In the proof of the algebraic Kunneth formula, they proof first the special case where every boundary map in $C$ is zero. This implies $\partial(c \otimes d) = (-1)^{i}c\otimes \partial d $.
I don't understand how this implies $C \otimes_{R} C' = \oplus_i C_i \otimes_R C' $ and hence $H_n ( C_i \otimes_R C')= C_i\otimes_R H_{n-i}(C'))$.
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$\begingroup$I think he wants you to think of $C_i$ as a chain complex concentrated in degree i. Write this chain complex as $\underline C_i$ (so it has $C_i$ in degree i and 0s elsewhere and trivial differentials). Then $(\underline C_i\bigotimes C')_n=C_i\otimes C'_{n-i}$, so $(C\bigotimes C')_n=\oplus_i \;C_i\otimes C'_{n-i}=\bigoplus_i(\underline C_i\bigotimes C')_n$ and both chain complexes have the same underlying $R$-modules in each degree.
Now since $C$ has trivial differential the differential on $C\otimes C'$ is $\hat\partial_n=\oplus_i(\partial_i\otimes 1_{n-i}+(-1)^i1_i\otimes \partial'_{n-i})=\oplus_i\;(-1)^i1_i\otimes \partial'_{n-i}$. But $(-1)^i1_i\otimes \partial'$ is just the differential on $\underline C_i\otimes C'$. Hence $\hat \partial=\bigoplus_i(-1)^i1_i\otimes \partial'$, and the differentials on $C\bigotimes C'$ and $\bigoplus_i\; \underline C_i\bigotimes C'$ also agree. That is, they are the same chain complexes.
The statement for homology follows since each $C_i$ is free, and have $\underline C_i$ is a projective chain complex. The operation of tensoring with a projective module commutes with the operation of taking homology, so you get
$H_n(C'\bigotimes C)=H_n(\bigoplus_i\;\underline C_i\bigotimes C')\cong \oplus_i\;H_n(\underline C_i\bigotimes C')\cong \oplus_i\; C_i\otimes H_{n-i}( C').$
I hope this may have been a little clearer (or at least a little more explicit).
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