Free cocompletion and preservation of colimits Let  $\mathcal{K}$ be a $\lambda$-accessible category and $\text{Pres}_{\lambda}(\mathcal{K})$ be the small category of its $\lambda$-presentables.

*

*It is well known that $\mathcal{K}$ is the $\lambda$-directed cocompletion of $\text{Pres}_{\lambda}(\mathcal{K})$.

Now, call

*

*$\hat{\mathcal{K}}$ the free directed cocompletion of $\mathcal{K}$.

*$\widehat{\text{Pres}_{\lambda}(\mathcal{K})}$ the free directed cocompletion of $\text{Pres}_{\lambda}(\mathcal{K})$.


Q:

*

*$\widehat{\text{Pres}_{\lambda}(\mathcal{K})} \cong \hat{\mathcal{K}} \ \ ? $


*Does the inclusion $\mathcal{K} \hookrightarrow \hat{\mathcal{K}}$ preserve directed colimits?


*I expect the answer to the previous question to be $\textit{no}$. Is there a category $\mathcal{D}$ which is a directed cocompletion of $\mathcal{K}$ and such that the inclusion preserves directed colimits?

 A: I'm not an expert on the subject, so please double-check this answer, and please correct me.

  
*
  
*$\widehat{\text{Pres}_{\lambda}(\mathcal{K})} \cong \hat{\mathcal{K}} \ \ ? $
  

No. Take $K := \omega+1 = \{0 \leq 1 \leq \ldots \leq \omega\}$ considered as a category.
Then $K$ is a Scott-domain (the finite elements are the finite ordinals, and every element is a directed join of finite elements). Therefore, it is a finitely accessible category. 
The free directed cocompletion of $K$ is $K' := \omega + 2 = \{0\leq1\leq\ldots\leq\omega\leq\omega+1\}$. 
We have that $\widehat{\mathrm{Pres}_\lambda K} =  K \not\simeq K' = \hat K$, as the terminal object in $K$ is not finitely presentable, whereas the terminal object in $K'$ is finitely presentable. 

  
*
  
*Does the inclusion $\mathcal{K} \hookrightarrow \hat{\mathcal{K}}$ preserve directed colimits?
  

No. As we saw earlier, the inclusion maps $\omega$ to $\omega + 1$ and so doesn't preserve directed colimits.

  
*
  
*I expect the answer to the previous question to be $\textit{no}$. Is there a category $\mathcal{D}$ which is a directed cocompletion of $\mathcal{K}$ and such that the inclusion preserves directed colimits?
  

Once you choose $\mathcal K$, its free directed cocompletion is determined (up to equivalence), so I assume you meant: is there a $\mathcal K$ whose inclusion into the directed cocompletion is finitely accessible?
In that case, the answer is 'yes', for example, taking $\mathcal K$ to be a finite poset. As all of its directed colimits are absolute, $\mathcal K$ is its own completion with the identity as inclusion, and the identity preserves directed colimits. 
