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$$0\to M\to Q_1\to Q_2\to\dots\to Q_i \to N\to 0$$ exact sequence, then $$H^n(N)\cong H^{n+i}(M)$$

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    Standard trick: break up a long exact sequence into lots of little short exact sequences.2012-11-08
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    As stated this is simply false. To get an example, just take $i$ large and let $Q_2=\cdots=Q_{i-1}=0$. At the very least, you should tell us what $M$ and $N$ and so on are, and what cohomology you have in mind!2012-11-08
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    You seem to have edited the question into something rather different to what it was at the time people wrote their answers. Please do not do that.2012-11-08

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