Is there a name for the following situation: I have a diagram of categories and functors DO(F) DO(A) ------> DO(B) | | | | H | |H | | v F v A ----------> B It does not commute, nor is there even a 2 cell in either direction. What I do have is illustrated below: DO(A) DO(A) /|\ /|\ / | \ / | \ / | \ / | \ H/ | \DO(F) H/ | \DO(F) / P \ / P' \ / | \ / | \ v | v v | v A <== | ==> DO(B) A ==> | <== DO(B) \ | / \ | / \ | / \ | / F\ | /H F\ | /H \ | / \ | / \ | / \ | / vvv vvv B B and, moreover, P -------> HF | | | | | | | | | | v v FH ------> P' commutes.
Incidentally, did you know that if Z and Z' are defined so that a d a' 0 --> Z ---> C ---> C ---> Z' ---> 0 is exact, then the homology is the image (= coimage) of a'.a: Z --> Z'? This is a triviality, but it gives a symmetric definition of homology. Notice that it defines something even when d.d is not 0. I guess it is Z mod Z meet ker(d). Mike
Yes, I like this symmetric presentation. ____ As to "Notice..." (where you mean ... meet im(d), of course), I would add the following. The condition dd = 0 is linked with aspects whose relevance is often ignored. Without it, many systems of use in homological algebra would loose any reasonable notion of "canonical isomorphism", and one should be extremely prudent in working with induced morphisms and Noether isomorphisms. Let us start (in Ab, or any abelian category) with a sublattice L of subobjects of a given object (necessarily modular) and consider the subquotients having numerator and denominator in L. Then, the canonical isomorphisms among these subquotients (induced by the identity) are closed under composition *if and only if* L is distributive. - Within this restriction, being "canonically isomorphic subquotients" has a precise meaning: there is a well-determined canonical isomorphism linking them. - Without this restriction, composing canonical isomorphisms can yield different isomorphisms between two given subquotients. Working up to canonical isomorphism, as commonly done in homological algebra, could easily lead to errors. (For instance, it is easy to construct such a situation for subquotiens of Z^2 (pairs of integers) in Ab - the classical example of an object whose lattice of subobjects is not distributive.) Concretely, the main systems of homological algebra giving rise to spectral sequences (filtered differential object, filtered complex, exact pairs, double complex) DO produce distributive lattices. Essentially, the proof is generally based on a crucial Birkoff theorem: the free modular lattice generated by two chains is distributive. But all this is no longer true without assuming dd = 0 (or something similar) in such systems: distributivity would fail. [[ It is easy to see the role of dd = 0 in the simplest case, the filtered differential object. We have a differential object (A, d) equipped with a consistent filtration (F_p A), so that every d(F_p A) is contained in F_p A. Then, one can prove that the sublattice of Sub(A) generated by the filtration and closed under d-images and d-preimages (written d*) is generated by two filtrations, the original one and 0 -> ... d(F_p A) ... -> dA -> d*0 -> ... d*(F_p A) ... -> A, using the inclusion dA -> d*0. ]] Zeeman was probably the first to recognise the importance of this fact: E.C. Zeeman, On the filtered differential group, Ann. Math. 66 (1957), 557-585. _____ The theory I am referring to can be seen in three papers of mine: M.G., On distributive homological algebra, I-III, Cahiers 25 (1984), 259-301; 25 (1984), 353-379; 26 (1985), 169-213. Recently, Francis Borceux and I have extended part of these results to a larger setting, including Grp (groups): F. Borceux - M. Grandis, Jordan-Holder, modularity and distributivity in non-commutative algebra, Dip. Mat. Univ. Genova, Preprint 474 (Feb 2003). http://www.dima.unige.it/~grandis/BGwe.dvi Marco Grandis On 30 Sep 2005, at 20:37, Michael Barr wrote: Incidentally, did you know that if Z and Z' are defined so that a d a' 0 --> Z ---> C ---> C ---> Z' ---> 0 is exact, then the homology is the image (= coimage) of a'.a: Z --> Z'? This is a triviality, but it gives a symmetric definition of homology. Notice that it defines something even when d.d is not 0. I guess it is Z mod Z meet ker(d). Mike
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