All $\mathsf{Set}$-valued presheaves on a cocomplete category are representable We often hear that, for a category $\mathcal{C}$, the presheaf category $\mathsf{PSh}(\mathcal{C})=[\mathcal{C}^{\text{op}},\mathsf{Set}]$ is (up to equivalence of categories) the free colimit completion of $\mathcal{C}$.
See, for example, the last point in this Wikipedia article.
If $\mathcal{C}$ is cocomplete then is it true that every presheaf is representable?
We can assume that $\mathcal{C}$ is locally small, or even small, if needs be.
This sounds very farfetched to me, and so in the likely event that it's wrong, what conditions can we place on $\mathcal{C}$ to ensure that every presheaf is representable?
Or is it the case that all such conditions are really so strong that we lose 'a lot' (whatever that might mean to you) of generality.

As a sketch inspiration (which I am not overly convinced is without mistakes), assume that $\mathcal{C}$ is bicomplete, total, and small.
We use freely some of the results from here.
Then it can be shown that the continuous presheaves $\mathcal{C}^\text{op}\to\mathsf{Set}$ are exactly the representable presheaves, and the fact that the Yoneda embedding $Y\colon\mathcal{C}\hookrightarrow\mathsf{Cont}(\mathcal{C}^\text{op},\mathsf{Set})$ has a left adjoint (by totality) tells us that it also has a right adjoint, and so $Y$ preserves colimits.
Any presheaf $F$ is the colimit of representable presheaves.
Say (for ease of notation) that $F=\mathrm{colim}_iF_i$ where $F_i=\mathrm{Hom}(-,c_i)$ for $c_i\in\mathcal{C}$.
Define (since $\mathcal{C}$ is cocomplete) $c=\mathrm{colim}_ic_i$.
Then
\begin{align*}
F &= \mathrm{colim}_iF_i\\
&= \mathrm{colim}_iY(c_i)\\
&= Y(\mathrm{colim}_ic_i) = Y(c),
\end{align*}
and so $F$ is representable.
 A: No, it is not true. For example take the presheaf $F$ on $\mathsf{Set}$ itself (which is bicomplete, locally small, everything we could want) given by $F(X) = \{0,1\}$ for all sets $X$ (and all maps are mapped to the identity). Then $F$ is not representable, because there is no set $Z$ such that $\operatorname{Map}(X,Z)$ has two elements for all sets $X$.
The flaw in your argument can be made evident here: $F$ itself is the coproduct (as a functor!) of the representable presheaf $\operatorname{Map}(-, *)$ with itself, where $* \in \mathsf{Set}$ is some singleton, i.e. $F \cong Y(*) \sqcup Y(*)$. But $F$ is not isomorphic to $Y(* \sqcup *)$. The fact is, $Y$ does not preserve colimits: it preserves limits. It's not true that $\operatorname{Hom}(-, \operatorname{colim}_i X_i) \cong \operatorname{colim}_i \operatorname{Hom}(-, X_i)$, but it's true that $\operatorname{Hom}(-, \lim_i X_i) \cong \lim_i \operatorname{Hom}(-, X_i)$.
A: No. The Yoneda embedding preserves almost no colimits: the colimits you get are "free" and so have nothing a priori to do with any colimits that already exist. Work out explicitly what a coproduct of representable presheaves looks like: it is almost never representable. 
Your argument cannot work for the simple reason that the category of presheaves is never (essentially) small. Also, Freyd showed that any cocomplete small category is a preorder, so even if your argument worked it would only apply, essentially, to suplattices. 
