Lemma 13.42.2. Let $\mathcal{D}$ be a triangulated category. Let

$A_ n \to B_ n \to C_ n \to A_ n[1]$

be an inverse system of distinguished triangles in $\mathcal{D}$. If $(A_ n)$ and $(C_ n)$ are essentially constant, then $(B_ n)$ is essentially constant and their values fit into a distinguished triangle $A \to B \to C \to A[1]$ such that for some $n \geq 1$ there is a map

$\xymatrix{ A_ n \ar[d] \ar[r] & B_ n \ar[d] \ar[r] & C_ n \ar[d] \ar[r] & A_ n[1] \ar[d] \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] }$

of distinguished triangles which induces an isomorphism $\mathop{\mathrm{lim}}\nolimits _{n' \geq n} A_{n'} \to A$ and similarly for $B$ and $C$.

Proof. After renumbering we may assume that $A_ n = A \oplus A'_ n$ and $C_ n = C \oplus C'_ n$ for inverse systems $(A'_ n)$ and $(C'_ n)$ which are essentially zero, see Lemma 13.42.1. In particular, the morphism

$C \oplus C'_ n \to (A \oplus A'_ n)[1]$

maps the summand $C$ into the summand $A[1]$ for all $n$ by a map $\delta : C \to A[1]$ which is independent of $n$. Choose a distinguished triangle

$A \to B \to C \xrightarrow {\delta } A[1]$

Next, choose a morphism of distingished triangles

$(A_1 \to B_1 \to C_1 \to A_1[1]) \to (A \to B \to C \to A[1])$

which is possible by TR3. For any object $D$ of $\mathcal{D}$ this induces a commutative diagram

$\xymatrix{ \ldots \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(C, D) \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(B, D) \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(A, D) \ar[r] \ar[d] & \ldots \\ \ldots \ar[r] & \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(C_ n, D) \ar[r] & \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(B_ n, D) \ar[r] & \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(A_ n, D) \ar[r] & \ldots }$

The left and right vertical arrows are isomorphisms and so are the ones to the left and right of those. Thus by the 5-lemma we conclude that the middle arrow is an isomorphism. It follows that $(B_ n)$ is isomorphic to the constant inverse system with value $B$ by the discussion in Categories, Remark 4.22.7. Since this is equivalent to $(B_ n)$ being essentially constant with value $B$ by Categories, Remark 4.22.5 the proof is complete. $\square$

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