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
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: