# The Stacks Project

## Tag 062A

Lemma 15.26.9. 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 $$C(f)_\bullet[1] \longrightarrow C(g)_\bullet$$

Proof. We first prove this if $A_\bullet$ is the complex consisting of $R$ placed in degree $0$. In this case the map we use is $$\xymatrix{ 0 \ar[r] \ar[d] & 0 \ar[r] \ar[d] & R \ar[r]^f \ar[d]^1 & R \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & R \ar[r]^g & R \ar[r] & 0 \ar[r] & 0 }$$ The cone of this is the chain complex consisting of $R \oplus R$ placed in degrees $1$ and $0$ and differential (15.26.6.1) $$\left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2}$$ We leave it to the reader to show this this chain complex is homotopic to the complex $fg : R \to R$. In general we write $C(f)_\bullet$ and $C(g)_\bullet$ as the total complex of the double complexes $$(R \xrightarrow{f} R) \otimes_R A_\bullet \quad\text{and}\quad (R \xrightarrow{g} R) \otimes_R A_\bullet$$ and in this way we deduce the result from the special case discussed above. Some details omitted. $\square$

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5671–5681 (see updates for more information).

\begin{lemma}
\label{lemma-cone-squared}
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
$$C(f)_\bullet[1] \longrightarrow C(g)_\bullet$$
\end{lemma}

\begin{proof}
We first prove this if $A_\bullet$ is the complex consisting of $R$ placed
in degree $0$. In this case the map we use is
$$\xymatrix{ 0 \ar[r] \ar[d] & 0 \ar[r] \ar[d] & R \ar[r]^f \ar[d]^1 & R \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & R \ar[r]^g & R \ar[r] & 0 \ar[r] & 0 }$$
The cone of this is the chain complex consisting of $R \oplus R$ placed in
degrees $1$ and $0$ and differential (\ref{equation-differential-cone})
$$\left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2}$$
We leave it to the reader to show this this chain complex is
homotopic to the complex $fg : R \to R$. In general we
write $C(f)_\bullet$ and $C(g)_\bullet$
as the total complex of the double complexes
$$(R \xrightarrow{f} R) \otimes_R A_\bullet \quad\text{and}\quad (R \xrightarrow{g} R) \otimes_R A_\bullet$$
and in this way we deduce the result from the special case discussed above.
Some details omitted.
\end{proof}

Comment #2773 by Darij Grinberg (site) on August 19, 2017 a 2:21 am UTC

"this this".

Comment #2776 by Darij Grinberg (site) on August 19, 2017 a 2:32 am UTC

This said, I really wouldn't mind a more mundane proof by explicit description of the map and of the quasi-isomorphisms... I'm not sure if I can follow the proof above.

## Add a comment on tag 062A

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).