The Stacks project

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

\[ C(f)_\bullet [1] \longrightarrow C(g)_\bullet \]

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 (15.28.6.1)

\[ \left( \begin{matrix} g & 1 \\ 0 & -f \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2} \]

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. $\square$


Comments (5)

Comment #2776 by on

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 on

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

Comment #2960 by on

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

Comment #3086 by on

@#2960 Either you can prove the required homotopy does not exist by giving a counter example and then of course the argument is wrong or you can just write one down. I've now checked this two times by writing it out on paper and each time it is obvious what you have to do, so I am going to leave it as is.


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