Every presheaf is a colimit of representables using point-wise computation of colimits Let $C$ be a small category and let $F \in \text{Fun}(C^{op}, \text{Set})$ be a presheaf. 

I'm trying to show it is a colimit of representables using the fact
  that colimits in functor categories are computed point-wise.


Define a comma category $D$ with objects pairs $(x,y)$ where $x \in C$ and $y \in F(x)$, and maps $(x_1, y_1) \to (x_2, y_2)$ are maps $f: x_1 \to x_2$ such that $F(f)(y_2) = y_1$.
Consider a functor $P: D \to C$ such that $P((x,y)) = x$.
We aim to show $\text{colim}_{D}(Y \circ P) \cong F$.
Define a functor $G: C^{op} \to \text{Fun}(D, \text{Set})$ by $G(x)((x_1,y_1)) = [Y\circ P]((x_1,y_1))(x) = \text{Hom}(x,x_1)$.
Since colimits are computed point-wise:
$\text{colim}_D(Y \circ P)(x) = \text{colim}_DG(x)$.
So, if we can $\text{colim}_DG(x) = F(x)$ for each $x \in C$ we'd be done.
Fix $x \in C$. It is very easy to show that $F(x)$ with maps $\psi_i: \text{Hom}(x, x_i) \to F(x)$ given by $\psi_i(g) = F(g)(y_i)$ for pairs $(x_i,y_i) \in D$ gives a cocone on $D$.
I can not find a way to prove it is universal. I've also considered the isomorphism $F(x) \cong \text{Hom}(Y(x), F)$ given by the Yoneda lemma, to no avail. Any suggestions?
 A: I've modified your notation slightly. $x_0$ is the point instead of $x$, and $\psi$ is indexed by pairs $x,y$ in $D$.
To show universality, you want to suppose that for any set $S$ with a cocone of maps $\psi_{x,y}' : \newcommand\Hom{\operatorname{Hom}}\Hom(x_0,x)\to S$, you can construct a unique map $\alpha : F(x_0)\to S$.
Then given $y\in F(x_0)$, define $\alpha(y)=\psi_{x_0,y}'(\newcommand\id{\operatorname{id}}\id_{x_0})$.
Then we check that $\alpha \circ \psi_{x,y} = \psi'_{x,y}$.
Suppose $g\in \Hom(x_0,x)$, then $\psi_{x,y}(g) = F(g)(y)$.
Then $$\alpha(\psi_{x,y}(g)) = \alpha(F(g)(y))=\psi'_{x_0,F(g)(y)}(\id_{x_0}).$$
Observe that given $g\in \Hom(x_0,x)$, $g$ defines a morphism between the pairs $(x,y)$ and $(x_0,F(g)(y))$ by definition, so we must have $\psi'_{x_0,F(g)(y)} = \psi'_{x,y}\circ g_*$.
Thus $\alpha(\psi_{x,y}(g)) = \psi'_{x,y}(g_*\id_{x_0}) = \psi'_{x,y}(g)$, as desired.
As for uniqueness, suppose $\beta$ were another appropriate map $F(x_0)$ to $S$.
Then since $y = \id_{F(x_0)}y = F(\id_{x_0})y = \psi_{x_0,y}(\id_{x_0})$, for any $y\in F(x_0)$, we have $\beta(y) = \beta(\psi_{x_0,y}(\id_{x_0}))=\psi'_{x_0,y}(\id_{x_0}) = \alpha(y)$. Thus $\alpha$ is unique.
