At the recent Ottawa conference I repeated, in passing, something I've been saying for over 30 years: given a subcategory of an abelian category the finite bi-closure is, of course, obtainable by alternately adjoining limits and colimits countably often; remarkably enough, one need only do two cycles (assuming direct sums that's kernels, cokernels, kernels, cokernels). The proof is a consequence of of the construction of free abelian categories. Susan asked me later if I had ever actually worked out a direct proof. Nope. But it became apparent to me, thinking about what I had had to say about free abelian categories that I didn't know 30 years ago, that it shouldn't take two cycles, just one: In the abelian setting, adjoining finite limits commutes with adjoining finite colimits. (No -- for those old enough to remember -- this is not still 1969 and Ottawa is not Seattle.) There are some finicky points one must deal with when talking about images of functors (in the case at hand, functors such as those that assign kernels to maps), so let me just cut here to a critical little lemma that says, in effect, "a kernel of a map between cokernels is a cokernel of a map between kernels:" LEMMA: In an abelian category an exact diagram of the form (all vertical arrows are downwards): c B --> A --> F --> O b | a | | B'--> A'--> F'--> O c' may be enlarged to an exact diagram of the form: O --> K'--> A + B'+ B --> A + A' | | | O --> K ----> A + B'------> A' | c B --> A --> F --> O b | a | | B'--> A'--> F'--> O c' The exactness of K'--> K --> F --> F' says, precisely, that the kernel of F --> F' is the cokernel of K'--> K. The middle vertical K --> F is K --> A + B'--> A --> F where A + B'--> A is the projection map. (For present purposes a projection map is one given by a matrix in which each entry is 1 if such fits else 0.) The two top-right vertical maps are also projection maps. The two top-right horizontal maps are given (as luck would have it) by matrices in which each entry is 1 or a single letter (a,b,c,c') if such fits, else 0. It's not hard to verify this lemma in the category of abelian groups, which -- given the exact representation theorem -- suffices for all abelian categories. As if often the case, the proof of the dual of the lemma harder in the concrete case: CO-LEMMA: In an abelian category an exact diagram of the form (all vertical arrows are downwards): c' O --> K'--> A'--> B' | | a | b O --> K --> A --> B c may be enlarged to an exact diagram of the form: c' O --> K'--> A'--> B' | | a | b O --> K --> A --> B c | A'------> A + B'----> F --> O | | | A'+ A --> A + B'+ B --> F'--> O For the record, note that "a kernel of a map between kernels is a kernel," that is, the same initial exact diagram may be enlarged to a diagram:with exact rows and exact left column: O | O ---> K''--> A'--> A + B; | | | c' O ---> K'---> A'----> B' | | a | b O ---> K ---> A ----> B c The two upper-right verticals are projection maps and the upper right horizontal map is given by a matrix in which each entry is a single letter (a,c'). Also for the record: the finite bi-completion of an additive category is not abelian. Start with the additive closure of a single map f:A --> B. In the finite bi-completion the canonical map from Cok(Ker(f)) to Ker(Cok(f)) will not be an isomorphism. In the finite _abelian_ finite bi-completion, of course, it must be. (It is precisely this isomorphism, invoked at the end of the Lemma in the case f = K --> F, that delivers the commutativity of finite left and right completions.)