Lemma 13.5.9. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Let $Z$ be an object of $\mathcal{D}$. The following are equivalent

$Q(Z) = 0$ in $S^{-1}\mathcal{D}$,

there exists $Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that $0 : Z \to Z'$ is an element of $S$,

there exists $Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that $0 : Z' \to Z$ is an element of $S$, and

there exists an object $Z'$ and a distinguished triangle $(X, Y, Z \oplus Z', f, g, h)$ such that $f \in S$.

If $S$ is saturated, then these are also equivalent to

the morphism $0 \to Z$ is an element of $S$,

the morphism $Z \to 0$ is an element of $S$,

there exists a distinguished triangle $(X, Y, Z, f, g, h)$ such that $f \in S$.

**Proof.**
The equivalence of (1), (2), and (3) is Homology, Lemma 12.8.3. If (2) holds, then $(Z'[-1], Z'[-1] \oplus Z, Z, (1, 0), (0, 1), 0)$ is a distinguished triangle (see Lemma 13.4.11) with “$0 \in S$”. By rotating we conclude that (4) holds. If $(X, Y, Z \oplus Z', f, g, h)$ is a distinguished triangle with $f \in S$ then $Q(f)$ is an isomorphism hence $Q(Z \oplus Z') = 0$ hence $Q(Z) = 0$. Thus (1) – (4) are all equivalent.

Next, assume that $S$ is saturated. Note that each of (5), (6), (7) implies one of the equivalent conditions (1) – (4). Suppose that $Q(Z) = 0$. Then $0 \to Z$ is a morphism of $\mathcal{D}$ which becomes an isomorphism in $S^{-1}\mathcal{D}$. According to Categories, Lemma 4.27.21 the fact that $S$ is saturated implies that $0 \to Z$ is in $S$. Hence (1) $\Rightarrow $ (5). Dually (1) $\Rightarrow $ (6). Finally, if $0 \to Z$ is in $S$, then the triangle $(0, Z, Z, 0, \text{id}_ Z, 0)$ is distinguished by TR1 and TR2 and is a triangle as in (7).
$\square$

## Comments (2)

Comment #8348 by Elías Guisado on

Comment #8956 by Stacks project on

There are also: