Lemma 12.24.13. Let $\mathcal{A}$ be an abelian category. Let $(K^\bullet , F)$ be a filtered complex of $\mathcal{A}$. Assume

1. for every $n$ there exist $p_0(n)$ such that $H^ n(F^ pK^\bullet ) = 0$ for $p \geq p_0(n)$,

2. for every $n$ there exist $p_1(n)$ such that $H^ n(F^ pK^\bullet ) \to H^ n(K^\bullet )$ is an isomorphism for $p \leq p_1(n)$.

Then

1. the spectral sequence associated to $(K^\bullet , F)$ is bounded,

2. the filtration on each $H^ n(K^\bullet )$ is finite,

3. the spectral sequence associated to $(K^\bullet , F)$ converges to $H^*(K^\bullet )$.

Proof. Fix $n$. Using the long exact cohomology sequence associated to the short exact sequence of complexes

$0 \to F^{p + 1}K^\bullet \to F^ pK^\bullet \to \text{gr}^ pK^\bullet \to 0$

we find that $E_1^{p, n - p} = 0$ for $p \geq \max (p_0(n), p_0(n + 1))$ and $p < \min (p_1(n), p_1(n + 1))$. Hence the spectral sequence is bounded (Definition 12.24.7). This proves (1).

It is clear from the assumptions and Definition 12.24.5 that the filtration on $H^ n(K^\bullet )$ is finite. This proves (2).

Next we prove that the spectral sequence weakly converges to $H^*(K^\bullet )$ using Lemma 12.24.10. Let us show that we have equality in (12.24.6.1). Namely, for $p + r > p_0(n + 1)$ the map

$d : F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) \to F^{p + r}K^{n + 1}$

ends up in the image of $d : F^{p + r}K^ n \to F^{p + r}K^{n + 1}$ because the complex $F^{p + r}K^\bullet$ is exact in degree $n + 1$. We conclude that $F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) = d(F^{p + r}K^ n) + \mathop{\mathrm{Ker}}(d) \cap F^ pK^ n$. Hence for such $r$ we have

$\mathop{\mathrm{Ker}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n} = F^ pK^{n} \cap d^{-1}(F^{p + r}K^{n + 1}) + F^{p + 1}K^{n}$

which proves the desired equality. To show that we have equality in (12.24.6.2) we use that for $p - r + 1 < p_1(n - 1)$ we have

$d(F^{p - r + 1}K^{n - 1}) = \mathop{\mathrm{Im}}(d) \cap F^{p - r + 1}K^ n$

because the map $F^{p - r + 1}K^\bullet \to K^\bullet$ induces an isomorphism on cohomology in degree $n - 1$. This shows that we have

$F^ pK^{n} \cap d(F^{p - r + 1}K^{n - 1}) + F^{p + 1}K^{n} = \mathop{\mathrm{Im}}(d) \cap F^ pK^{n} + F^{p + 1}K^{n}$

for such $r$ which proves the desired equality.

To see that the spectral sequence abuts to $H^*(K^\bullet )$ using Lemma 12.24.10 we have to show that $\bigcap _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Im}}(d) \cap K^ n$ and $\bigcup _ p (\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n) = \mathop{\mathrm{Ker}}(d) \cap K^ n$. For $p \geq p_0(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Im}}(d) \cap K^ n$ and for $p \leq p_1(n)$ we have $\mathop{\mathrm{Ker}}(d) \cap F^ pK^ n + \mathop{\mathrm{Im}}(d) \cap K^ n = \mathop{\mathrm{Ker}}(d) \cap K^ n$. Combining weak convergence, abutment, and boundedness we see that (2) and (3) are true. $\square$

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