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 13.9.6. 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

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

such that

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

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

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 L^\bullet $.

Proof. We set

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

and we define

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

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

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

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

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

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

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

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

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

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


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