4 Oct 2004 2:30 - 4:00 M Barr Absolute homology Abstract: What does it mean to for a functor to preserve homology? There is no universal mapping property that homology has, but mere isomorphism is not enough. Let A be an abelian category and DO(A) denote the category of differential objects over A: pairs (C,d) s.t. dd = 0. Any additive functor F: A --> B induces DO(F): DO(A) ---> DO(B) given by DO(F)(C,d) = (FC,Fd). (Actually, you don't need F additive to give this definition; it need only preserve 0, but additivity is required to say anything useful about it.) If H is the homology functor, we thus get a square 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 will show we have is illustrated below: DO(A) DO(A) /|\ /|\ / | \ / | \ / | \ / | \ DO(F)/ | \H H/ | \DO(F) / Phi \ / Phi' \ / | \ / | \ v | v v | v DO(B) <== | ==> A A ==> | <== DO(B) \ | / \ | / \ | / \ | / H\ | /F F\ | /H \ | / \ | / \ | / \ | / vvv vvv B B and, moreover, Phi ------> HF | | | | | | | | | | v v FH -----> Phi' commutes. The two diagrams are dual to each other so what I am about to say applies to either one. We will say that F preserves the homology of (C,d) when both 2 cells are ismomorphisms, when evaluated there. I will show that the left 2 cell is an isomorphism for all (C,d) iff F is left exact and the right one is an isomorphism iff F is right exact. The same is true of the right hand diagram (but note that that diagram is mirror image reversed from the first).