# On the dual version of an isomorphism of spectral sequence term (from Cartan and Eilenberg)

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 $$u+v=n, u and for $$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

• I think it is better to understand the main definitions and theorems you don't need the proofs.
– ali
Aug 5 '20 at 7:59
• 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
– ali
Aug 5 '20 at 8:10