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).