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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04
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The zero-th right derived functor of a left exact functor recovers the functor

Statement

Assume the Axiom of Dependent Choice.

Let I be a supplied injective resolution datum on a class D, and let F:A→B be an additive left exact functor between abelian categories. Then for every A∈D there is a canonical isomorphism RI0F(A)→ ∼ F(A), natural in A.

Facts & Assumptions

Given: An object A∈D.

[L1]

The chosen injective resolution of A is an exact coaugmented complex 0→A→I0(A)→I1(A)→⋯ (Injective resolutions in an abelian category).

[L2]

The zeroth cohomology object is the quotient of the kernel of d0:F(I0(A))→F(I1(A)) by the zero-th coboundary, which is 0 (Cohomology object of a cochain complex).

[L3]

Left exactness means that F preserves the kernel at the beginning of the displayed injective resolution (Left exact and right exact functors).

[L4]

The assignments A↦RI0F(A) are already functorial (Right derived functors relative to supplied data are additive functors).

Proof

technique · direct
1.1L1L3givenalgebra

By [L1], the morphism 0→A→I0(A)→I1(A) is exact. Applying F and using [L3] gives an exact sequence 0→F(A)→F(I0(A))→F(I1(A)). Therefore F(A)=ker⁡(F(I0(A))→F(I1(A))).

2.1L2step 1.1

By [L2], the zeroth cohomology of F(I(A)del) is that kernel, because the zero-th coboundary object is 0. Hence RI0F(A)=H0(F(I(A)del))≅F(A).

3.1L4step 2.1∎

Step 2.1 is natural in A, and [L4] records the functoriality of RI0F. Therefore the displayed isomorphism is natural.

Depends on

Used by

Dependency tree · two levels

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Sources