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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?

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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.

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