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 12.8.3. Let $\mathcal{C}$ be an additive category. Let $S$ be a multiplicative system. Let $X$ be an object of $\mathcal{C}$. The following are equivalent

  1. $Q(X) = 0$ in $S^{-1}\mathcal{C}$,

  2. there exists $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that $0 : X \to Y$ is an element of $S$, and

  3. there exists $Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that $0 : Z \to X$ is an element of $S$.

Proof. If (2) holds we see that $0 = Q(0) : Q(X) \to Q(Y)$ is an isomorphism. In the additive category $S^{-1}\mathcal{C}$ this implies that $Q(X) = 0$. Hence (2) $\Rightarrow $ (1). Similarly, (3) $\Rightarrow $ (1). Suppose that $Q(X) = 0$. This implies that the morphism $f : 0 \to X$ is transformed into an isomorphism in $S^{-1}\mathcal{C}$. Hence by Categories, Lemma 4.26.21 there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves (1) $\Rightarrow $ (3). Similarly, (1) $\Rightarrow $ (2). $\square$


Comments (2)

Comment #332 by arp on

Typo: In the statement of the Lemma, it shouldn't say "compatible with the triangulated structure."

Comment #468 by Nuno on

Also, should be: Let be an object of .


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