TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Canonical truncations fit a distinguished triangle

Statement

Every short exact sequence $0 \to A \overset{i}{\to} B \overset{q}{\to} C \to 0$ of cochain complexes gives a natural distinguished triangle $A \to B \to C \to A [1]$ in $D (A)$ . In particular, for every integer $n$ there are canonical distinguished triangles

$τ^{\leq n} X \to X \to τ^{\geq n + 1} X \to (τ^{\leq n} X) [1],$

$τ^{\leq n} X \to τ^{\leq n + 1} X \to H^{n + 1} (X) [- n - 1] \to (τ^{\leq n} X) [1],$

$H^{n} (X) [- n] \to τ^{\geq n} X \to τ^{\geq n + 1} X \to H^{n} (X) [- n + 1] .$

Facts & Assumptions

Given: Every short exact sequence $0 \to A \overset{i}{\to} B \overset{q}{\to} C \to 0$ of cochain complexes gives a natural distinguished triangle $A \to B \to C \to A [1]$ in $D (A)$ . In particular, for every integer $n$ there are canonical distinguished triangles

$τ^{\leq n} X \to X \to τ^{\geq n + 1} X \to (τ^{\leq n} X) [1],$

$τ^{\leq n} X \to τ^{\leq n + 1} X \to H^{n + 1} (X) [- n - 1] \to (τ^{\leq n} X) [1],$

$H^{n} (X) [- n] \to τ^{\geq n} X \to τ^{\geq n + 1} X \to H^{n} (X) [- n + 1] .$

[F1]

Canonical truncations preserve the stated cohomology degrees and give natural truncation maps (Canonical truncation is a complex and has the claimed cohomology).

[F2]

The derived category is triangulated, its localization is exact, and images of cone triangles are distinguished (The derived category inherits a triangulated structure).

[F3]

A short exact sequence of complexes in an abelian category gives a long exact homology sequence (The long exact sequence in homology).

Proof

1.1F3algebra

Define $e : Cone (i) \to C$ by $e (b, a) = q (b)$ . It is a termwise epimorphic complex map. Its kernel identifies with $Cone (1_{A})$ via $(a^{'}, a) \mapsto (i (a^{'}), a)$ ; the homotopy $h (a^{'}, a) = (0, a^{'})$ contracts this kernel, including when $A = 0$ . The long exact sequence for kernel, cone and quotient makes $e$ a quasi-isomorphism.

2.1F2step 1.1algebra

The cone triangle is distinguished and $Q (e)$ is invertible. Transporting it gives the short-exact-sequence triangle with connecting map $Q (p) Q (e)^{- 1}$ , where $p (b, a) = a$ maps to $A [1]$ . A map of short exact sequences induces $(b, a) \mapsto (v b, u a)$ on cones, commuting with $e$ and $p$ ; this proves naturality with the stated signs.

3.1F1step 2.1algebra

Apply this construction to $0 \to τ^{\leq n} X \to X \to R \to 0$ . The quotient has $R^{n} = X^{n} / \ker d^{n} ≅ im d^{n}$ , $R^{i} = X^{i}$ for $i > n$ and zero below. The natural $R \to τ^{\geq n + 1} X$ is zero at $n$ , the quotient at $n + 1$ , and identity above. Its kernel is the two-term identity complex on $im d^{n}$ , so it is a quasi-isomorphism. This yields the first triangle.

4.1F1step 3.1algebra∎

Apply the first triangle to $τ^{\leq n + 1} X$ at cut $n$ ; its lower tail has just $H^{n + 1} (X)$ in degree $n + 1$ . Apply it to $τ^{\geq n} X$ at cut $n$ ; its upper head is just $H^{n} (X)$ in degree $n$ . The cohomology formulas and natural comparison maps identify the remaining truncations, giving the second and third triangles.

Depends on

Dependency tree · 2 levels

Sources

Remark 13.12.4 and its three triangles