The Stacks project

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 (4). $\square$