CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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An exact functor preserves quasi-isomorphisms and reflects them when it is conservative

Statement

Let $F : A \to B$ be an exact functor between abelian categories. Then $F$ preserves quasi-isomorphisms. If $F$ is also conservative, then it reflects quasi-isomorphisms.

Facts & Assumptions

Given: An exact functor $F : A \to B$ and a chain map $f : C_{∙} \to D_{∙}$ .

[L1]

A quasi-isomorphism is a chain map inducing isomorphisms on all homology objects (Quasi-isomorphism).

[L2]

Exact functors commute with homology (An exact functor commutes with homology).

[L3]

A conservative functor reflects isomorphisms (Conservative functor).

Proof

technique · direct

1.1L1L2

If $f$ is a quasi-isomorphism, then [L1] says each $H_{n} (f)$ is an isomorphism. By [L2], the induced map on the homology of $F (f)$ is identified with $F (H_{n} (f))$ , hence is an isomorphism. Therefore $F (f)$ is a quasi-isomorphism by [L1].

2.1L1L2L3∎

Now assume $F$ is conservative and $F (f)$ is a quasi-isomorphism. Then [L1] makes each induced map on $H_{n} (F (f))$ an isomorphism. By [L2], those are the morphisms $F (H_{n} (f))$ . Since $F$ reflects isomorphisms by [L3], each $H_{n} (f)$ is an isomorphism, so $f$ is a quasi-isomorphism by [L1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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, Section 12.7, Lemma 12.7.2 (standard reference, not scraped)
Romyar Sharifi, Homological Algebra, Lemma 2.5.2 (standard reference, not scraped)