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Can we characterize entirely for which categories $C$ the end on $C$ preserves filtered colimits, in a sense that the natural map

$$ \operatorname*{colim}_{i \in I} \int_{x \in C} A_i (x,x) \to \int_{x \in C} \operatorname*{colim}_{i \in I} A_i (x,x)$$

is an isomorphism for every filtered diagram of $A_i \in \text{Set}^{C^\text{op} \times C}$ ?

It is easy to see that this will be the case when $C$ is (equivalent to) a finitely generated category, but are there some non-finitely generated categories for which this is also true?

Of course, if one replace ends by limits then there are plenty of example: for any category with an initial objects $C$, limits along $C$ are just evaluation at the initial object and these preserves all colimits, and more generally it is enough for the category to admit a cofinal finitely generated subcategory but I'm not aware of a similar phenomenon for ends...

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    $\begingroup$ It is necessary and sufficient that $\textrm{Hom}_\mathcal{C} : \mathcal{C}^\textrm{op} \times \mathcal{C} \to \textbf{Set}$ be a finitely presentable object in the category $[\mathcal{C}^\textrm{op} \times \mathcal{C}, \textbf{Set}]$, but I suppose that is not a satisfactory answer for you? $\endgroup$
    – Zhen Lin
    Commented Mar 11, 2023 at 1:19
  • $\begingroup$ Yes, at the minimum, I'd like to know if it is equivalent to $C$ being (equivalent to a category which is) finitely generated or not. $\endgroup$
    – Simon Henry
    Commented Mar 11, 2023 at 1:22
  • $\begingroup$ I should even, Cauchy completion of a finitely generated category in fact, as such a cauchy completion can have an infinite number of objects, and that doesn't change the End $\endgroup$
    – Simon Henry
    Commented Mar 11, 2023 at 4:27
  • $\begingroup$ In a different context: replace sets by the stable $\infty$-category of modules over a commutative ring $k$ and $C$ the $\infty$-category of perfect complexes on a derived scheme $X$. This property is then equivalent to smoothness of $X$ over $k$. If you take a finite commutative monoid $M$ with associated $k$-algebra $k[M]$ that is non smooth, this will mean that $C=$ the category with one object associated to $M$ is a non-example. E.g. the group with $2$ elements. Therefore, the property of being of finite presentation (or even finite) is not sufficient. $\endgroup$
    – D.-C. Cisinski
    Commented Mar 11, 2023 at 14:30
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    $\begingroup$ @MaximeRamzi You are right, $M$ can not be finitely presentable in the $\infty$-category of spaces. That means my comment about a counter example was written too quickly. $\endgroup$
    – D.-C. Cisinski
    Commented Mar 12, 2023 at 11:25

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