Lemma 13.4.6. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. If

\[ \xymatrix{ Z \ar[r]_ h \ar[d]_ c & X[1] \ar[d]^{a[1]} \\ Z \ar[r]^ h & X[1] } \]

is commutative and $a^2 = a$, $c^2 = c$, then there exists a morphism $b : Y \to Y$ with $b^2 = b$ such that $(a, b, c)$ is an endomorphism of the triangle $(X, Y, Z, f, g, h)$.

**Proof.**
By TR3 there exists a morphism $b'$ such that $(a, b', c)$ is an endomorphism of $(X, Y, Z, f, g, h)$. Then $(0, (b')^2 - b', 0)$ is also an endomorphism. By Lemma 13.4.5 we see that $(b')^2 - b'$ has square zero. Set $b = b' - (2b' - 1)((b')^2 - b') = 3(b')^2 - 2(b')^3$. A computation shows that $(a, b, c)$ is an endomorphism and that $b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0$.
$\square$

## Comments (1)

Comment #321 by arp on

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