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EDIT Title has been edited.

Let $C$ be a category, and $$\hat{C} = [C^{op}, (Set)]$$ be its free cocompletion. Despite its name, the free cocompletion of free cocompletion is not equivalent to the free cocompletion in general. Namely, $\hat{C} \not\simeq \hat{\hat{C}}$. For example, take $C = \{*\}$. [1].

There is a better cocompletion, called the Cauchy completion $\bar{C}$. If $C$ is small, then we have $$ C \hookrightarrow \bar{C} \hookrightarrow \hat{C}.$$

By theorem 1 in [2], it is better in the sense that $$\bar{C} \simeq \bar{\bar{C}},$$ so $\bar{C}$ is actually a cocompletion, and also that $$\hat{C} \simeq \hat{\bar{C}},$$ so $\bar{C}$ provides what $C$ needs without changing it too much. After all, in many cases it's better to view $C$ as $\hat{C}$ [3].

Question

Is $\bar{C}$ the largest category between $C$ and $\hat{C}$ whose free cocompletion is $\hat{C}$? More precisely, among all categories $D$ with $\hat{C} \simeq \hat{D}$ and $$C \hookrightarrow D \hookrightarrow \hat{C},$$ is $\bar{C}$ the universal one?

Reference

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    $\begingroup$ Better according to what criteria? It depends on what you want to do with completions. $\endgroup$
    – Andrej Bauer
    Commented Dec 9, 2020 at 8:09
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    $\begingroup$ You are doing yourself a disservice with the title and all the talk about "better", since the actual question has nothing to do with it. May I suggest that you change the title to "The largest cocomplete category between a category and its free cocompletion", or some such? And maybe tone down the talk of "better" and "right". $\endgroup$
    – Andrej Bauer
    Commented Dec 9, 2020 at 8:12
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    $\begingroup$ What you're calling "Cauchy completion" is more commonly referred to as "Karoubi envelope" (as Andrej Bauer says) or "idempotent completion". It is indeed the subcategory of all tiny objects of $\mathrm{Psh}(\mathcal{C})$ (as can be proven via the usual argument that every tiny object in $\mathrm{Psh}(\mathcal{C})$ is a colimit of objects in $\mathcal{C}$, and so a retract of one of them). In particular, since all objects of $\mathcal{D}$ are tiny in $\mathrm{Psh}(\mathcal{D})$ it satisfies the required property. $\endgroup$
    – Denis Nardin
    Commented Dec 9, 2020 at 8:21
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    $\begingroup$ The language of this question is very strange. The Cauchy completion is not cocomplete! And I agree with Andrej that all this talk of "better" is very distracting and has nothing to do with your actual question. The answer to your actual question is yes, and you can see, for example, this blog post: qchu.wordpress.com/2015/05/07/tiny-objects $\endgroup$
    – Qiaochu Yuan
    Commented Dec 9, 2020 at 9:06
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    $\begingroup$ The OP seems to be of the opinion that (co)completion should be an idempotent operation. I don't agree – in fact, the qualifier "free" is there to tell you that the opposite is true. $\endgroup$
    – Zhen Lin
    Commented Dec 9, 2020 at 10:57

1 Answer 1

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The answer is positive.

I found a published account with details to be chapter 6 and 7 of Handbook of Categorical Algebra 1 by Francis Borceux.

Thanks to the comments, useful links that summarize how it goes are

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