The Stacks project

Proof. If $\mathcal{A}$ has countable direct sums and if countable direct sums are exact, then this follows from the discussion above. In general, we proceed as follows; we strongly suggest the reader skip this proof. Consider the object $A = (F^{p + 1}K)$ of $\text{Gr}(\mathcal{A})$, i.e., we put $F^{p + 1}K$ in degree $p$ (the funny shift in numbering to get numbering correct later on). We endow it with a differential $d$ by using $d$ on each component. Then $(A, d)$ is a differential object of $\text{Gr}(\mathcal{A})$. Consider the map

which is given in degree $p$ by the inclusions $F^{p + 1}A \to F^ pA$. This is clearly an injective morphism of differential objects $\alpha : (A, d) \to (A, d)[-1]$. Hence, we can apply Remark 12.22.6 with $S = \text{id}$ and $T = [1]$. The corresponding spectral sequence $(E_ r, d_ r)_{r \geq 0}$ in $\text{Gr}(\mathcal{A})$ is the spectral sequence we are looking for. Let us unwind the definitions a bit. First of all we have $E_ r = (E_ r^ p)$ is an object of $\text{Gr}(\mathcal{A})$. Then, since $T^ rS = [r]$ we have $d_ r : E_ r \to E_ r[r]$ which means that $d_ r^ p : E_ r^ p \to E_ r^{p + r}$.

To see that the description of the graded pieces hold, we argue as above. Namely, first we have $E_0 = \mathop{\mathrm{Coker}}(\alpha : A \to A[-1])$ and by our choice of numbering above this gives $E_0^ p = \text{gr}^ pK$. The first differential is given by $d_0^ p = \text{gr}^ pd : E_0^ p \to E_0^ p$. Next, the description of the boundaries $B_ r$ and the cocycles $Z_ r$ in Remark 12.22.6 translates into a straightforward manner into the formulae for $Z_ r^ p$ and $B_ r^ p$ given above. $\square$