Lemma 13.4.5. Let $\mathcal{D}$ be a pre-triangulated category. Let

$(0, b, 0), (0, b', 0) : (X, Y, Z, f, g, h) \to (X, Y, Z, f, g, h)$

be endomorphisms of a distinguished triangle. Then $bb' = 0$.

Proof. Picture

$\xymatrix{ X \ar[r] \ar[d]^0 & Y \ar[r] \ar[d]^{b, b'} \ar@{..>}[ld]^\alpha & Z \ar[r] \ar[d]^0 \ar@{..>}[ld]^\beta & X[1] \ar[d]^0 \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }$

Applying Lemma 13.4.2 we find dotted arrows $\alpha$ and $\beta$ such that $b' = f \circ \alpha$ and $b = \beta \circ g$. Then $bb' = \beta \circ g \circ f \circ \alpha = 0$ as $g \circ f = 0$ by Lemma 13.4.1. $\square$

Comment #320 by arp on

Typo: I think the reference should be to Lemma 13.4.2 (tag 0149) not 13.4.3.

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

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