# 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 complex $C(f)_\bullet$ is the complex $$\ldots \to 0 \to R \xrightarrow{f} R \to 0 \to \ldots$$ with $R$ placed in (homological) degrees $1$ and $0$. The map of complexes 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 $R^{\oplus 2}$ placed in degrees $1$ and $0$ and differential (15.26.6.1) $$\left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2}$$ To see this chain complex is homotopic to $C(fg)_\bullet$, i.e., to $R \xrightarrow{fg} R$, consider the maps of complexes $$\xymatrix{ R \ar[d]_{(1, -g)} \ar[r]_{fg} & R \ar[d]^{(0, 1)} \\ R^{\oplus 2} \ar[r] & R^{\oplus 2} } \quad\quad \xymatrix{ R^{\oplus 2} \ar[d]_{(1, 0)} \ar[r] & R^{\oplus 2} \ar[d]^{(f, 1)} \\ R \ar[r]^{fg} & R }$$ 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 $$(R \xrightarrow{f} R) \otimes_R A_\bullet \quad\text{and}\quad (R \xrightarrow{g} R) \otimes_R A_\bullet$$ 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 5774–5784 (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 complex $C(f)_\bullet$ is the
complex
$$\ldots \to 0 \to R \xrightarrow{f} R \to 0 \to \ldots$$
with $R$ placed in (homological) degrees $1$ and $0$. The map
of complexes 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 $R^{\oplus 2}$ placed in
degrees $1$ and $0$ and differential (\ref{equation-differential-cone})
$$\left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2}$$
To see this chain complex is homotopic to
$C(fg)_\bullet$, i.e., to $R \xrightarrow{fg} R$,
consider the maps of complexes
$$\xymatrix{ R \ar[d]_{(1, -g)} \ar[r]_{fg} & R \ar[d]^{(0, 1)} \\ R^{\oplus 2} \ar[r] & R^{\oplus 2} } \quad\quad \xymatrix{ R^{\oplus 2} \ar[d]_{(1, 0)} \ar[r] & R^{\oplus 2} \ar[d]^{(f, 1)} \\ R \ar[r]^{fg} & R }$$
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.

\medskip\noindent
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
$$(R \xrightarrow{f} R) \otimes_R A_\bullet \quad\text{and}\quad (R \xrightarrow{g} R) \otimes_R A_\bullet$$
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.

Comment #2882 by Johan (site) on October 6, 2017 a 7:05 pm UTC

Thanks for "this this". Also added a few more lines. See here.

Comment #2960 by Darij Grinberg (site) on October 15, 2017 a 7:42 am UTC

This looks a lot clearer now, but I still can't find "the required homotopy" in the R -> R case...

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