Lemma 13.4.18. Let $F : \mathcal{D} \to \mathcal{D}'$ be a fully faithful exact functor of pre-triangulated categories. Then a triangle $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ is distinguished if and only if $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished in $\mathcal{D}'$.

**Proof.**
The “only if” part is clear. Assume $(F(X), F(Y), F(Z))$ is distinguished in $\mathcal{D}'$. Pick a distinguished triangle $(X, Y, Z', f, g', h')$ in $\mathcal{D}$. By Lemma 13.4.7 there exists an isomorphism of triangles

Since $F$ is fully faithful, there exists a morphism $c : Z \to Z'$ such that $F(c) = c'$. Then $(1, 1, c)$ is an isomorphism between $(X, Y, Z)$ and $(X, Y, Z')$. Hence $(X, Y, Z)$ is distinguished by TR1. $\square$

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