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Let $(C,\partial)$ be a chain complex where each $C_i$ is an $R$-module (R being a given ring). We know that the quotients $H_i(C,\partial)=\ker(\partial_i)/Im(\partial_{i+1}$ are also $R$-modules. I wonder if the family $\{H_i\}_{i\geq 0}$ forms a new chain complex, I mean under what conditions, there exists a family of $R$-module homomorphisms $d_i:H_i(C, \partial)\rightarrow H_{i-1}(C, \partial)$ such that $d_{i}\circ d_{i+1}=0$

Now if such a chain complex $(H(C,\partial),d)$ exists, then we can take its $i$th homology $R$-module:
$H_i(H(C,\partial),d))$, does this "homology of homology" has an algebraic/topological meaning?

What happens if we take a cochain complex instead of a chain complex, is the situation similar?

palio
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    You are scratching on the door of spectral sequences. With appropriate conventions on gradings, you can iterate this process to calculate successive approximations of a filtered complex. – Sammy Black Nov 10 '13 at 07:16
  • Have a look at the introduction to Allen Hatcher's book on spectral sequences: http://www.math.cornell.edu/~hatcher/SSAT/SSATpage.html – Sammy Black Nov 10 '13 at 07:18
  • I looked into the first chapter and it does not seem to answeer my question, but rather talking about exact couples and things like that.. thank you for your help! – palio Nov 10 '13 at 08:30
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    Of course you can make any sequence of modules into a chain complex: set $d = 0$. – Zhen Lin Nov 10 '13 at 13:04
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    1)Is there any other conditions on the chain complex to be able to construct a family of non trivial $R$-module homomorphisms $d_i:H_i(C,\partial)\rightarrow H_{i-1}(C,\partial)$ 2) If a family of $R$-modules is endowed with a trivial homomorphism $\partial =0$ then the complex obtained is both a chain complex and a cochain complex and the associated homology and cohomology coincide, is this correct ? – palio Nov 10 '13 at 14:02

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