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

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Commutative monoids are semiadditive and not additive

Statement refuted

Refuted claim: the category $C M o n$ of commutative monoids is additive.

It is only semiadditive.

Facts & Assumptions

Given: The category $C M o n$ and the monoid $N$ .

[L1]

A semiadditive category has finite biproducts (Semiadditive category).

[L2]

A semiadditive category is preadditive exactly when every morphism has an additive inverse (A semiadditive category is preadditive exactly when every morphism has an additive inverse).

[L3]

There is a zero-kernel, non-monic morphism in a merely semiadditive category (A zero kernel does not force monicity in a merely semiadditive category).

Counterexample

technique · direct

1.1

Finite products and finite coproducts of commutative monoids are both given by finite Cartesian products: for finitely many summands, every tuple automatically has finite support. So $C M o n$ has finite biproducts and is semiadditive by [L1].

2.1

The identity endomorphism of $N$ has no additive inverse under pointwise addition, since $f + g = 0$ in $End (N)$ would force $f (1) = 0$ . Therefore [L2] says $C M o n$ is not preadditive and hence not additive. The witness [L3] records the same failure through kernels and monicity.

L2L3step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Peter Freyd, Abelian Categories, Exercise 2A (standard reference, not scraped)