Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
This category is more properly called the "ind-completion of $\mathcal{C}$". –  Zhen Lin Oct 19 '12 at 13:25

1 Answer 1

up vote 1 down vote accepted

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).

share|improve this answer
    
You are right; So we also should pose the question of finite coproducts for $Ind(\mathcal{C})$. –  Sasha Oct 19 '12 at 12:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.