PropositionStatement: AI-adaptedProof: 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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Morphisms into a homotopically injective complex need no roof

Statement

For a K-injective complex $I$ and any complex $X$ , $Q : {Hom}_{K} (X, I) \to {Hom}_{D} (X, I)$ is bijective. Moreover, if $s : I \to J$ is a quasi-isomorphism, its cone triangle is split in $K$ : $J ≅ I \oplus Cone (s)$ with $s$ corresponding to the inclusion.

Facts & Assumptions

Given: For a K-injective complex $I$ and any complex $X$ , $Q : {Hom}_{K} (X, I) \to {Hom}_{D} (X, I)$ is bijective. Moreover, if $s : I \to J$ is a quasi-isomorphism, its cone triangle is split in $K$ : $J ≅ I \oplus Cone (s)$ with $s$ corresponding to the inclusion.

[F1]

K-injectivity annihilates Hom from every acyclic complex into each shift of $I$ (Homotopically injective bounded below complex).

[F2]

The cone of a quasi-isomorphism is acyclic (A chain map is a quasi-isomorphism exactly when its cone is acyclic).

[F3]

Both representable Hom sequences of a distinguished triangle are exact (Long exact Hom sequences of a distinguished triangle).

[F4]

The derived category has both roof presentations and denominator detection of equality (Derived category of an abelian category).

Proof

1.1F1F2F3

For any quasi-isomorphism $s : U \to V$ , the cone and its shifts are acyclic. Apply ${Hom}_{K} (-, I)$ to its triangle: the two adjacent cone Hom groups vanish by K-injectivity. Thus precomposition by $s$ gives a bijection ${Hom}_{K} (V, I) \to {Hom}_{K} (U, I)$ . This includes all zero objects.

2.1F4step 1.1algebra

Given a left roof $X \overset{s}{\leftarrow} U \overset{f}{\to} I$ , the bijection gives a unique $b : X \to I$ with $b s = f$ in $K$ , so the roof equals $Q (b)$ . An equality $Q (b) = Q (c)$ is witnessed by $b s = c s$ for a denominator $s$ and the same bijection gives $b = c$ . Equivalently a right-roof denominator out of $I$ has a retraction and therefore also eliminates the roof.

2.2F1F3step 1.1algebra

For $s : I \to J$ the bijection gives $r : J \to I$ with $r s = 1_{I}$ in $K$ . Let $C = Cone (s)$ and let $p : C \to I [1]$ be its projection. K-injectivity gives $p = 0$ in $K$ . Hom exactness supplies $v : C \to J$ with $i v = 1_{C}$ , where $i : J \to C$ . Replace $v$ by $v - s r v$ , so also $r v = 0$ . Then $1_{J} - s r - v i$ is killed by $i$ and hence factors as $s u$ by Hom exactness. Applying $r$ gives $u = 0$ . Consequently $(s, v) : I \oplus C \to J$ and $(r, i)$ are inverse in $K$ .

3.1step 2.2algebra∎

The cone signs can also be checked on matrices. Choose a representative homotopy $r s - 1_{I} = d_{I} h + h d_{I}$ . The degree-minus-one map $H : C^{n} \to I [1]^{n - 1}$ given by $H (j, x) = r j + h x$ satisfies $d_{I [1]} H + H d_{C} = p$ . Thus the split connecting map is zero with the specified cone convention, rather than after an unrecorded change of sign.

Depends on

Used by

Dependency tree · two levels

21 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

13.18.3–13.18.8; W 10.4.8 for the equivalence (standard reference, not scraped)