Let $\Lambda $ be a fixed set. For each $\alpha\in\Lambda $ is $\{C_n^{\alpha}\}_{n\in\mathbb{Z}}$ a chain complex with boundary homomorphism $\partial^{\alpha}$.
Prove that there is an isomorphism $$\phi_n:H_n(C_*)\to\bigoplus_{\alpha\in\Lambda}H_n(C_*^{\alpha}) $$
I have thought about doing the following:
Let's define $\phi_n(\overline{(x_{\alpha})})=(\overline{x_{\alpha}})$, to see that this is an R-homomorphism, let's take $r,s\in R$ and $\overline{(x_{\alpha})}, \overline{(y_{\alpha})}\in H_n(C_*)$ with which $\phi_n(r\overline{(x_{\alpha})}+s\overline{(y_{\alpha})})=\phi_n(\overline{(rx_{\alpha})}+\overline{(sy_{\alpha})})=\phi_n(\overline{(rx_{\alpha}+sy_{\alpha})})=(\overline{rx_{\alpha}+sy_{\alpha}})=r(\overline{x_{\alpha}})+s(\overline{y_{\alpha}})=r\phi_n(\overline{(x_{\alpha})})+s\phi_n(\overline{(y_{\alpha})})$.
This function is subjective because if $(\overline{x_{\alpha}})\in \bigoplus_{\alpha\in\Lambda}H_n(C_*^{\alpha})$ then for each $\alpha\in\Lambda$ take $x_{\alpha}$ as a representative for $\overline{x_{\alpha}}$, then $\overline{(x_{\alpha})}$ is a preimage of $(\overline{x_{\alpha}})\in \bigoplus_{\alpha\in\Lambda}H_n(C_*^{\alpha})$
This function is injective because if $\phi_n(\overline{(x_{\alpha})})=(\overline{x_{\alpha}})=(\overline{0})$ then $\overline{x_{\alpha}}=\overline{0}$ for all $\alpha\in\Lambda$ but I do not know what to do to show that $\overline{(x_{\alpha})}=\overline{(0)}$. Could someone help me please? Thank you.
You have $H_n(C_*) = \ker(\partial_n) /\text{im}(\partial_{n+1})$. But now $\partial_k = \bigoplus \partial_k^\alpha$ and therefore $\ker(\partial_n) = \bigoplus \ker(\partial_n^\alpha)$ and $\text{im}(\partial_{n+1}) = \bigoplus \text{im}(\partial_{n+1}^\alpha)$.
Note that $\phi_n(\overline{(x_{\alpha})})$ is defined precisely for the elements ${(x_{\alpha})} \in \ker(\partial_n) = \bigoplus \ker(\partial_n^\alpha)$, i.e. for the elements with $x_\alpha \in \ker(\partial_n^\alpha)$ for all $\alpha$.
$\overline{x_\alpha} = \overline{0}$ means that $x_\alpha \in \text{im}(\partial_{n+1}^\alpha)$, thus $(x_\alpha) \in\bigoplus \text{im}(\partial_{n+1}^\alpha) = \text{im}(\partial_{n+1}) $. Hence $\overline{(x_\alpha)} = \overline{(0)}$.