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Suppose we have a map of spectral sequences $\{E_{p,q}^r,d^r\}\to \{{E'}_{p,q}^r,d'^r\}$, both generated from total chain complexes, $C$ and $C'$ respectively, such that for some $r$ the map between the $r$-th pages $E_{p,q}^r\to {E'}_{p,q}^r$ is an isomorphism. Using the five lemma, we can show that for any $r'>r$, we also have an isomorphism of the $r'$ pages. Can we deduce anything about the homology of the filtered chain complexes associated with these spectral sequences?

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  • $\begingroup$ Is the map of spectral sequences induced by a chain map between $C$ and $C'$? This MO question might interest you (it deals with filtered complexes though). At the very least if both spectral sequences converge, then both homologies have a graduation such that the two associated graded modules are isomorphic... $\endgroup$ – Najib Idrissi Feb 10 '15 at 10:21
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There is this comparison theorem in Weibel's book An introduction to homological algebra:

Comparison Theorem 5.2.12 Let $\{ E^r_{pq} \}$ and $\{E^{'r}_{pq} \}$ converge to $H_*$ and $H'_*$, respectively. Suppose given a map $h : H_* \to H_*'$ compatible with a morphism $f: E \to E'$ of spectral sequences. If $f^r : \{E^r_{pq}\} \to \{E^{'r}_{pq} \}$ is an isomorphism for all $p$ and $q$ and some $r$ (hence for $r = \infty$ by the Mapping Lemma), then $h : H_* \to H'_*$ is an isomorphism.

I think that's probably the best kind of result you're going to get. Of course if the spectral sequences do not converge, I don't believe it will be possible to compare the homologies. Now if they do converge, this Comparison Theorem together with your hypotheses directly imply that the homologies are isomorphic.

Note that you need the map $h : H_* \to H'_*$ from the start; so for example if your map of spectral sequences is not induced by a chain map $C \to C'$, I don't know if it will be possible to compare $H_*$ and $H'_*$.


The Exercise 5.4.4 in the aforementioned book might be useful too, and an even better result given your hypotheses actually. Suppose $f : C \to C'$ is a chain map and let $C_f$ be its mapping cone. Then the Exercise tells you that you can get a spectral sequence out of $C_f$, say $\{ E^r_p(C_f) \}$, and that it is the mapping cone of the induced map $E^r_p \to E'^r_p$. If $E^r_p \to E'^r_p$ is an isomorphism, then its mapping cone $E^r(C_f)_p$ is zero, hence $E^\infty(C_f)$ is identically zero. This implies that the mapping cone $C_f$ has trivial homology (we get convergence for free because $E^\infty = 0$!), hence that $f$ was a quasi-isomorphism.

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