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I am currently reading Künneth spectral sequence, which is given below.

Let $R$ be a ring and A$=\big\{A_n,d_n:A_n\longrightarrow A_{n-1}\big|d_{n-1}\circ d_n=0\big\}_{n\in \Bbb Z}$ be a chain complex of right $R$-modules with each $A_n$ is flat and $A_n=0$ when $n<0$. Let C$=\big\{C_n,\partial_n:C_n\longrightarrow C_{n-1}\big|\partial_{n-1}\circ \partial_n=0\big\}_{n\in \Bbb Z}$ be a chain complex of left $R$-modules with $C_n=0$ if $n<0$. Then, We have two spectral sequences $^\text{I}E,\ ^\text{II}E$ both converging to $H_n(\textbf{A}\otimes_R\textbf{C})$ such that $$^\text{II}E^2_{p,q}\simeq\bigoplus_{l+k=q}\text{Tor}^R_p\bigg(H_k(\textbf{A}), H_l(\textbf{C})\bigg)\text{ and }^\text{I}E^2_{p,q}=\begin{cases} H_p(\textbf{ A}\otimes_R\textbf{ C}) & \text{ if }q=0,\\0 & \text{ if }q\not=0.\end{cases}$$

Now, the proof based on Künneth formula for homology. Also, these spectral sequences can give an alternative proof of Künneth formula for homology, when chain complexes are positive. So, the argument is somewhat cyclic. I am looking for an alternative proof of Künneth spectral sequence. Is it possible? Any help will be appreciated. Thanks in advance.

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    $\begingroup$ I know a simple proof of the Künneth SS that does not use the Künneth formula. However, it uses the derived tensor product and the derived categories. Would that be satisfactory? $\endgroup$
    – Denis Nardin
    Commented Jun 7, 2020 at 20:13
  • $\begingroup$ Could you tell me some more, like, where it can be found? $\endgroup$
    – Sumanta
    Commented Jun 7, 2020 at 20:18
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    $\begingroup$ It's essentially the proof of theorem 8.1 in Elmendorf, Kriz, May rephrased in the language of complexes instead of spectra (or, if you want, just by taking modules over the Eilenberg-MacLane spectrum $HR$) $\endgroup$
    – Denis Nardin
    Commented Jun 7, 2020 at 20:54

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