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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:

$F^n H^i(K^\bullet) = H^i (K^\bullet)$ if $i <n$, $F^n H^i(K^\bullet) = 0$ 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?

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I have figured out what my problem was. The components of the filtration on the cohomology are the images of the cohomology of the filtration components on $K^\bullet$ under complex inclusion, and this is not generally an injection. Therefore, trivially intersecting (in some term) subcomplexes can give rise to non-trivially intersecting subgroups of the cohomology. – Dima Sustretov Aug 21 '13 at 13:22
how does one mark the question as answered without adding an answer? I cannot answer my own question. – Dima Sustretov Aug 21 '13 at 13:22

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