I just read the Wikipedia-article about filtered category, and now I wonder why it mentions filtered colimits and cofiltered limits, but not filtered limits. On the nLab-article titled filtered limit a filtered limit is even defined as a limit over a cofiltered category.

Is there a reason for this? The first thing that comes to my mind when I try to think of a limit over a filtered (but not cofiltered) category is a pullback. So apart from this simple case, are limits over filtered categories not very useful, or well-behaved, or whatever?

  • 2
    $\begingroup$ Filtered colimits have special properties in many concrete categories of interest, and cofiltered limits are their formal dual. I do not know of any good properties of filtered limits. $\endgroup$ – Zhen Lin Jul 16 '13 at 0:43
  • 4
    $\begingroup$ The special property Zhen alludes to is that in $\text{Set}$, filtered colimits are precisely the ones which commute with finite limits. This implies various other nice things. There is no dual property of filtered limits, to my knowledge. $\endgroup$ – Qiaochu Yuan Jul 16 '13 at 0:47

If a category $C$ has a terminal object, then, as far as limits in $C$ are concerned, "filtered" would be useless information. To see this, consider any functor into $C$, say $F:I\to C$. Define $I^+$ to be the category obtained by adjoining a terminal object to $I$. (If $I$ already has a terminal object, ignore it and adjoin a new one.) So the objects of $I^+$ are those of $I$ plus the new terminal object $1$, and the morphisms of $I^+$ are those of $I$ and, for each object of $I^+$, a single morphism from that object to $1$. Extend $F$ to a functor $F^+:I^+\to C$ by sending the new $1$ of $I^+$ to the terminal object of $C$ and defining $F$ on the new morphisms of $I^+$ in the only possible way. Then, unless I've stupidly overlooked something, $I^+$ is filtered, and the limit of $F^+$ agrees with that of $F$. In other words, any limit at all (like that of $F$) can be turned into a filtered limit (like that of $F^+$) by a trivial modification.

  • 1
    $\begingroup$ I suppose this is to say that filtered limits aren't special at all since each limit in a category with terminal object can be turned into a filtered limit :-) Thanks a lot for your answer! $\endgroup$ – Stefan Hamcke Jul 16 '13 at 10:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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