Let $C_*$ be a chain complex such that every $C_i$ is a torsion-free finitely generated abelian group, with $C_i=0$ for every $i<0$ and every $i>N$ for some sufficiently large integer $N$. If every homology group $H_i(C_*)$ is a torsion group, then $$\bigoplus_{i \text{ even}}C_i\cong\bigoplus_{i \text{ odd}}C_i.$$

I honestly have no clue of how to begin proving this. Any help would be appreciated


1 Answer 1


Since the $C_i$ are free, only the ranks matter, and we may tensor with $\mathbb{Q}$. Then the complex becomes exact because the homology was assumed to be torsion. Now use that the Euler characteristic of an exact complex vanishes.

  • $\begingroup$ Sorry, but I am suposed to work with almost the mere definitions. It was homework after a short introduction to homological algebra taught in an algebraic topology course, so we don't have all that machinery. I would be very grateful if you can try to prove it in such a basic context. $\endgroup$
    – Tomás
    Commented Sep 27, 2014 at 2:49
  • $\begingroup$ It's easy enough to unpack the proof Martin has outlined, if you really have to. $\endgroup$
    – Zhen Lin
    Commented Sep 27, 2014 at 9:21
  • $\begingroup$ The key (the only!?) ingredient from this that you need at the level of f.g. abelian groups is that if $0\to A'\to A\to A''\to0$ is a short exact secuence, then the ranks of the free parts add up, i.e. $r(A)=r(A')+r(A'')$. Do this for all the SES involving chains, cycles, boundaries, homology. Massive cancellation will take place! $\endgroup$ Commented Sep 27, 2014 at 20:50

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