# Spectral sequence page isomorphism

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?

• 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... – Najib Idrissi Feb 10 '15 at 10:21

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.
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.