If $\mathcal{C}$ is an essentially small category, is it true that $Ind(\mathcal{C})$, the full subcategory of $Fun(\mathcal{C},Set^{op})$ consisting of functors which can be expressed as filtered colimits of representable ones, is closed under small colimits?

I think that it should be equivalent to be closed under equalizers, since $Ind(\mathcal{C})$ is closed under filtered colimits, and hence under coproducts.

Thank you, Sasha

  • $\begingroup$ This category is more properly called the "ind-completion of $\mathcal{C}$". $\endgroup$
    – Zhen Lin
    Oct 19, 2012 at 13:25
  • 1
    $\begingroup$ If $\mathcal{C}$ admits finite colimits, then $\mathrm{Ind}(\mathcal{C})$ admits small colimits, see Proposition 6.1.18 of the book "Categories and sheaves" of M. Kashiwara and P. Schapira. $\endgroup$
    – felix
    Sep 9, 2021 at 16:38

1 Answer 1


Consider the category $\mathbf{Set}_{10}$ of sets of order at most $10$. This category has all filtered colimits but is not cocomplete (it does not contain the coproduct of two sets of order ten).

  • $\begingroup$ You are right; So we also should pose the question of finite coproducts for $Ind(\mathcal{C})$. $\endgroup$
    – Sasha
    Oct 19, 2012 at 12:58

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