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Oct 29, 2015 at 16:55 review Low quality posts
Oct 29, 2015 at 17:46
Nov 26, 2012 at 2:33 comment added Dylan Wilson Whoops! Silly me (comment removed, as well as the fake proof)
Nov 26, 2012 at 2:01 comment added Martin Brandenburg -1. The full subcategory of abelian groups consisting of torsion abelian groups is coreflective, hence complete (see also the other comments). By the same reason, many other potential examples you first think of don't work. It is a common false belief (which also gets repeated again and again on mathoverflow) that limits and colimits are reflected/preserved/created by forgetful functors, see also mathoverflow.net/questions/23478/…
Nov 26, 2012 at 1:18 comment added Todd Trimble @William: Weibel is wrong here. To anyone who has submitted an answer which has not resolved the question: please consider deleting your answer, because this silly MO mechanism will automatically award an "accepted answer" to the highest voted answer after the bounty is over and the OP hasn't accepted anything. In this case, that would be the present answer. :-) I might also mention that I am trying to email people who might well be able to answer this. (Pity that there aren't a huge number of categorists who tune in here.)
Nov 26, 2012 at 0:55 comment added Tom Goodwillie People keep making this same mistake. The fact that the product of some torsion abelian groups in the category of all abelian groups is not torsion does not imply that there is no product of these same objects in the subcategory. In fact, in the product of all the groups Z/n the subgroup consisting of torsion elements is a product (in the category of torsion abelian groups).
Nov 26, 2012 at 0:12 history answered William Harrison CC BY-SA 3.0