Show that $\bigoplus_{i \text{ even}}C_i=\bigoplus_{i\text{ odd}}C_i$ 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.
 A: 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.
