I am trying to understand a proof that, for any small category $\mathcal{C}$, the category $\widehat{C} = [\mathcal{C}^\mathrm{op}, \textbf{Set}]$ is the free cocompletion of $\mathcal{C}$. In particular, given a functor $F:\mathcal{C}\to\mathcal{D}$ to a cocomplete category $\mathcal{D}$, I want to extend it to a functor $\widehat{F}:\widehat{\mathcal{C}}\to\mathcal{D}$ that preserves colimits. By "extend" $F$ to $\widehat{F}$, I mean that $F = \widehat{F}\circ Y$, where $Y:\mathcal{C}\to\widehat{\mathcal{C}}$ is the Yoneda functor $x\mapsto \operatorname{Hom}_\mathcal{C}(-, x)$.
Given a presheaf $P \in \widehat{C}$, we know that $P = \operatorname{colim}_{(x,y)\in(*\Rightarrow P)^\mathrm{op}} Y(x)$, where $(*\Rightarrow P)^\mathrm{op}$ is the category with objects $(x,y)$ for $x \in \mathcal{C}$ and $y\in P(x)$, and with morphisms $(x,y)\to(x',y')$ given by morphisms $f:x'\to x$ in $\mathcal{C}$ such that $f(y') = y$.
Therefore, in order for $\widehat{F}$ to extend $F$ and be cocontinuous, we had better have $$ \widehat{F}(P) = \operatorname{colim}_{(x,y)\in(*\Rightarrow P)^\mathrm{op}} F\circ Y(x). $$ I have two problems:
- Why is $\widehat{F}$ a functor?
- Why is $\widehat{F}$ cocontinuous?
My Attempts
In the sources I have found, $(1)$ is either described as obvious, or as following because all of the assignments are functorial. I think I have managed to construct a natural map $\widehat{F}(A) \to \widehat{F}(B)$ for any natural transformation $A\Rightarrow B$ of presheaves. I won't go through the details, but my idea is to construct a cocone $Y(x) \to \operatorname{colim}_{(*\Rightarrow B)^\mathrm{op}} Y(x)$ on the diagram of which $\widehat{F}(A)$ is the colimit. This cocone must factor through $\widehat{F}(A)$, which gives us a natural map $\widehat{F}(A) \to \widehat{F}(B)$. This seems to me less obvious than any author implied, because (a) I am not even sure it works and (b) the details are quite unpleasant.
As for $(2)$, I have no idea how to show it. My lecture notes observe that $$ \widehat{F}(\operatorname{colim}_i P_i) = \operatorname{colim}_{(x,y)\in(*\Rightarrow \operatorname{colim}_iP_i)^\mathrm{op}}F\circ Y(x) $$ and $$ \operatorname{colim}_i \widehat{F}(P_i) = \operatorname{colim}_i \operatorname{colim}_{(x,y) \in (*\Rightarrow P_i)^\mathrm{op}}F\circ Y(x), $$ and then assert that these colimits are obviously the same. To me, this is not obvious. I have also seen the justification that this works because we can interchange the order of colimits. Based on context, I think this refers to the "Fubini Theorem" that $$ \operatorname{colim}_I \operatorname{colim}_J = \operatorname{colim}_J \operatorname{colim}_I $$ for small categories $I$ and $J$. As far as I can tell, this doesn't apply here because the shape of the diagram for the inner colimit depends on the outer colimit, and we can only interchange the order of colimits when the shapes of the diagrams do not depend on one another.