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Feb 21, 2022 at 12:37 comment added Paul Taylor To amplify @TimCampion's comment, the forgetful functor from the category of models of a finitary (essentially) algebraic theory (eg categories) creates filtered colimits. So they work smoothly, without additing new algebraic structure. Why this should apply to cofiltered limits, I don't understand.
Feb 21, 2022 at 8:36 history edited Matthieu Romagny CC BY-SA 4.0
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May 4, 2017 at 18:37 comment added Tim Campion It should be mentioned that since $Pro(C) = Ind(C^{op})^{op}$, the question can be viewed as equivalent to asking why the Ind-construction is important, and here there are lots of familiar examples to illustrate this. For example, a $k$-algebra is the same thing as an ind-object in finitely-presentable $k$-algebras. Dually, an affine scheme over $Spec(k)$ is the same as a pro-object in affine schemes of finite type over over $Spec(k)$. On the ubiquity of the Ind-construction, see e.g. here.
May 2, 2017 at 21:19 vote accept Zhaoting Wei
May 2, 2017 at 15:42 answer added Tim Porter timeline score: 7
May 1, 2017 at 17:47 comment added Dylan Wilson Another relevant comment is that $Pro(C)$ views all the elements of $C$ as cocompact (maps into them from a cofiltered limit factor through some finite stage), whereas $Fun(C, Sets)^{op}$ requires the objects of $C$ to have a much stronger requirement: maps into them commutes with arbitrary limits. Such objects rarely appear in nature.
May 1, 2017 at 17:40 comment added Dylan Wilson I dunno if "lost" is the right word. You've just defined a different category, which you already know by another name: it's $Fun(C, Sets)^{op}$. Maybe one thing you lose is control. The inclusion $C \to Pro(C)$ preserves finite limits, which is not true of the inclusion $C \to Fun(C, Sets)^{op}$ in general.
May 1, 2017 at 16:07 history asked Zhaoting Wei CC BY-SA 3.0