# Are Indizations cocomplete

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

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This category is more properly called the "ind-completion of $\mathcal{C}$". –  Zhen Lin Oct 19 '12 at 13:25
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).
You are right; So we also should pose the question of finite coproducts for $Ind(\mathcal{C})$. –  Sasha Oct 19 '12 at 12:58