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 13.16.4. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{I} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ be a subset containing $0$ such that every object of $\mathcal{A}$ is a subobject of an element of $\mathcal{I}$. Let $a \in \mathbf{Z}$.

  1. Given $K^\bullet $ with $K^ n = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet $ with $K^ n \to I^ n$ injective and $I^ n \in \mathcal{I}$ for all $n$ and $I^ n = 0$ for $n < a$,

  2. Given $K^\bullet $ with $H^ n(K^\bullet ) = 0$ for $n < a$ there exists a quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^ n \in \mathcal{I}$ and $I^ n = 0$ for $n < a$.

Proof. Proof of part (1). Consider the following induction hypothesis $IH_ n$: There are $I^ j \in \mathcal{I}$, $j \leq n$ almost all zero, maps $d^ j : I^ j \to I^{j + 1}$ for $j < n$ and injective maps $\alpha ^ j : K^ j \to I^ j$ for $j \leq n$ such that the diagram

\[ \xymatrix{ \ldots \ar[r] & K^{n - 1} \ar[d]^\alpha \ar[r] & K^ n \ar[d]^\alpha \ar[r] & K^{n + 1} \ar[r] & \ldots \\ \ldots \ar[r] & I^{n - 1} \ar[r] & I^ n & & } \]

is commutative, such that $d^ j \circ d^{j - 1} = 0$ for $j < n$ and such that $\alpha $ induces isomorphisms $H^ j(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ j)/\mathop{\mathrm{Im}}(d^{j - 1})$ for $j < n$. Note that this implies
\begin{equation} \label{derived-equation-new-star} \alpha (\mathop{\mathrm{Im}}(d_ K^{n - 1})) \subset \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)) \cap \mathop{\mathrm{Im}}(d^{n - 1}) \subset \alpha (K^ n) \cap \mathop{\mathrm{Im}}(d^{n - 1}). \end{equation}

If these inclusions are not equalities, then choose an injection

\[ I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1}) \longrightarrow I \]

with $I \in \mathcal{I}$. Denote $\alpha ' : K^ n \to I$ the map obtained by composing $\alpha \oplus 1 : K^ n \to I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ with the displayed injection. Denote $d' : I^{n - 1} \to I$ the composition $I^{n - 1} \to I^ n \to I$ of $d^{n - 1}$ by the inclusion of the first summand. Then $\alpha '(K^ n) \cap \mathop{\mathrm{Im}}(d') = \alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$ simply because the intersection of $\alpha '(K^ n)$ with the first summand of $I^ n \oplus K^ n/\mathop{\mathrm{Im}}(d_ K^{n - 1})$ is equal to $\alpha '(\mathop{\mathrm{Im}}(d_ K^{n - 1}))$. Hence, after replacing $I^ n$ by $I$, $\alpha $ by $\alpha '$ and $d^{n - 1}$ by $d'$ we may assume that we have equality in Equation ( Once this is the case consider the solid diagram

\[ \xymatrix{ K^ n/\mathop{\mathrm{Ker}}(d_ K^ n) \ar[r] \ar[d] & K^{n + 1} \ar@{..>}[d] \\ I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n))) \ar@{..>}[r] & M } \]

The horizontal arrow is injective by fiat and the vertical arrow is injective as we have equality in ( Hence the push-out $M$ of this diagram contains both $K^{n + 1}$ and $I^ n/(\mathop{\mathrm{Im}}(d^{n - 1}) + \alpha (\mathop{\mathrm{Ker}}(d_ K^ n)))$ as subobjects. Choose an injection $M \to I^{n + 1}$ with $I^{n + 1} \in \mathcal{I}$. By construction we get $d^ n : I^ n \to I^{n + 1}$ and an injective map $\alpha ^{n + 1} : K^{n + 1} \to I^{n + 1}$. The equality in Equation ( and the construction of $d^ n$ guarantee that $\alpha : H^ n(K^\bullet ) \to \mathop{\mathrm{Ker}}(d^ n)/\mathop{\mathrm{Im}}(d^{n - 1})$ is an isomorphism. In other words $IH_{n + 1}$ holds.

We finish the proof of by the following observations. First we note that $IH_ n$ is true for $n = a$ since we can just take $I^ j = 0$ for $j < a$ and $K^ a \to I^ a$ an injection of $K^ a$ into an element of $\mathcal{I}$. Next, we note that in the proof of $IH_ n \Rightarrow IH_{n + 1}$ we only modified the object $I^ n$, the map $d^{n - 1}$ and the map $\alpha ^ n$. Hence we see that proceeding by induction we produce a complex $I^\bullet $ with $I^ n = 0$ for $n < a$ consisting of objects from $\mathcal{I}$, and a termwise injective quasi-isomorphism $\alpha : K^\bullet \to I^\bullet $ as desired.

Proof of part (2). The assumption implies that the morphism $K^\bullet \to \tau _{\geq a}K^\bullet $ (Homology, Section 12.14) is a quasi-isomorphism. Apply part (1) to find $\tau _{\geq a}K^\bullet \to I^\bullet $. The composition $K^\bullet \to I^\bullet $ is the desired quasi-isomorphism. $\square$

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