In Weibel's book, a spectral sequence $E^r_{p,q}$ is said to weakly converge to a graded object $H_{\ast}$ if for every $n$ there exists a filtration $\dots \subset F_{r}H_{n} \subset F_{r-1}H_{\ast} \subset \dots H_{n}$ such that $E^{\infty}_{p,q} \simeq \text{gr}_{p}H_{p+q}$. Moreover, a filtration is called exhaustive if $\cup_r F_{r}H_n = H_n$, and it is called Hausdorff if $\cap_r F_r H_n = 0 $. A spectral sequence which weakly converges and that is both exhaustive and Hausdorff is said to approach $H_{\ast}$.

Later on in the book the filtrations for a double complex are introduced, and the two spectral sequences that arise from these filtrations are studied. For example, the filtration by columns of a double complex is defined as $$ (\tau^{\leq n}C_{\ast,\ast})_{p,q} = \left\{ \begin{array}{lr} C_{p,q} & p \leq n\\ 0 & p > n \end{array} \right. $$

If we consider the filtration by columns of a double complex $C_{p,q}$ which is zero in the fourth quadrant, we get a spectral sequence that weakly converges to $H_{\ast}(\text{Tot}^{\tiny\prod}C_{\ast, \ast})$. What I don't understand is why the spectral sequence doesn't approach $H_{\ast}(\text{Tot}^{\tiny\prod}C_{\ast,\ast})$. In the book it is said that the filtration on the total complex is exhaustive, but to me it seems it is also Hausdorff. Am I getting something wrong?

I think that even if it is not true in full generality (strange categories might be badly behaved), it is nevertheless true for categories of modules or categories of sheaves, where we have an explicit construction for direct sums and direct products.

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    $\begingroup$ The issue here, I believe, is the failure of inverse limits to be exact. As a result, even if the filtration on the total complex is Hausdorff, the filtration on homology might not be. This is discussed at length in Boardman's "Conditionally convergent spectral sequences." $\endgroup$ – Tyler Lawson May 5 '20 at 17:31
  • $\begingroup$ Yeah, searching among old questions I found an answer that said exactly the same thing, thank you. $\endgroup$ – Federico Barbacovi May 5 '20 at 17:36
  • $\begingroup$ And that implies it doesn't work even for modules or sheaves of modules, right? $\endgroup$ – Federico Barbacovi May 5 '20 at 19:05
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    $\begingroup$ Just to give a short example: A homology class lies in $F_rH_*$, if it has a represetative in $F_rC_*$ Now look at the complex of vector spaces over $\mathbb{F}_2$ given by $0\rightarrow F[t]\rightarrow F[t]\rightarrow 0$ ($F[t]$ denotes the polynomial ring) and let the differential be given by multiplication with $1+t$. This complex has a filtration given by $F_{-r}C_* =t^rC_*$. Now the Kokernel of the differential is just an 1-dimensional vector space, and the nontrivial class is represented given by $[1]=[t]=[t^2]=\ldots$ and hence that class lives in $\bigcap_r F_rH_*$. $\endgroup$ – HenrikRüping May 5 '20 at 19:43

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