposition 3.7.3 of Borceux's book (Handbook of categorical algebra vol. 1), a sufficient condition is that $F:\mathcal{A}\to \mathcal{B}$ is full and faithful.

EDIT: by Exercice 3.9.5 of Borceux's book (Handbook of categorical algebra vol. 1), this condition is not necessary. Another sufficient condition is $F$ full and $\forall A,A'\in \mathcal{A}$ and any parallel arrows $f,f':A\rightrightarrows A', Gf=Gf' \Rightarrow Ff=Ff$ (there is a typo by the way in the book, $A$ is written instead of $A'$)

## 1 Answer

$F$ being full and faithful is a necessary and sufficient condition for this to be true for all $\mathcal{K}$ and all $G$. To see the necessity, take $\mathcal{K} = \mathrm{Set}$ and $G = \mathcal{A}(a,-)$ for some object $a$. then $\mathrm{Lan}_F G(b) = \mathcal{B}(F(a),b)$, and so $(\mathrm{Lan}_F \circ F)(a') = \mathcal{B}(F(a),F(a'))$, with the unit $G \Rightarrow \mathrm{Lan}_F \circ F$ being the action of $F$ on arrows. Thus, if this is an isomorphism for all $a$, then $F$ is full and faithful.

For fixed $\mathcal{K}$ and/or $G$, weaker conditions can suffice. For instance, if $\mathcal{K}=1$, then there is only one possible $G$, and the condition always holds no matter what $F$ is.

• It's a very interesting post but $G$ is fixed. I edit my question. Jun 2 '18 at 0:32