Lemma 13.4.11. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle.

1. If $h = 0$, then there exists a right inverse $s : Z \to Y$ to $g$.

2. For any right inverse $s : Z \to Y$ of $g$ the map $f \oplus s : X \oplus Z \to Y$ is an isomorphism.

3. For any objects $X', Z'$ of $\mathcal{D}$ the triangle $(X', X' \oplus Z', Z', (1, 0), (0, 1), 0)$ is distinguished.

Proof. To see (1) use that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Z) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, X[1])$ is exact by Lemma 13.4.2. By the same token, if $s$ is as in (2), then $h = 0$ and the sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Z) \to 0$

is split exact (split by $s : Z \to Y$). Hence by Yoneda's lemma we see that $X \oplus Z \to Y$ is an isomorphism. The last assertion follows from TR1 and Lemma 13.4.10. $\square$

There are also:

• 13 comment(s) on Section 13.4: Elementary results on triangulated categories

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).