Let $\mathcal{C}$ and $\mathcal{D}$ be categories, and suppose $F\colon\mathcal{C}\to\mathcal{D}$ is a functor. It induces two adjoint pairs between $Set~^{\mathcal{C}}$ and $Set~^{\mathcal{D}}$; one is denoted $(F^\star,F\_\star)$ and one is denoted $(F\_!,F^\star)$. One proves easily that the counit to $(F^\star,F\_\star)$ is a natural isomorphism of functors $\mathcal{C}\to Set$ if and only if $F$ is fully faithful.


I am interested in the counit of the other adjunction $F\_!:Set~^{\mathcal{C}}\Longleftrightarrow Set^{\mathcal{D}}:F^\*$. Lets denote it by $$\epsilon\_F\colon F\_!F^\*\to \operatorname {id}\_{Set^{\mathcal{D}}}.$$


Question: Under what conditions on $F$ is $\epsilon\_F^~$ a natural isomorphism?



When is the $(F_!,F^*)$ counit a natural isomorphism?
It appears to me that the condition on $F:\cal C\to\cal D$ would be:


For any morphism $s: a\to b$ of $\cal D$, the following category is connected:


An object consists of a $\cal C$-object $c$ and a factorization $a\to F(c)\to b$ of $s$.


A morphism $c\_1\to c\_2$ is a $\cal C$-morphism such that the induced map $F(c\_1)\to F(c\_2)$ is compatible with the maps from $a$ and to $b$.


I don't recall ever having run into this sort of 'two-sided comma category' before, but it seems to be the answer.


I got this by choosing $G$ from $\cal D$ to Set to be represented by the object $a$ and thinking about the fiber of the map $\epsilon: (F\_!F^\*G)(b)\to G(b)$ over the element $s$.


