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.
Best Answer
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.