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3 formulas corrected

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

$$0\to{\bar{E}}^{r+1}_{n+1,0} 0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^r_{n+1,0}\to E^{r+1}_{n-r+1,r}\to\bar{E}^{r+1}_{n-r+1,r}\to0$$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+1}_{n+1,0} $0\to{\bar{E}}^{r+2}_{n+1,0} \to\bar{E}^r_{n+1,0}\to E^{\infty}_{n-r+1,r}\to\bar{E}^{\infty}_{n-r+1,r}\to0$$to\bar{E}^{r+1}_{n+1,0}\to E^{\infty}_{n-r,r}\to\bar{E}^{\infty}_{n-r,r}\to0$$ Putting$r=n+1$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.

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

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

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

Putting $r=n+1$ it becomes

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

So far so good.

1

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 $\bar{H}_*$.