Lemma 15.28.7. Let $R$ be a ring. Let $\varphi : E \to R$ be an $R$-module map. Let $f \in R$. Set $E' = E \oplus R$ and define $\varphi ' : E' \to R$ by $\varphi $ on $E$ and multiplication by $f$ on $R$. The complex $K_\bullet (\varphi ')$ is isomorphic to the cone of the map of complexes

**Proof.**
Denote $e_0 \in E'$ the element $1 \in R \subset R \oplus E$. By our definition of the cone above we see that

where in the last $=$ we map $(0, e_1 \wedge \ldots \wedge e_{n - 1})$ to $e_0 \wedge e_1 \wedge \ldots \wedge e_{n - 1}$ in $\wedge ^ n(E')$. A computation shows that this isomorphism is compatible with differentials. Namely, this is clear for elements of the first summand as $\varphi '|_ E = \varphi $ and $d_{C(f)}$ restricted to the first summand is just $d_{K_\bullet (\varphi )}$. On the other hand, if $e_1 \wedge \ldots \wedge e_{n - 1}$ is in the second summand, then

and on the other hand

which is the image of the result of the previous computation. $\square$

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## Comments (2)

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