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

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

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

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.
$\square$

## Comments (2)

Comment #9835 by Elías Guisado on

Comment #9836 by Elías Guisado on

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