False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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FALSE: Every algebra for a monad is free

Statement

False claim: every algebra for every monad is a free algebra.

The free-monoid monad has a two-element algebra that is not free.

Facts & Assumptions

Given: The free-monoid monad on $S e t$ .

[L1]

Its algebras are monoids, while its free algebras are free word monoids (The free-monoid monad has monoids as its Eilenberg–Moore algebras; Free algebra for a monad).

[L2]

A monoid has associative multiplication and a two-sided identity (Semigroup and monoid).

Refutation

technique · direct

1.1

By [L1], it suffices to exhibit a monoid that is not isomorphic to a free word monoid.

1.2

Let $M = {1, e}$ with $1$ an identity and $e^{2} = e$ . Its multiplication is associative by inspection, so it is a monoid and hence an algebra for the free-monoid monad.

L1L2

2.1

A free word monoid on the empty set has one element. A free word monoid on a nonempty set contains the distinct words $[], [x], [x, x], \dots$ for any chosen generator $x$ , so it is infinite. Therefore no free word monoid has two elements, and $M$ is not free.

L1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Riehl, Category Theory in Context, 2nd ed., Definitions 5.2.4 and 5.2.8 (standard reference, not scraped)