Kan extensions and the yoneda embedding. [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 $D(-,fc)$?

 A: I know its good "explanation".
Let $X, Y$ be sets, and $f:X\to Y$ be a map. Then, it induces a inverse image map $f^{-1}:P(Y)\to P(X)$. Regarding $P(X),P(Y)$ as categories by its inclusion order, $f^{-1}$ can be regard as a functor. Then, this functor has a left and right adjoint. The left adjoint functor is just an image $f_*:P(X)\to P(Y)$, and right adjoint $f_!:P(X)\to P(Y)$ is called "dual image, small image" defined by $f_!(U)=Y-f(X-U)$. It is easy to check they are actually adjoint.
This is a special case of Kan extensions. Because by taking $2=\{0\to 1\}$ and regarding $X$ as discrete category, we can describe $P(X)$ as just a functor category $2^X$. And the "composite with $f$ functor" is nothing but the inverse image map. 
Furthermore, these categories are $2$-enriched, so the contravariant Hom-functor $Hom(-,x):X^{op}=X\to 2$ is seen as an object of $2^X$. But except for $y=x$, $Hom(y,x)$ is empty. This means it's a "characteristic functor" of $\{x\}\in P(X)$. So, $\{-\}:X\to P(X), x\mapsto \{x\}$ is seen as a "$2$-enriched yoneda embedding".
In this case, the restriction of $f_*$ to X along the yoneda embedding is $f$, composed with the yoneda embedding. However, it is clear that there is no nice connection between $f_!$ and the yoneda embedding. I think this is a good analogy to explain the advantage of left Kan extension in combination with yoneda embedding.
A: As Martin stated, the answer is no in general. However, the answer is yes if f is full and faithful and everything in $D$ can be written as a colimit of something in $C$ (i.e. you want that $D \to Set^{C^{op}}$ via $d \mapsto Hom(f(blank),d)$ to be fully faithful as well).
In this case, $f_+$ (which is classicaly denoted $f_*$) agrees with f on representables. An example of such a situation is when $f$ is the inclusion of compact Hausdorff spaces into compactly generated Hausdorff spaces.
A: 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 $C^{op}=D=Set, f = Hom(-,2), x = 0, c = 1$.
A: Let $U: \mathscr{A} \to \mathscr{C}$. If $\mathscr{E}$  has (enought large)  limits resp. colimits then the functor  $U^*: CAT(\mathscr{C}^{op}, \mathscr{E} ) \to  CAT(\mathscr{A}^{op}, \mathscr{E} ): P \mapsto   P\circ U^{op}$   has a left adjoin $U_!=Lan_{ U^{op}}$ (puntual Kan extention) resp. a right adjoint  $U_*=Ran_{ U^{op}  }$ and $U_! \dashv U^* \dashv U_*$ with  $U_!(P)(X):= {\underrightarrow{lim}} (P\circ \pi^{op}: (X\downarrow U)^{op}\to  \mathscr{A} ^{op}\to \mathscr{E})$
$U_*(P)(X):={\underleftarrow{lim}} (P\circ \pi^{op}: (U\downarrow X)^{op}\to  \mathscr{E} $ ). 
Let  $\mathscr{E} =Set$ and $\mathscr{A}, \mathscr{B}$ small (we can have more general conditions for the existence of puntual Kan extentions) ,  we have 
$U_!(P)$= $Lan_{ h_-} (h_U)(P)$ = 
$\underrightarrow{lim}$$_{(A, a)\in \mathscr{A} \downarrow P }$ $h_{ U(A)}$, 
$U_*(P)(X)  =\mathscr{A} ^>(h^U_X, P)$ 
indeed:  
$(\underrightarrow{lim}$$_{(A, a)}$ $h_{U(A)}, Q)\cong$, 
$ {\underleftarrow{lim}}_{(A, a)}  QU(A)\cong $
$({\underrightarrow{lim}}_{(A,a)} h_A, Q\circ U)$ ;  
$(Q\circ U, P) \cong  ({\underrightarrow{lim}}_{(X, x)\in \mathscr{C}\downarrow Q } h^U_X , P) \cong  {\underleftarrow{lim}}_{(X, x)}  (h_X, \mathscr{A} ^>(h^U_-, P)) \cong   (Q(-), \mathscr{A}^>(h^U_-, P))$.
Then (answere for you): $U_∗(h_Y)(X)=(h^U_X,h_Y)$.
