Lemma 15.95.8. Let $A$ be a ring and let $f \in A$ be a nonzerodivisor. Let $M^\bullet$ be a complex of $f$-torsion free $A$-modules. If $\mathop{\mathrm{Ker}}(d^ i \bmod f^2)$ surjects onto $\mathop{\mathrm{Ker}}(d^ i \bmod f)$, then the canonical map

$(1, d^ i) : (\eta _ fM)^ i / f(\eta _ fM)^ i \longrightarrow f^ iM^ i/f^{i + 1}M^ i \oplus f^{i + 1}M^{i + 1}/f^{i + 2}M^{i + 1}$

identifies the left hand side with a direct sum of submodules of the right hand side.

Proof. With notation as in Remark 15.95.5 we define a map $t^{-1} : Z^ i \to (\eta _ fM)^ i / f(\eta _ fM)^ i$. Namely, for $x \in M^ i$ with $d^ i(x) = f^2y$ we send the class of $x$ in $Z^ i$ to the class of $f^ ix$ in $(\eta _ fM)^ i / f(\eta _ fM)^ i$. We omit the verification that this is well defined; the assumption of the lemma exactly signifies that the domain of this operation is all of $Z^ i$. Then $t \circ t^{-1} = \text{id}_{Z^ i}$. Hence $t^{-1}$ defines a splitting of the short exact sequence in Remark 15.95.5 and the resulting direct sum decomposition

$(\eta _ fM)^ i / f(\eta _ fM)^ i = Z^ i \oplus B^{i + 1}$

is compatible with the map displayed in the lemma. $\square$

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