PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A distinguished triangle with zero first map is split

Statement

If $X \overset{0}{\to} Y \overset{g}{\to} Z \overset{h}{\to} X [1]$ is distinguished, then it is isomorphic to the right rotation of a canonical split triangle, namely $X \overset{0}{\to} Y \overset{(\binom{1}{0})}{\to} Y \oplus X [1] \overset{(0 1)}{\to} X [1] .$

Facts & Assumptions

Given: A distinguished triangle whose first map is zero.

Proof

1.1

In the left rotation $Y \overset{g}{\to} Z \overset{h}{\to} X [1] \to Y [1]$ , the final arrow is zero. Exactness of $T (X [1], -)$ therefore gives $s : X [1] \to Z$ with $h s = 1_{X [1]}$ .

given

2.1

Exactness of $T (Z, -)$ factors $1_{Z} - s h$ as $g t$ for some $t : Z \to Y$ ; exactness of $T (Y, -)$ and $f = 0$ give $t g = 1_{Y}$ . Put $t^{'} = t - t s h .$ Since consecutive triangle maps compose to zero, $h g = 0$ ; since $h s = 1$ , one obtains $t^{'} g = 1$ , $g t^{'} = 1 - s h$ , and $t^{'} s = 0$ .

step 1.1given

3.1

The identities in step 2.1 show that $(t^{'}, h) : Z \to Y \oplus X [1]$ and $(g, s) : Y \oplus X [1] \to Z$ are inverse. They intertwine all three displayed maps, so they give the claimed isomorphism of triangles.

step 2.1given∎

Depends on

Used by

A distinguished triangle is split up to rotation exactly when one map vanishes Proposition

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Lemma 13.4.11 (standard reference, not scraped)