This is an attempt to prove that direct product of chain complexes commutes with homology (exercise in Weibel's book). I've had some success since I've proved that $Z_n(\prod_{\alpha \in A} C_{\alpha \cdot}) = \prod_{\alpha \in A} Z_n(C_{\alpha \cdot})$. In other words that direct product commutes with the $n$-cycles of a chain complex. My attempt at the $n$-boundary part:
$B_n(\prod_{\alpha \in A} C_{\alpha \cdot}) = \text{im } d_{n+1} = \{\prod_{\alpha\in A}d_{\alpha, n+1}(x_{\alpha}) : x_{\alpha} \in C_{\alpha, n+1}\} = \prod_{\alpha} B_n(C_{\alpha \cdot})$. The last step I thinks is clear.
Now how do I throw together the rest of the proof (showing that direct product therefore commutes with homology $H_n$)? Except in this case the two product formulas are only isomorphic is what Weibel's book seems to imply, where as the formulas we found above for cycles and boundaries have equality.
By the way, these are chain complexes of $R$-modules.
Best Answer
To treat this exercise in the most general setup, you can say that in a category in which limits exists and the limit functor is exact (AB4* means exactly that since limit is a right adjoint functor so it is always left exact, see the Appendix in Weibel's book), the homology functor of any degree does commute with direct product. Fortunately the category of $R$-modules does verify AB4* (and AB4 actually). The proof is simple: take the following exact sequence $$0\to \ker d_n\to \text{im}\ d_{n+1}\to H_n(A_n^{i})\to 0$$ where $A^{i}$ is a complex for $i\in I$. Apply the exact functor direct product. You immediately get what you want since by exactness direct product commutes with $\ker$ (always) and $\text{im}$ (AB4*).