Lemma 13.5.2. Let $\mathcal{D}$ be a pre-triangulated category. Let $S \subset \text{Arrows}(\mathcal{D})$.

1. If $S$ contains all identities and MS6 holds (Definition 13.5.1), then every isomorphism of $\mathcal{D}$ is in $S$.

2. If MS1, MS5 (Categories, Definition 4.27.1) and MS6 hold, then MS2 holds.

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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