The two double complex spectral sequences have the same abutment but not the same pages

Both convergence theorems identify the unfiltered target in degree $n$ with $H_n(T)$ for the same total complex $T$ and the same differential $h+v$. They make different specified filtrations on that object; equality of targets alone asserts neither equality of those filtrations nor equality of the pages.

For the early-page witness, take $C_{1,0}=C_{0,0}=k$, $h_{1,0}=1_k$, and every other component and arrow zero. The only row complex is $k\stackrel{1}{\to }k$, whose kernel in degree one and cokernel in degree zero are both zero. Thus the row $E^1$ page is zero. Each nonzero column is a single $k$, so column $E_{1,0}^1=E_{0,0}^1=k$, with $d_{1,0}^1=1_k$. Hence its $E^1$ page is nonzero but its $E^2$ page is zero. The total complex is also $k\stackrel{1}{\to }k$ and has zero homology.

For the filtration witness, instead take just $C_{1,0}=k$ with all arrows zero. Then $H_1(T)=k$. The row filtration has $F_0^{\mathrm{r}\mathrm{o}\mathrm{w}}{H}_1(T)=k$, since vertical index zero is already included. The column filtration has $F_0^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1(T)=0$ and $F_1^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1(T)=k$. Thus the filtrations on the same nonzero target differ; the jumps occur at different filtration degrees.

These witnesses establish the two possible failures while step 1.1 proves the common-target assertion. All witnesses have finite support and specified zero or identity maps, with no representative choices. The zero complex would give equal zero pages and filtrations, which is consistent with the claim that differences can occur.