Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Hom is left exact in each variable

Statement

Let A be an abelian category and X an object of A.

  1. If 0→A→B→C is exact, then 0→A(X,A)→A(X,B)→A(X,C) is exact in Ab.
  2. If A→B→C→0 is exact, then 0→A(C,X)→A(B,X)→A(A,X) is exact in Ab.

Thus Hom is left exact in each variable.

Facts & Assumptions

Given: An abelian category A and an object X of A.

[L1]

In an abelian category, exactness at the left end means that the first displayed map is a kernel of the second (Degenerate exactness criteria).

[L2]

Abelian categories have all finite limits, and representable functors preserve existing small limits (An abelian category has all finite limits and all finite colimits, Every covariantly representable functor to Set preserves all existing small limits).

[L3]

The covariant and contravariant Hom assignments are the representable functors A(X,−) and Aop(X,−) (The covariant and contravariant hom-assignments and the hom-bifunctor of a locally small category, The opposite of an abelian category is abelian).

[L4]

The target category of these Hom functors is Ab, which is abelian (Abelian groups form an abelian category).

Proof

technique · direct
1.1L1L2L3L4assume-hyp

Assume 0→A→B→C is exact. By [L1], the map A→B is a kernel of B→C. By [L2] and [L3], the representable functor A(X,−) preserves that kernel. Therefore 0→A(X,A)→A(X,B)→A(X,C) is exact in Ab by [L1] applied inside the abelian category [L4].

2.1L2L3L4step 1.1

Passing to the opposite category, the exact sequence A→B→C→0 becomes a left-exact sequence in Aop. Applying step 1.1 there to the representable functor Aop(X,−)=A(−,X) gives 0→A(C,X)→A(B,X)→A(A,X) exact in Ab.

3.1step 1.1step 2.1∎

Hence Hom is left exact in each variable.

Depends on

Used by

Dependency tree · two levels

29 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