There is a comparison theorem for spectral sequnces in Weibel's book (5.2.12) stating;


Assume $E\_{p,q}$ and $\bar E\_{p,q}$ converge to $H\_\* $ $\bar H\_\*$ respectively. Furthermore we have given a map $h: H\_{\*} \to \bar H\_{\*} $ compatible with a morphism $f$ of spectral sequences. 


If $f^r: E^r\_{p,q} \to \bar E^r\_{p,q}$ is an isomorphism for all $p,q$ and some $r$ then $h$ is an isomorphism.


What I want to ask is what happens if we have a milder situation than isomorphism. For example if they just differ on the border? 


To be precise let $E^2\_{p,q}$ and $\bar E^2\_{p,q}$ are two first quadrant spectral sequences converging to $H\_\* $ $\bar H\_\*$ respectively. Also there is a map $h$ compatible with a morphism $f$ of spectral sequences as above. Assume $E^2\_{p,q} \cong \bar E^2\_{p,q}$ if $q\neq0$ and $E^2\_{p,0}$ vanishes. Can we calculate kernel and cokernel of $h$? 


Thanks for your help.



Comparing Spectral Sequences
The cokernel is entirely due to $\bar{E}^\infty\_{\*,0}$ but the kernel is more mysteriuous.


First observe that if $g:C\_\cdot\to D\_\cdot$ is a chain map and $i$ is such that $g\_i$ is surjective and $g\_{i-1}$ is injective, then 
$H\_i(g)$ is surjective and $H\_{i-1}(g)$ is injective. (Exercise.)


Using this, one sees by induction on $r$ that $f^r\_{p,q}$ is surjective for $q\geq1$ and injective (hence bijective) for $q\geq r-1\geq1$.


Now take $r=\infty$. One sees that $h$ hits all layers except the top one in the filtration of $\bar{H}\_\*$. 


Let me include Grilo's formulas as I now believe they should read: 
We have exact sequences


$$0\to{\bar{E}}^{r+2}\_{n+1,0} \to\bar{E}^{r+1}\_{n+1,0}\to E^{r+2}\_{n-r,r}\to\bar{E}^{r+2}\_{n-r,r}\to0$$


and then
$$0\to{\bar{E}}^{r+2}\_{n+1,0} \to\bar{E}^{r+1}\_{n+1,0}\to E^{\infty}\_{n-r,r}\to\bar{E}^{\infty}\_{n-r,r}\to0$$


Putting $r=n$ it becomes


$$0\to{\bar{E}}^{\infty}\_{n+1,0} \to\bar{E}^{n+1}\_{n+1,0}\to E^{\infty}\_{0,n+1}\to\bar{E}^{\infty}\_{0,n+1}\to0$$


or


$$H\_{n+1}\to \bar{H}\_{n+1} \to\bar{E}^{n+1}\_{n+1,0}\to E^{\infty}\_{0,n+1}\to\bar{E}^{\infty}\_{0,n+1}\to0$$


So far so good.


