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.

## 13.42 Essentially constant systems

Some preliminary lemmas on essentially constant systems in triangulated categories.

**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.
$\square$

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

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

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

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

Next, choose a morphism of distingished triangles

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

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. $\square$

Lemma 13.42.3. Let $\mathcal{A}$ be an abelian category. Let $A_ n$ be an inverse system of objects of $D(\mathcal{A})$. Assume

there exist integers $a \leq b$ such that $H^ i(A_ n) = 0$ for $i \not\in [a, b]$, and

the inverse systems $H^ i(A_ n)$ of $\mathcal{A}$ are essentially constant for all $i \in \mathbf{Z}$.

Then $A_ n$ is an essentially constant system of $D(\mathcal{A})$ whose value $A$ satisfies that $H^ i(A)$ is the value of the constant system $H^ i(A_ n)$ for each $i \in \mathbf{Z}$.

**Proof.**
By Remark 13.12.4 we obtain an inverse system of distinguished triangles

Of course we have $\tau _{\leq a}A_ n = H^ a(A_ n)[-a]$ in $D(\mathcal{A})$. Thus by assumption these form an essentially constant system. By induction on $b - a$ we find that the inverse system $\tau _{\geq a + 1}A_ n$ is essentially constant, say with value $A'$. By Lemma 13.42.2 we find that $A_ n$ is an essentially constant system. We omit the proof of the statement on cohomologies (hint: use the final part of Lemma 13.42.2). $\square$

Lemma 13.42.4. Let $\mathcal{D}$ be a triangulated category. Let

be an inverse system of distinguished triangles. If the system $C_ n$ is pro-zero (essentially constant with value $0$), then the maps $A_ n \to B_ n$ determine a pro-isomorphism between the pro-object $(A_ n)$ and the pro-object $(B_ n)$.

**Proof.**
For any object $X$ of $\mathcal{D}$ consider the exact sequence

Exactness follows from Lemma 13.4.2 combined with Algebra, Lemma 10.8.8. By assumption the first and last term are zero. Hence the map $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (B_ n, X) \to \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (A_ n, X)$ is an isomorphism for all $X$. The lemma follows from this and Categories, Remark 4.22.7. $\square$

Lemma 13.42.5. Let $\mathcal{A}$ be an abelian category.

be an inverse system of maps of $D(\mathcal{A})$. Assume

there exist integers $a \leq b$ such that $H^ i(A_ n) = 0$ and $H^ i(B_ n) = 0$ for $i \not\in [a, b]$, and

the inverse system of maps $H^ i(A_ n) \to H^ i(B_ n)$ of $\mathcal{A}$ define an isomorphism of pro-objects of $\mathcal{A}$ for all $i \in \mathbf{Z}$.

Then the maps $A_ n \to B_ n$ determine a pro-isomorphism between the pro-object $(A_ n)$ and the pro-object $(B_ n)$.

**Proof.**
We can inductively extend the maps $A_ n \to B_ n$ to an inverse system of distinguished triangles $A_ n \to B_ n \to C_ n \to A_ n[1]$ by axiom TR3. By Lemma 13.42.4 it suffices to prove that $C_ n$ is pro-zero. By Lemma 13.42.3 it suffices to show that $H^ p(C_ n)$ is pro-zero for each $p$. This follows from assumption (2) and the long exact sequences

Namely, for every $n$ we can find an $m > n$ such that $\mathop{\mathrm{Im}}(\beta _ m)$ maps to zero in $H^ p(C_ n)$ because we may choose $m$ such that $H^ p(B_ m) \to H^ p(B_ n)$ factors through $\alpha _ n : H^ p(A_ n) \to H^ p(B_ n)$. For a similar reason we may then choose $k > m$ such that $\mathop{\mathrm{Im}}(\delta _ k)$ maps to zero in $H^{p + 1}(A_ m)$. Then $H^ p(C_ k) \to H^ p(C_ n)$ is zero because $H^ p(C_ k) \to H^ p(C_ m)$ maps into $\mathop{\mathrm{Ker}}(\delta _ m)$ and $H^ p(C_ m) \to H^ p(C_ n)$ annihilates $\mathop{\mathrm{Ker}}(\delta _ m) = \mathop{\mathrm{Im}}(\beta _ m)$. $\square$

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