Lemma 13.4.1. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. Then $g \circ f = 0$, $h \circ g = 0$ and $f[1] \circ h = 0$.

**Proof.**
By TR1 we know $(X, X, 0, 1, 0, 0)$ is a distinguished triangle. Apply TR3 to

Of course the dotted arrow is the zero map. Hence the commutativity of the diagram implies that $g \circ f = 0$. For the other cases rotate the triangle, i.e., apply TR2. $\square$

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