Remark 12.22.6 (Variant). Let $\mathcal{A}$ be an abelian category and let $S, T : \mathcal{A} \to \mathcal{A}$ be shift functors, i.e., isomorphisms of categories. Assume that $TS = ST$ as functors. Consider pairs $(A, d)$ consisting of an object $A$ of $\mathcal{A}$ and a morphism $d : A \to SA$ such that $d \circ S^{-1}d = 0$. The category of these objects is abelian. We define $H(A, d) = \mathop{\mathrm{Ker}}(d)/\mathop{\mathrm{Im}}(S^{-1}d)$ and we observe that $H(SA, Sd) = SH(A, d)$ (canonical isomorphism). Given a short exact sequence

$0 \to (A, d) \to (B, d) \to (C, d) \to 0$

we obtain a long exact homology sequence

$\ldots \to S^{-1}H(C, d) \to H(A, d) \to H(B, d) \to H(C, d) \to SH(A, d) \to \ldots$

(note the shifts in the boundary maps). Since $ST = TS$ the functor $T$ defines a shift functor on pairs by setting $T(A, d) = (TA, Td)$. Next, let $\alpha : (A, d) \to T^{-1}(A, d)$ be injective with cokernel $(Q, d)$. Then we get an exact couple as in Remark 12.21.5 with shift functors $TS$ and $T$ given by

$(H(A, d), S^{-1}H(Q, d), \overline{\alpha }, f, g)$

where $\overline{\alpha } : H(A, d) \to T^{-1}H(A, d)$ is induced by $\alpha$, the map $f : S^{-1}H(Q, d) \to H(A, d)$ is the boundary map and $g : H(A, d) \to TH(Q, d) = TS(S^{-1}H(Q, d))$ is induced by the quotient map $A \to TQ$. Thus we get a spectral sequence as above with $E_1 = S^{-1}H(Q, d)$ and differentials $d_ r : E_ r \to T^ rSE_ r$. As above we set $E_0 = S^{-1}Q$ and $d_0 : E_0 \to SE_0$ given by $S^{-1}d : S^{-1}Q \to Q$. If according to our conventions we define $B_ r \subset Z_ r \subset E_0$, then we have for $r \geq 1$ that

1. $SB_ r$ is the image of

$(T^{-r + 1}\alpha \circ \ldots \circ T^{-1}\alpha )^{-1} \mathop{\mathrm{Im}}(T^{-r}S^{-1}d)$

under the natural map $T^{-1}A \to Q$,

2. $Z_ r$ is the image of

$(S^{-1}T^{-1}d)^{-1} \mathop{\mathrm{Im}}(\alpha \circ \ldots \circ T^{r - 1}\alpha )$

under the natural map $S^{-1}T^{-1}A \to S^{-1}Q$.

The differentials can be described as follows: if $x \in Z_ r$, then pick $x' \in S^{-1}T^{-1}A$ mapping to $x$. Then $S^{-1}T^{-1}d(x')$ is $(\alpha \circ \ldots \circ T^{r - 1}\alpha )(y)$ for some $y \in T^{r - 1}A$. Then $d_ r(x) \in T^ rSE_ r$ is represented by the class of the image of $y$ in $T^ rSE_0 = T^ rQ$ modulo $T^ rSB_ r$.

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