The Stacks project

Proof. If we have a second morphism of triangles $(a, b', c)$ then $(0, b - b', 0)$ is a morphism of triangles. Hence we have to show: the only morphism $b : Y \to Y'$ such that $X \to Y \to Y'$ and $Y \to Y' \to Z'$ are zero is $0$. We will use Lemma 13.4.2 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition $g' \circ b : Y \to Y' \to Z'$ is zero, then $b$ lifts to a morphism $Y \to X'$ which has to be zero. This proves (1).

The proof of (2) and (4) are dual to this argument.

Assume (5). Consider the diagram

We may choose $\epsilon $ such that $b = \epsilon \circ g$. Then $g' \circ \epsilon \circ g = 0$ which implies that $g' \circ \epsilon = \delta \circ h$ for some $\delta \in \mathop{\mathrm{Hom}}\nolimits (X[1], Z')$. Since $\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = 0$ we conclude that $g' \circ \epsilon = 0$. Hence $\epsilon = f' \circ \gamma $ for some $\gamma \in \mathop{\mathrm{Hom}}\nolimits (Z, X')$. Since $\mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$ we conclude that $\epsilon = 0$ and hence $b = 0$ as desired. $\square$