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 4.27.21 there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves (1) $\Rightarrow$ (3). Similarly, (1) $\Rightarrow$ (2). $\square$

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 $X$ be an object of $\mathcal{C}$.

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