[Edit: For a category $C$ let $C^\wedge$ denote the category of presheaves on $C$.]


Let $f:C\to D$ be a functor. Precomposition with $f^{op}$ induces a functor
$f^\wedge:D^\wedge \to C^\wedge$. This functor has both a left- and a right adjoint, called *left-* and *right kan extension*:


$f\_\wedge \dashv f^\wedge \dashv f\_+$.


Now for $c\in C$ we get $D^\wedge(D(-,fc),Y)=Y(fc)=f^\wedge Y(c)=C^\wedge(C(-,c),f^\wedge Y)$. This gives us the restriction of $f\_\wedge$ to $C$ along the yoneda embedding: It is $f$ (composed with the yoneda embedding).


Now here's my question:



> 
> What is the restriction of $f\_+$ to $C$ along the yoneda embedding?
> 
> 
> 


It seems not to agree with $f$ but:



> 
> Is there a nice connection between $f\_+C(-,c)$ and $D(-,fc)$?
> 
> 
> 



Kan extensions and the yoneda embedding.
For all $Z \in C^\wedge, Y \in D^\wedge$, we have $C^\wedge(f^\wedge Y,Z)=D^\wedge(Y,f\_+ Z)$. If we put $Y = D(-,d), Z = C(-,c)$, we get


$(f\_+ C(-,c))(d) = C^\wedge(f^\wedge D(-,d),C(-,c)) = C^\wedge(D(f-,d),C(-,c))$


There seems to be no connection between $f\_+ C(-,c)$ and $D(-,fc)$ (only when $f$ is an equivalence). For example,


$D^\wedge(D(-,fc),f\_+ C(-,c)) = C^\wedge(f^\wedge D(-,fc),C(-,c)) = C^\wedge(D(f-,fc),C(-,c))$


and it is possible to construct an example where there is no natural transformation $D(f-,fc) \to C(-,c)$ at all. For example if $D(fx,fc)$ is nonempty, but $C(x,c)$ is empty. Take $C^{op}=D=Set, f = Hom(-,2), x = 0, c = 1$.


