Lemma 13.4.9. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent

1. $f$ is an isomorphism,

2. $(X, Y, 0, f, 0, 0)$ is a distinguished triangle, and

3. for any distinguished triangle $(X, Y, Z, f, g, h)$ we have $Z = 0$.

Proof. By TR1 the triangle $(X, X, 0, 1, 0, 0)$ is distinguished. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. By TR3 there is a map of distinguished triangles $(1, f, 0) : (X, X, 0) \to (X, Y, Z)$. If $f$ is an isomorphism, then $(1, f, 0)$ is an isomorphism of triangles by Lemma 13.4.3 and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism. $\square$

Comment #379 by Fan on

More precisely, (1) implies (2) is by TR1 applied to the isomorphism $(1, f, 0): (X, X, 0, 1, 0, 0) \to (X, Y, 0, f, 0, 0$.

(2) implies (3) is by Lemma 13.4.6, which shows that $(1, 1, 0): (X, Y, 0, f, 0, 0) \to (X, Y, Z, f, g, h)$ is an isomorphism.

(3) implies (1) is by Lemma 13.4.3 applied to the morphism $(1, f, 0): (X, X, 0, 1, 0, 0) \to (X, Y, 0, f, 0, 0$.

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