Lemma 15.28.9. Let $R$ be a ring. Let $A_\bullet $ be a complex of $R$-modules. Let $f, g \in R$. Let $C(f)_\bullet $ be the cone of $f : A_\bullet \to A_\bullet $. Define similarly $C(g)_\bullet $ and $C(fg)_\bullet $. Then $C(fg)_\bullet $ is homotopy equivalent to the cone of a map

**Proof.**
We first prove this if $A_\bullet $ is the complex consisting of $R$ placed in degree $0$. In this case the complex $C(f)_\bullet $ is the complex

with $R$ placed in (homological) degrees $1$ and $0$. The map of complexes we use is

The cone of this is the chain complex consisting of $R^{\oplus 2}$ placed in degrees $1$ and $0$ and differential (15.28.6.1)

To see this chain complex is homotopic to $C(fg)_\bullet $, i.e., to $R \xrightarrow {fg} R$, consider the maps of complexes

with obvious notation. The composition of these two maps in one direction is the identity on $C(fg)_\bullet $, but in the other direction it isn't the identity. We omit writing out the required homotopy.

To see the result holds in general, we use that we have a functor $K_\bullet \mapsto \text{Tot}(A_\bullet \otimes _ R K_\bullet )$ on the category of complexes which is compatible with homotopies and cones. Then we write $C(f)_\bullet $ and $C(g)_\bullet $ as the total complex of the double complexes

and in this way we deduce the result from the special case discussed above. Some details omitted. $\square$

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