filtration on the cohomology of a complex

filtration on the cohomology of a complex



Let $K^\bullet$ be a complex and let $F_I$ and $F_{II}$ be two filtrations

on it. suppose $F_I^i K^n$ intersects $F_{II}^i K^n$ trivially. It then

follows that in the induced filtration on the cohomology of the complex

$F_I^i H^n$ intersects $F_{II}^i H^n$ trivially.

However, I have encountered a situation where this doesn't seem to hold.

In order to construct the spectral sequence of a composition of functors

one makes use of a certain double complex called Cartan-Eilenberg

resolution, call it $C^{\bullet\bullet}$, and considers the spectral

sequence of a filtered complex for the total complex of it, $K^\bullet$.

There are two filtrations on $K^\bullet$, vertical and horizontal. One of

the properties of $C^{\bullet\bullet}$ is that the filtration on the

cohomology of the total complex induced by the vertical filtration is

stupid:

$F^n H^i = H^i$ if $i <n$, $F^n H^i = 0$ if $i \geq n$.

But the filtration on the cohomology induced by the horizontal filtration

can be anything, in particular horizontal $F^i_h H^n$ can intersect

non-trivially the vertical $F^j_v H^n$ even if $i+j > n$ and the

corresponding elements of the filtration on $K^\bullet$ do intersect

trivially.

What is wrong in the above reasoning?