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I deleted the second condition for halfexactness, because it was wrong. If a functor fulfills the property, that from A->B->C is exact follows F(A)->F(B)->F(C) is exact, then the functor is already exact. Just apply this condition to sequences like 0->A->B and B->C->0. 192.52.0.198 ( talk) 14:56, 5 December 2014 (UTC)
Isn't it necessary that exact functors between abelian categories are additive? Otherwise the equivalence of the two different definitions and the left-exactness probably would not hold. -- -oo- 7 July 2005 13:44 (UTC)
Short exact sequences redirects to exact sequences. Isn't this a problem?
A pedantic point: the first paragraph is a little sloppy, in that it fixes a short exact sequence 0->A->B->-0, whereas obviously we should quantify over all such. Artie p 10:09, 24 May 2006 (UTC)
i know the definition by commuting with direct/inverse limits. it can be shown to be equivalent. should this be mentioned? -- Ibotty 14:10, 3 December 2006 (UTC)
From the article:
It seems it should be instead:
(note added word finite). VictorPorton ( talk) 17:29, 17 October 2014 (UTC)