CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27

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A covariant hom functor on an additive category need not preserve cokernels

Statement refuted

Refuted claim: every covariant hom-functor on an additive category preserves cokernels.

Take the additive category $A b$ and the covariant hom-functor $A b (Z / 2, -)$ .

Facts & Assumptions

Given: The sequence $Z \overset{2}{\to} Z \to Z / 2$ in $A b$ .

[L1]

In a preadditive category, hom-functors take values in abelian groups (The hom-bifunctor of a preadditive category takes values in abelian groups).

[L2]

For modules, postcomposition gives the induced maps on Hom groups (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

Counterexample

technique · direct

1.1

The cokernel of multiplication by $2$ on $Z$ is $Z / 2$ . Applying $A b (Z / 2, -)$ gives $Hom (Z / 2, Z) \overset{(2)_{*}}{\to} Hom (Z / 2, Z) \to Hom (Z / 2, Z / 2)$ , with the induced maps described by [L2].

2.1

Every homomorphism $Z / 2 \to Z$ is zero, while $Hom (Z / 2, Z / 2) ≅ Z / 2$ . So the image sequence is $0 \to 0 \to Z / 2$ , whose first cokernel is $0$ , not $Z / 2$ .

L1step 1.1

3.1

Therefore the covariant hom-functor $A b (Z / 2, -)$ does not preserve this cokernel.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

A. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.14 and 3.15 (standard reference, not scraped)