# 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?

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

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.