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Tag 062A

Chapter 15: More on Algebra > Section 15.26: The Koszul complex

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}

    Comments (4)

    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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