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Let $E^{pq}_2\implies H^{p+q}$ and $\hat{E}^{pq}_2\implies \hat{H}^{p+q}$ be first quadrant spectral sequences of abelian groups. Suppose we are given a morphism $h:H^*\to \hat{H}^*$ compatible with a morphism $f:E\to \hat{E}$ of the spectral sequences. If $f_2:E^{pq}_2\to \hat{E}^{pq}_2$ is the zero morphism for all $p,q$, does it follow that $h$ is the zero morphism?

Let me illustrate what I've done so far assuming $E^{pq}_2=\hat{E}^{pq}_2=0$ for all $p\geq 3$ (which is actually the case I'm interested in). Here we have filtrations $0\subset A^2\subset A^1\subset H^q$ and $0\subset B^2\subset B^1\subset \hat{H}^q$ and the edge maps fit in a commutative diagram

\begin{matrix} H^q &\to &E_2^{0q} \\ \downarrow & &\downarrow\\ \hat{H}^q &\to &\hat{E}^{0q}_2 \end{matrix}

The right vertical arrow is zero hence the image of $H^q \to\hat {H}^q$ is contained in $B^1$. Similarly the image of $A^1\to B^1$ is contained in $B^2$, and $A^2\to B^2$ is zero. But I don't see how to conclude that $H^q\to \hat{H}^q$ is zero. I feel like there is something about spectral sequences that I am not using...

Remark: It is a textbook result (e.g. in Weibel Comparison Theorem 5.2.12) that if $f_2$ is an isomorphism for all $p,q$ then $h$ is an isomorphism. I tried to mimic the proof of this statement but it doesn't work because the 5-lemma is not true for zero morphisms (i.e. a morphism of short exact sequences whose 4 outer arrows are zero doesn't necessarily have its middle arrow also zero).

Marco Flores
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    I think the problem you mention about the "$0$-five lemma" is insurmountable. $f_2=0$ implies that all $f_r$ are $0$, and so that $f_\infty=0$, and there is no more compatibility with $h$ than on the $\infty$-page. In other words, you might as well start on the $E_\infty$-page, and then you are faced with your issue. So to construct a counterexample, start with a counterexample for the "$0$-five lemma", define a spectral sequence with no differential so that the $E_\infty$-page looks like the quotients of your counterexample, and define the $H^*$ as appropriate – Maxime Ramzi Jul 05 '19 at 20:58
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    Here is an easy example following @Max advice. Consider a complex $C^.:A\xrightarrow{d} B$ (and $0$ everywhere else) with $d$ mono. Consider the filtrations $F$ and ${'F}$ given by $F^0=C^., F^1=B[-1]$ and $F^2=0$ for $F$ and ${'F}^0=C^.$ and ${'F}^1=0$ for ${'F}$. The spectral sequences associated to these two filtered complexes degenerates at $E_2$ (even $E_1$ for the first one). There is an obvious inclusion ${'F}\to F$ which induces a map of spectral sequences. This map is zero at $E_2$ and at every higher page, but not on the abutment where this is an isomorphism. – Roland Jul 06 '19 at 09:51

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