Remark 15.95.5. 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. For every $i$ set $\overline{M}^ i = M^ i/fM^ i$. Denote $B^ i \subset Z^ i \subset \overline{M}^ i$ the boundaries and cocycles for the differentials on the complex $\overline{M}^\bullet = M^\bullet \otimes _ A A/fA$. We claim that there exists a commutative diagram

with exact rows. Here are the constructions of the maps

If $x \in (\eta _ fM)^ i$ then $x = f^ ix'$ with $d^ i(x') = 0$ in $\overline{M}^{i + 1}$. Hence we can define the map $t$ by sending $x$ to the class of $x'$.

If $y \in M^{i + 1}$ has class $\overline{y}$ in $B^{i + 1} \subset \overline{M}^{i + 1}$ then we can write $y = fy' + d^ i(x)$ for $y' \in M^{i + 1}$ and $x \in M^ i$. Hence we can define the map $s$ sending $\overline{y}$ to the class of $f^{i + 1}x$ in $(\eta _ fM)^ i /f(\eta _ fM)^ i$; we omit the verification that this is well defined.

If $x \in M^ i$ has class $\overline{x}$ in $B^ i \subset \overline{M}^ i$ then we can write $x = fx' + d^{i - 1}(z)$ for $x' \in M^ i$ and $z \in M^{i - 1}$. We define the map $s'$ by sending $\overline{x}$ to the class of $f^ i d^{i - 1}(z)$ in $(\eta _ fM)^ i/f(\eta _ fM)^ i$. This is well defined because if $fx' + d^{i - 1}(z) = 0$, then $f^ ix'$ is in $(\eta _ fM)^ i$ and consequently $f^ id^{i - 1}(z)$ is in $f(\eta _ fM)^ i$.

We omit the verification that the lower row in the displayed diagram is a short exact sequence of modules. It is immediately clear from these constructions that we have commutative diagrams

where the upper horizontal arrow is given by the identification of the summands $B^{i + 1}$ in source and target. In other words, we have found an acyclic subcomplex of $\eta _ fM^\bullet / f(\eta _ fM^\bullet ) = \eta _ fM^\bullet \otimes _ A A/fA$ and the quotient by this subcomplex is a complex whose terms $Z^ i/B^ i$ are the cohomology modules of the complex $\overline{M}^\bullet = M^\bullet \otimes _ A A/fA$.

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