Lemma 13.5.2. Let \mathcal{D} be a pre-triangulated category. Let S \subset \text{Arrows}(\mathcal{D}).
Proof. Assume S contains all identities and MS6 holds. Let f : X \to Y be an isomorphism of \mathcal{D}. Consider the diagram
\xymatrix{ 0 \ar[r] \ar[d]^1 & X \ar[r]_1 \ar[d]^1 & X \ar[r] \ar@{-->}[d] & 0[1] \ar[d]^{1[1]} \\ 0 \ar[r] & X \ar[r]^ f & Y \ar[r] & 0[1] }
The rows are distinguished triangles by Lemma 13.4.9. By MS6 we see that the dotted arrow exists and is in S, so f is in S.
Assume MS1, MS5, MS6. Suppose that f : X \to Y is a morphism of \mathcal{D} and t : X \to X' an element of S. Choose a distinguished triangle (X, Y, Z, f, g, h). Next, choose a distinguished triangle (X', Y', Z, f', g', t[1] \circ h) (here we use TR1 and TR2). By MS5, MS6 (and TR2 to rotate) we can find the dotted arrow in the commutative diagram
\xymatrix{ X \ar[r] \ar[d]^ t & Y \ar[r] \ar@{..>}[d]^{s'} & Z \ar[r] \ar[d]^1 & X[1] \ar[d]^{t[1]} \\ X' \ar[r] & Y' \ar[r] & Z \ar[r] & X'[1] }
with moreover s' \in S. This proves LMS2. The proof of RMS2 is dual. \square
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