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Let's $(K^{\bullet}, d^{\bullet})$ is the complex over field $A$ (i.e. all $K^{i}$ are vector spaces over this field) and $(L^{\bullet}, {\delta}^{\bullet})$ such that $$L^{i}=H^{i}(K)~\text{and all}~{\delta}^{i}=0.$$ Why this two complexes $K$ and $L$ are quasi-isomorphic? Why it's wrong for complex over ring?

Thanks a lot!

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3 Answers 3

The result you mention is true for complexes of modules over a ring of global dimension at most one. It is not entirely trivial to prove, see section 2.5 in these lecture notes by Keller. Note that fields have global dimension zero.

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To show that $K$ and $L$ are quasi-isomorphic, try splitting each vector space $K^n$ as $\operatorname{im} d^{n-1}\oplus H^n(K)\oplus (K^n/\ker d^n)$, such that $d^n$ acts in an appropriate way. Then show that the maps $H^n(K)\to K^n$ define a quasi-isomorphism.

To show that this fails over a ring, you need to find a chain complex that doesn't split in this way.

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but $\operatorname{im} d^n \subset \ker d^{n+1}$ –  Aspirin Jul 19 '12 at 20:00
    
Aspirin: yes, that was wrong. It's something like that though –  Colin McQuillan Jul 19 '12 at 20:07

The general statement for why this may be false over a ring goes like this: for two $R$-modules $A,B$ consider $\mathrm{Ext}^2(A,B)$. This parameterizes extensions $0\to A\to X\to Y \to B\to 0$ under Baer sum. It is straightforward to check that for a non-zero class in $\mathrm{Ext}$ given by $0\to A\to X\to Y\to B\to 0$, $0\to A\to B \to 0$ with 0 differential has the same homology as $0\to X\to Y\to 0$ with the same differential is in the 4-term sequence, but it may happen that there are no maps between $A$ and $X$, and so there cannot be a quasi-isomorphism. Unfortunately, an illuminating example does not spring to mind right now.

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