The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 13.9.8. Let $\mathcal{A}$ be an additive category. Let $\alpha : K^\bullet \to L^\bullet $ be a morphism of complexes of $\mathcal{A}$. There exists a factorization

\[ \xymatrix{ K^\bullet \ar[r]^ i \ar@/_1pc/[rr]_\alpha & \tilde K^\bullet \ar[r]^{\tilde\alpha } & L^\bullet } \]

such that

  1. $\tilde\alpha $ is a termwise split surjection (see Definition 13.9.4),

  2. there is a map of complexes $s : \tilde K^\bullet \to K^\bullet $ such that $s \circ i = \text{id}_{K^\bullet }$ and such that $i \circ s$ is homotopic to $\text{id}_{\tilde K^\bullet }$.

Moreover, if both $K^\bullet $ and $L^\bullet $ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so is $\tilde K^\bullet $.

Proof. Dual to Lemma 13.9.6. Take

\[ \tilde K^ n = K^ n \oplus L^{n - 1} \oplus L^ n \]

and we define

\[ d^ n_{\tilde K} = \left( \begin{matrix} d^ n_ K & 0 & 0 \\ 0 & - d^{n - 1}_ L & \text{id}_{L^ n} \\ 0 & 0 & d^ n_ L \end{matrix} \right) \]

in other words $\tilde K^\bullet = K^\bullet \oplus C(1_{L^\bullet [-1]})$. Moreover, we set

\[ \tilde\alpha = \left( \begin{matrix} \alpha & 0 & \text{id}_{L^ n} \end{matrix} \right) \]

which is clearly a split surjection. It is also clear that it defines a morphism of complexes. We define

\[ i = \left( \begin{matrix} \text{id}_{K^ n} \\ 0 \\ 0 \end{matrix} \right) \]

so that clearly $\tilde\alpha \circ i = \alpha $. We set

\[ s = \left( \begin{matrix} \text{id}_{K^ n} & 0 & 0 \end{matrix} \right) \]

so that $s \circ i = \text{id}_{K^\bullet }$. Finally, let $h^ n : \tilde K^ n \to \tilde K^{n - 1}$ be the map which maps the summand $L^{n - 1}$ of $\tilde K^ n$ via the identity morphism to the summand $L^{n - 1}$ of $\tilde K^{n - 1}$. Then it is a trivial matter to prove that

\[ \text{id}_{\tilde K^\bullet } - i \circ s = d \circ h + h \circ d \]

which finishes the proof of the lemma. $\square$


Comments (1)

Comment #293 by arp on

Hmm, I don't think is a morphism of complexes.

However everything seems OK in the corresponding Lemma 13.8.6 for split injections.


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