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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