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As the title says, I am confused about this statement. Let me explain why:

Let $\mathcal{F} = \cdots \to F^i \to F^{i + 1} \to \cdots$ be a cochain complex. I'm reading a blog post which defines the canonical filtration to be $\mathrm{Fil}^i_c \mathcal{F} := \tau_{\leq -i} \mathcal{F}$, where $\tau_{\leq i}$ is the truncation filtration:

$$ \tau_{\leq i}\mathcal{F} = \begin{cases} F^j & \text{ if } j < i \\ \mathrm{Ker}(F^i \to F^{i + 1}) & \text{ if } j = i \\ 0 & \text{ if } j > i. \end{cases}$$

So then, it would seem that

$\mathrm{Fil}^0_c\mathcal{F} = \cdots \to F^{-2} \to F^{-1} \to \mathrm{Ker}(F^0 \to F^1) \to 0 \to \cdots$, and $\mathrm{Fil}^1_c\mathcal{F} = \cdots \to F^{-2} \to \mathrm{Ker}(F^{-1} \to F^0) \to 0$, meaning that

$\mathrm{gr}^0_c\mathcal{F} = \mathrm{Fil}^0_c\mathcal{F}/\mathrm{Fil}^1_c\mathcal{F} = \cdots \to 0 \to 0 \to \cdots \to F^{-1}/(\mathrm{Ker}(F^{-1} \to F^0) \to \mathrm{Ker}(F^0 \to F^1) \to 0 \to \cdots$,

but the correct answer is supposed to have $H^{0}(\mathcal{F})$ in degree 0. Am I making a mistake somewhere?

ufabao
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    Note however that you have a canonical quasi-isomorphism $\operatorname{gr}_c^0\mathcal{F}\overset{\sim}{\rightarrow} H^0(\mathcal{F})$. This is maybe what is meant in your blog (or an isomorphism in the derived category). – Roland Jan 04 '18 at 08:30
  • @Roland Ahh yeah I can't believe I missed that. Thanks so much! – ufabao Jan 04 '18 at 09:13

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