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S Jun 12, 2015 at 11:52 history suggested Arrow CC BY-SA 3.0
LaTeX'd the answer
Jun 12, 2015 at 11:40 review Suggested edits
S Jun 12, 2015 at 11:52
Oct 28, 2009 at 16:35 comment added Agustí Roig Thanks. In fact, looking at what I have written, I see a possible and right criticism. In order to prove that R^nF is a universal delta-functor, from the universal property of the total derived functor RF, I use the fact that R^nT^0 is already a universal delta-functor. So it is not quite a "universal delta-functor" free argument. :-( (But at least, I don't use that R^nF is a universal delta-functor in order to prove that it is a universal delta-functor!)
Oct 28, 2009 at 15:28 comment added Andrew Critch Thanks! Parts (2) and (3) were the ones causing me the most trouble :)
Oct 28, 2009 at 15:26 vote accept Andrew Critch
Oct 28, 2009 at 15:04 history answered Agustí Roig CC BY-SA 2.5