I have the following problem as part of my exam preporation and I need an idea how to approach it at least:
Let $\mathfrak C$ be a small category and $F: \mathfrak C^{opp} \rightarrow \mathfrak S et$ be a presheaf. Lets constract new category $\mathfrak F$ as follows: $\operatorname{Ob} \mathfrak F = \sqcup_{X \in \operatorname{Ob} \mathfrak C} F(X) $ (disjoint union of all object images under the action of functor $F$) and for all $x \in F(X)$ and $y \in F(Y)$ $\operatorname{Hom}_\mathfrak F (x, y)$ consitst of all arrows $\phi \in \operatorname{Hom}_{\mathfrak C} (X, Y)$ such that $F \phi (y) = x$.
Find colimit of the diagram $D: \mathfrak F \rightarrow \mathfrak F unc(\mathfrak C^{opp}, \mathfrak Set)$ which maps object $x$ (which is also an element of $F(X)$) to presheaf $\operatorname{Hom}_\mathfrak C (-, X)$ and arrow $\phi$ from $\operatorname{Hom}_\mathfrak F (x, y)$ to natural transformation $ \phi_*$ of left multiplication on $\phi$.
Thanks!
It is going to turn out that the colimit of the diagram $D$ is just going to be $F$ itself, due to the Yoneda lemma.
We define a co-cone $\alpha$ under $D$ by setting $\alpha_{(x,X)}:D(x,X)\to F$ to be the unique natural transformation $\hom(-,X)\to F$ that sends $id_X$ to $x\in F(X)$. To show that this is natural, consider that for $f:(x,X)\to (y,Y)$, we need $$\alpha_{(y,Y)}\circ\hom(-,f)=\alpha_{(x,X)},$$ and by the Yoneda lemma it is sufficient to say that the left hand composite sends $id_X$ to $x$. First, $\hom(X,f)(id_X)=f\in\hom(X,Y)$; then because $$(\alpha_{(y,Y)})_Y(id_Y)=y$$ it must be the case that $$(\alpha_{(y,Y)})_X(f)=F(f)(y).$$ But since $f$ was a morphism $(x,X)\to (y,Y)$ in $\mathfrak{F}$, this means that $F(f)(y)=x$. This shows that $\alpha$ is a co-cone as desired.
(Because it's easy to get lost in the subscripts, remember that $\alpha_{(x,X)}$ is a natural transformation $\hom(-,X)\to F$, which has as its component at each $Y\in\mathfrak{C}^{op}$ the function $(\alpha_{(x,X)})_Y:\hom(Y,X)\to F(Y)$.)
To show that this is a colimit, assume that you have a co-cone $\beta$ under $D$ to another presheaf $G$. It will again be the case that each component $\beta_{(x,X)}:\hom(-,X)\to G$ picks out a unique $$g_{x,X}=(\beta_{(x,X)})_X(id_X)\in G(X)$$ by the Yoneda lemma. What you wish to show is that the collection of functions $\{\gamma_X\}_{X\in\mathfrak{C}^{op}}$ defined by $$\gamma_X(x)=g_{x,X}$$ defines a natural transformation $\gamma:F\to G$, and does so uniquely. In this case, "uniquely" means that for each $(x,X)$, the morphisms $\gamma_X\circ\alpha_{(x,X)}$ are equal to $\beta_{(x,X)}$; and conversely, given any natural transformation $\chi:F\to G$, that the co-cone $D\to G$ whose components at $(x,X)$ are $\chi_X\circ\alpha_{(x,X)}$, simply gives back $\chi$ when fed to the above construction. This is a mouthful, but it's actually quite simple (though not short) to show.
Hopefully this helps!