Is there an example that in some abelian category $A$, direct limit always exist, but filtered colimit does not always exist?


No. Every filtered category admits a cofinal functor from a directed category, so the existence of directed colimits implies that of filtered colimits.

The construction of this directed category is slightly technical. It can be read in the first part of Adamek and Rosicky's monograph on locally presentable and accessible categories.

EDIT: An error in the A-R proof has been pointed out on MSE before and was recalled by @user12580 in the comments, together with a link to a correct proof, which interestingly precedes the A-R book by quite some time.

  • $\begingroup$ So there's really no difference between filtered colimit and directed colimit, except when we define the colimit, filtered colimit is more convenient? $\endgroup$ – Qixiao Apr 13 '17 at 15:08
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    $\begingroup$ It's a bit much to say there's no difference, but in some sense, that's right. We generally prefer filtered colimits because they come out of natural constructions, such as the canonical diagram of the finitely presentable objects in a locally finitely presentable category. $\endgroup$ – Kevin Carlson Apr 13 '17 at 18:25
  • $\begingroup$ I don't know who "we" is, but I prefer directed colimits (often called "direct limits" although they are no limits, sigh). $\endgroup$ – Martin Brandenburg Apr 14 '17 at 0:45
  • $\begingroup$ I think the proof in Adámek’s book is not quite right, as described in math.stackexchange.com/questions/2886797/… $\endgroup$ – user12580 Oct 13 '18 at 2:16
  • $\begingroup$ For example, consider a category freely generated by two morphisms between two objects (so that we have loops of arbitrary length). Then add one terminal object to this category. This category has the property that any finite subcategory can be extended to a finite subcategory with a unique terminal object. But the two finite subcategories, with each one of them contains exactly one of the generating morphism, admits a union which is not a finite subcategory. $\endgroup$ – user12580 Oct 13 '18 at 2:22

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