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
\[ \xymatrix{ X \ar[r] \ar[d]^1 & X \ar[r] \ar[d]^ f & 0 \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{1[1]} \\ X \ar[r]^ f & Y \ar[r]^ g & Z \ar[r]^ h & X[1] } \]
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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