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Let $C_*$ be a chain complex such that each $C_i$ is a torsion-free, finite-range abelian group with $C_i=0$ for all $i<0$. Suppose that $C_i=0$ for all $i$ is sufficiently large and that for all $i$ there exists an integer $N_i\geq1$ such that $N_i(H_i(C_*))=0$, that is, each $H_i(C_*)$ is a torsion group. Show that $$\bigoplus_{i \text{ even}}C_i=\bigoplus_{i\text{ odd}}C_i$$

I do not know how to solve this problem, I would appreciate any suggestions. We have to $$...\overset{\partial}{\rightarrow}C_{n+1}\overset{\partial}{\rightarrow}C_{n}\overset{\partial}{\rightarrow}C_{n-1}\overset{\partial}{\rightarrow}...$$

where each $C_i$ has a base with a finite amount of elements, say $B_i$ and we also have $\partial(\partial(c_i))=0$ for all $c_i\in B_i$. How can I use all this to define an isomorphism between $\bigoplus_{i \text{ even}}C_i$ and $\bigoplus_{i\text{ odd}}C_i$? Thank you.

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    $\begingroup$ It's a standard fact in homological algebra that the alternating sum of the ranks of the chain group in a finite chain complex of f.g. abelian groups is equal to the alternating sum of the ranks of the homology groups. This is algebraic meaning of homology-invariance of Euler characteristic, if you like. You can try to prove/use this. $\endgroup$
    – Balarka Sen
    May 18, 2019 at 23:51
  • $\begingroup$ @BalarkaSen Yes, I know this fact and here I found a proof of this math.stackexchange.com/questions/2807224/…, but I do not know how to use this to solve the problem $\endgroup$
    – user425181
    May 19, 2019 at 1:00

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Your complex is of the form

$$ 0 \to C_n \to \cdots \to C_1 \to C_0 \to 0 $$

and each $C_i$ is free as a $\mathbb{Z}$-module, i.e. for each $i$ we have a non-negative integer $r_i$ such that $C_i \simeq \mathbb{Z}^{r_i}$ via some isomorphism $f_i : C^i \to \mathbb{Z}^{r_i}$. If we define $d_i := f_{i-1}\partial_if_i^{-1}$, it is easy to see that $f_\bullet : (C_\bullet,\partial) \to (\mathbb{Z}^{r_\bullet},d_\bullet)$ is a chain complex isomorphism. Hence we can reduce this to the case

$$ 0 \to \mathbb{Z}^{r_n} \to \cdots \to \mathbb{Z}^{r_1} \to \mathbb{Z}^{r_0} \to 0. $$

Note that by a rank argument, we will have $\bigoplus_{\text{$k$ odd}}\mathbb{Z}^{r_k} \simeq \bigoplus_{\text{$k$ even}}\mathbb{Z}^{r_k}$ if and only if $\sum_{\text{$k$ odd}}{r_k}$ = $\sum_{\text{$k$ even}}{r_k}$. This is equivalent to having that

$$ \chi(C_\bullet) = \chi(\mathbb{Z}^{r_\bullet}) = \sum_{k \geq 0}(-1)^k \operatorname{rk}\mathbb{Z}^{r_k} = \sum_{k \geq 0}(-1)^k r_k = 0. $$

As said in the comments, it is a well known fact that the alternating sum of ranges of each $C_i$ coincides with the alternating sum of the ranges of each $H_i(C_\bullet)$. Thus, what we are being asked to prove is equivalent to proving $\sum_{k \geq 0}(-1)^k \operatorname{rk}{H}_k(C_\bullet) = 0$, and the latter follows from the fact that each homology group is of torsion.

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  • $\begingroup$ Hola Guido, yo creo que $d_i=f_{i-1}\partial_i f_i^{-1}$ en vez de lo que pusiste, estoy en lo correcto? La verdad no entiendo por que $\bigoplus_{\text{$k$ odd}}\mathbb{Z}^{r_k} \simeq \bigoplus_{\text{$k$ even}}\mathbb{Z}^{r_k}$ if and only if $\sum_{\text{$k$ odd}}{r_k}$ = $\sum_{\text{$k$ even}}{r_k}$, tienes algun argumento para probar esto? por que esto es verdad? $\endgroup$
    – user425181
    May 19, 2019 at 1:21
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    $\begingroup$ You're right about the first equality, my bad. I have fixed it. As for the second fact, this is the same argument one makes when proving that two $\mathbb{R}$-vector spaces are isomorphic iff they have the same dimension (cont.) $\endgroup$
    – qualcuno
    May 19, 2019 at 1:31
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    $\begingroup$ Note that $\bigoplus_{j}\mathbb{Z}^{r_j} = \mathbb{Z}^{(\sum_j r_j)}$, we simply map each $e_i$ in the direct sum to $e_i$ in the latter. Now, in general $\mathbb{Z}^r \simeq \mathbb{Z}^s$ iff $d = s$. Sufficiency is evident as in that case, we have equality. As for necessity, if these modules are isomorphic, then their ranks coincide, which is precisely to say that $r = s$. (I've tried to keep this in English as the site rules mandate, but feel free to ask if anything is not clear, conceptually or otherwise). $\endgroup$
    – qualcuno
    May 19, 2019 at 1:34
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    $\begingroup$ To expand on the equality of the last comment: an explicit map is to send the $i$-th canonical vector of $\mathbb{Z}^{r_j}$ to the $(\sum_{k=1}^{j-1}r_k + i)$-canonical vector of $\mathbb{Z}^{\sum_j r_j}$. This induces a map into the latter from the direct sum, which is bijective. $\endgroup$
    – qualcuno
    May 19, 2019 at 1:38
  • $\begingroup$ Thank you very much for being so explicit, I value that much, I only have one last question and that is because if the homology group is torsion, then why $\sum_{k \geq 0}(-1)^k \operatorname{rk}{H}_k(C_\bullet) = 0$? (Do you have social networks so we can talk better and solve doubts?) $\endgroup$
    – user425181
    May 19, 2019 at 3:44

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