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I'm trying to take spectral sequences as a black box for application in commutative algebra and I admit that I haven't really gone through (or understand) all the proofs of all the isomorphisms between various terms of a spectral sequence and respective cohomologies assuming some term of the sequences are zero.

So in this post, my question is about the dual (homology) version of Proposition 5.6 (c) , Chapter XV, of Homological algebra by Cartan & Eilenberg. Let me first recall (for the case $r=2$ only) that it says that: if $E_2^{p,q}$ is a spectral sequence converging to $ H^{p+q}$ and we have $E_2^{u,v}=0$ for $u+v=n-1, u\le p-2 $ and for $u+v=n, u<p$ and for $u+v=n+1, u\ge p+2$ and for $u+v=n, u>p$, then we have an isomorphism $E_2^{p,n-p}\cong H^n$ .

Now my question is: What is the dual version of this result (for first quadrant spectral sequence $E^2_{p,q}$ and $H_n$ ) ?

My naive idea: Since $E^2_{p,q}=E_2^{-p, -q}$ and $H_n=H^{-n}$, so if I just change all the $u,v,n,p,q$ in the conditions to it's negative and then rewrite the conditions, then do I get a valid dual result ?

Thanks

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  • $\begingroup$ I think it is better to understand the main definitions and theorems you don't need the proofs. $\endgroup$
    – ali
    Aug 5, 2020 at 7:59
  • $\begingroup$ the conditions in the theorem means that all the boundary maps coming to or going out of $E^{p,n-p}$ are zero and also all other elements on the diagonal $u+v=n$ are zero. just look at the definition of the boundary map in homology and rewrite the conditons $\endgroup$
    – ali
    Aug 5, 2020 at 8:10

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