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 distinguished 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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