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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

Student
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    Better according to what criteria? It depends on what you want to do with completions. – Andrej Bauer Dec 09 '20 at 08:09
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    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". – Andrej Bauer Dec 09 '20 at 08:12
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    Anyway, the answer to the question is Yes, it's the Morita theory of the presheaf construction. Two categories have equivalent presheaves iff their Cauchy completions are equivalent. – Ivan Di Liberti Dec 09 '20 at 08:14
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    Hmm, is the OP using "Cauchy completion" for "Karoubi envelope"? – Andrej Bauer Dec 09 '20 at 08:15
  • @AndrejBauer Thanks for your comment. But $\bar{C}$ isn't cocomplete.. is it? By the paper [2], it only includes the small absolute colimits. And yes, there are many names for this term (https://ncatlab.org/nlab/show/Cauchy+complete+category). – Student Dec 09 '20 at 08:19
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    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. – Denis Nardin Dec 09 '20 at 08:21
  • @IvanDiLiberti would you mind providing more details or pointers? – Student Dec 09 '20 at 08:25
  • @DenisNardin does your argument shows that $\bar{C}$ is the largest? – Student Dec 09 '20 at 08:54
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    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: https://qchu.wordpress.com/2015/05/07/tiny-objects/ – Qiaochu Yuan Dec 09 '20 at 09:06
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    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. – Zhen Lin Dec 09 '20 at 10:57
  • @QiaochuYuan Thank you for your wonderful write-up. I think it is almost clear.. my main crux lies in the fact that there are many characterization of the Cauchy completion. By using the definition you provided (using tiny presheaves), I have reduced my problem to an easy exercise. Embarrassingly enough, I struggled to resolve it.. perhaps I missed some other basic things again. --- The direction I'm still having trouble with is that $(\Phi: \hat{D_1} \simeq \hat{D_2})$ implies $\bar{D_1} \simeq \bar{D_2}$. I will have to construct a map from $\bar{D_1}$ to $\bar{D_2}$ from $\Phi$. But this.. – Student Dec 09 '20 at 19:30
  • But this requires the fact that for any small categories $A, B$ and any equivalence $\Phi: \hat{A} \simeq \hat{B}$, one has $((\Phi(a))(b) \simeq (\Phi^{-1}(b))(a)$ naturally in $a \in A$ and $b \in B$. Is this true? Don't we need extra condition on $\Phi$? – Student Dec 09 '20 at 19:33
  • @Student: you just use the fact that an equivalence on presheaf categories induces an equivalence on tiny presheaves, since being tiny is a categorical property. – Qiaochu Yuan Dec 09 '20 at 19:34
  • @QiaochuYuan I think Mike Shulman's answer in this post suggests that the equivalence $\bar{D_1} \simeq \bar{D_2}$ is in the sense of profunctors, not usual functors. This seems to be pretty close to my confusion above, in which I formulated using letters $A, B$. – Student Dec 09 '20 at 19:48
  • @Student: what Mike says is that if $\text{Psh}(A) \cong \text{Psh}(B)$ then $A, B$ (not their categories of presheaves) are equivalent in the profunctor (bi)category. This follows from the universal property of presheaves as the free cocompletion. But the equivalence $\text{Psh}(A) \cong \text{Psh}(B)$ is just a regular ol' equivalence of categories. – Qiaochu Yuan Dec 09 '20 at 19:52
  • Found a published account: chapter 6 and 7 of Handbook of Categorical Algebra 1 by Francis Borceux. Useful links that summarize how it goes are (https://mathoverflow.net/questions/85768/morita-equivalence-for-categories) (https://mathoverflow.net/questions/163638/morita-equivalence-via-kan-extension) and @QiaochuYuan's wonderful write up (https://qchu.wordpress.com/2015/05/07/tiny-objects/). Thanks for all the help.. – Student Dec 10 '20 at 09:47
  • The text of the question is still misleading. By its logic, the "best" "cocompletion" of $C$ would be $C$ itself. – Reid Barton Dec 10 '20 at 16:08
  • @ReidBarton I have changed the subject to "Is Cauchy completion the largest extension with the same free cocompletion?" yesterday. Not sure if it has taken effect on your machine. – Student Dec 10 '20 at 16:10
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    I'm referring to phrases like "$\bar C$ is a cocompletion" (it is not) and "Despite its name, the free cocompletion of free cocompletion is not equivalent to the free cocompletion" (why would one think that?), which have been pointed out already by others, as well as "$\bar C$ provides what $C$ needs" (why does $C$ need anything in particular?) – Reid Barton Dec 10 '20 at 16:15
  • @ReidBarton I agree with you. There are many bad terms here. $\bar{C}$ is not a "cocompletion", but it's named Cauchy completion any way.. it is indeed trying to complete $C$ in that it adds all small absolute colimits. – Student Dec 10 '20 at 21:40

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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

Student
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