CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Hom is not exact

Statement refuted

For every object $X$ in an abelian category, the functor $A (-, X)$ is exact.

Facts & Assumptions

Given: In $A b$ , the short exact sequence $0 \to Z \overset{\times 2}{\to} Z \to Z / 2 \to 0.$

[L1]

The category $A b$ is abelian (Abelian groups form an abelian category).

[L2]

Hom is left exact, but no right exactness was asserted (Hom is left exact in each variable).

Counterexample

technique · direct

1.1L1L2given

Apply $A b (-, Z)$ to the given short exact sequence. This yields $0 \to Hom (Z / 2, Z) \to Hom (Z, Z) \overset{(\times 2)^{*}}{\to} Hom (Z, Z)$ , which is left exact by [L2].

2.1step 1.1algebra

The group $Hom (Z, Z)$ is $Z$ , and precomposition with $\times 2$ is multiplication by $2$ on that copy of $Z$ . This map is not surjective, so the Hom sequence is not exact at the right-hand term.

3.1step 2.1∎

Therefore the contravariant Hom functor need not be exact.

Depends on

Used by

Nothing in the library uses this result yet.

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, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)