Lemma 13.42.1. Let $\mathcal{D}$ be a triangulated category. Let $(A_ i)$ be an inverse system in $\mathcal{D}$. Then $(A_ i)$ is essentially constant (see Categories, Definition 4.22.1) if and only if there exists an $i$ and for all $j \geq i$ a direct sum decomposition $A_ j = A \oplus Z_ j$ such that (a) the maps $A_{j'} \to A_ j$ are compatible with the direct sum decompositions and identity on $A$, (b) for all $j \geq i$ there exists some $j' \geq j$ such that $Z_{j'} \to Z_ j$ is zero.

**Proof.**
Assume $(A_ i)$ is essentially constant with value $A$. Then $A = \mathop{\mathrm{lim}}\nolimits A_ i$ and there exists an $i$ and a morphism $A_ i \to A$ such that (1) the composition $A \to A_ i \to A$ is the identity on $A$ and (2) for all $j \geq i$ there exists a $j' \geq j$ such that $A_{j'} \to A_ j$ factors as $A_{j'} \to A_ i \to A \to A_ j$. From (1) we conclude that for $j \geq i$ the maps $A \to A_ j$ and $A_ j \to A_ i \to A$ compose to the identity on $A$. It follows that $A_ j \to A$ has a kernel $Z_ j$ and that the map $A \oplus Z_ j \to A_ j$ is an isomorphism, see Lemmas 13.4.12 and 13.4.11. These direct sum decompositions clearly satisfy (a). From (2) we conclude that for all $j$ there is a $j' \geq j$ such that $Z_{j'} \to Z_ j$ is zero, so (b) holds. Proof of the converse is omitted.
$\square$

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