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Let $C$ be any category. $\text{Ind}(C)$ is the free completion under all filtered colimits.

Can you help me in computing the Ind-completion of the terminal category?

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  • $\begingroup$ That is not the correct definition of Ind(C). Filtered-colimit preserving functors from Ind(C) to an inductive category correspond to arbitrary functors from C to an inductive category, not filtered-colimit preserving functors. $\endgroup$ Commented Mar 30, 2018 at 21:28
  • $\begingroup$ Sorry, I edited. I copy and pasted from somewhere in which was wrong. $\endgroup$ Commented Mar 30, 2018 at 21:35

1 Answer 1

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The Ind-completion may be constructed as the closure of the representables in $\widehat{C}$ under filtered colimits. But the subcategory of representables in $\widehat{*}$ is a terminal category, so every functor into it is constant. Any filtered category $J$ is connected, so the colimit of a constant $J$-indexed functor at the representable $y*$ is just $y*$ itself. That is, the terminal category is its own Ind-completion.

To confirm, we can check the universal property. Given any functor $F:*\to D$, where $D$ has filtered colimits, the claim is that there exists a unique functor $*\to D$ extending $F$ and preserving filtered colimits; in other words the claim is simply that $F$ automatically preserves filtered colimits. But this follows by the same argument as above: namely, that the filtered colimit of a constant diagram on some object $x$ is simply $x$.

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  • $\begingroup$ On the other hand the free completion under colimits is Set, am I correct? $\endgroup$ Commented Mar 30, 2018 at 22:31
  • $\begingroup$ Yes, that's right. $\endgroup$ Commented Mar 30, 2018 at 22:37

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