Has anyone seen this before, or can anyone give a principled explanation. Begin with the following observation. Suppose del del ... -----> (C_n,d) -----> (C_{n-1},d) ---> ... ---> (C_0,d) --> 0 is a chain complex of differential abelian groups, except that we assume d.del = -del.d. Suppose (C_n,d) is exact for each n. Let C be the direct sum of the C_n with boundary given by the matrix ( d 0 0 ... ) ( del d 0 ... ) ( 0 del d ... ) ( . . . . ) ( . . . . ) ( . . . . ) The finite truncations F^n = C_0 +...+ C_n are readily shown to be exact: F^0 = C_0 is by hypothesis and there is an exact sequence 0 --> F^{n-1} --> F^n --> C_n --> 0. AB5 now shows that the direct limit, which is C, is exact. Of course, this is going to be false for a category, such as compact abelian groups, that is not AB5. Except it is true. Using duality, we can translate it to the following statement for abelian groups. Suppose del del ... <----- (C_n,d) <----- (C_{n-1},d) <--- ... <--- (C_0,d) <-- 0 is a cochain complex of differential abelian groups, except that we assume d.del = -del.d. Suppose (C_n,d) is exact for each n. Let C be the direct product of the C_n with boundary given by the matrix ( d del 0 ... ) ( 0 d del ... ) ( 0 0 d ... ) ( . . . . ) ( . . . . ) ( . . . . ) then C is exact. The proof is by using elements and although AB4 (at least countable) is clearly a necessary condition for this (it is the case that all the del's are 0), it sure doesn't look sufficient. BTW, both are false if you change direct sum to product in the first or product to sum in the second. Michael