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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)[reply]

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)[reply]

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)[reply]

ah, i see. its in the examples. i think a more general definition should not be a part of examples. --Ibotty 14:14, 3 December 2006 (UTC)[reply]

From the article:

Let C be a category with finite projective (resp. inductive) limits. Then a functor u from C to another category C′ is left (resp. right) exact if it commutes with projective (resp. inductive) limits.

It seems it should be instead:

Let C be a category with finite projective (resp. inductive) limits. Then a functor u from C to another category C′ is left (resp. right) exact if it commutes with finite projective (resp. inductive) limits.

(note added word finite). VictorPorton (talk) 17:29, 17 October 2014 (UTC)[reply]