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May 26, 2011 at 21:16 answer added Henning Krause timeline score: 17
Sep 30, 2010 at 18:31 comment added Matthias Künzer Verdier refers to Freyd for this universal construction. What you can't do with the mere Frobenius abelian structure is taking a cone of a morphism of bijectives. You can take a cokernel, embed it into a bijective, but this attempt doesn't give anything unique up to isomorphism, for you can add further bijectives. Heller answered the question which extra structure a Frobenius abelian category needs in order that the full subcategories of bijectives be a Puppe triangulated category in Stable homotopy categories, Bull. Am. Math. Soc. 74, p. 28-63, 1968.
Feb 13, 2010 at 0:16 vote accept Shizhuo Zhang
Feb 12, 2010 at 0:40 comment added Shizhuo Zhang Yes,that is correct. We made assumptions that the original triangulated category is Karoubian. Then the Beck's theorem is pretty nice. But if it is not, it might be subtle though there is also Beck's theorem
Feb 12, 2010 at 0:20 comment added Greg Stevenson A minor point but if your original triangulated category is not idempotent complete then there are more projectives in the abelianization than just the representables.
Feb 11, 2010 at 21:27 answer added Greg Stevenson timeline score: 22
Feb 11, 2010 at 15:30 comment added Shizhuo Zhang Des Catégories Dérivées des Catégories Abéliennes.
Feb 11, 2010 at 15:27 comment added babubba Could you please give a reference where to find this abelianization you are writing about?
Feb 11, 2010 at 14:10 history edited Shizhuo Zhang CC BY-SA 2.5
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Feb 11, 2010 at 12:24 history edited Shizhuo Zhang CC BY-SA 2.5
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Feb 11, 2010 at 12:16 history asked Shizhuo Zhang CC BY-SA 2.5