Suppose to have a short exact sequence of chain complexes of $\mathbb{Z}$-modules: $$0\to A^\bullet\to B^\bullet\to C^\bullet\to 0$$ such that $A^k,B^k,C^k$ are non zero for $k=0,1,2$. Moreover, suppose that $H^j(C^\bullet)$ is torsion free and non-zero, for all $j=0,1,2$. Is it true that if $H^2(A^\bullet)$ is torsion free, then also $H^2(B^\bullet)$ is torsion free? What about the converse?
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$\begingroup$ Of course not, take $A = Z$ (in degree 0), $B = Z \stackrel{2}\to Z$ (in degrees $-1$ and 0), $C = Z$ (in degree $-1$). $\endgroup$– SashaCommented Feb 24, 2015 at 11:32
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$\begingroup$ you are right, I forgot a couple of assumptions. $\endgroup$– user53075Commented Feb 24, 2015 at 11:37
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$\begingroup$ Add to $A$ and $C$ arbitrary acyclic complexes and to $B$ their direct sum. $\endgroup$– SashaCommented Feb 24, 2015 at 11:38
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$\begingroup$ I am thinking of chain complexes with $A^k=0$, for $k<0$. Does this make any difference? $\endgroup$– user53075Commented Feb 24, 2015 at 11:41
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$\begingroup$ @user53075 You can shift Sasha's example up so that $A$ is concentrated in degree 2, $B$ is concentrated in degrees $1$ and $2$, and $C$ is concentrated in degree $1$. $\endgroup$– Tyler LawsonCommented Feb 24, 2015 at 18:14
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