ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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.

The Kleisli adjunction for the maybe monad is monadic but not strictly monadic

Example

For the maybe monad $T (X) = X ⨿ {⊥}$ on $S e t$ , the comparison from its Kleisli category to its Eilenberg–Moore category is an equivalence but not an isomorphism under the fixed concrete coproduct encoding. Hence the Kleisli right adjoint is monadic but not strictly monadic.

Facts & Assumptions

Given: The maybe monad with unit the inclusion and multiplication collapsing either occurrence of $⊥$ .

[L1]

A Kleisli arrow $X \to Y$ is an ordinary map $X \to T (Y)$ (Kleisli category of a monad).

[L2]

An equivalence consists of quasi-inverse functors and natural isomorphisms between their composites and the identity functors (Equivalence, quasi-inverse, and adjoint equivalence of categories).

[L3]

An isomorphism of categories is bijective on objects and morphisms (A functor is an isomorphism of categories exactly when its object and morphism maps are bijective).

[L4]

A right adjoint with induced monad $T$ is monadic when its comparison functor $K$ is an equivalence of categories, and strictly monadic when $K$ is an isomorphism of categories (Monadic and strictly monadic functors).

[L5]

The comparison functor is the unique $K$ with $U^{T} K = U$ , $K F = F^{T}$ and $K (ε_{d}) = ε_{K d}^{T}$ , namely $K (d) = (U d, U ε_{d})$ and $K (h) = U (h)$ (The comparison functor to the Eilenberg–Moore category exists and is unique).

[L6]

For a monad $(T, η, μ)$ on $C$ there is an adjunction $F_{T} : C ⇄ C_{T} : U_{T}$ whose induced monad is $(T, η, μ)$ on the nose (The Kleisli adjunction induces the given monad).

[L7]

For a monad $T$ on $C$ the canonical comparison $M : C_{T} \to C^{T}$ sends $A$ to the free algebra $(T A, μ_{A})$ and is fully faithful, and its strict image is exactly the full subcategory of free $T$ -algebras (The comparison from the Kleisli category is fully faithful with image the free algebras).

Verification

technique · direct

1.1

A map $X \to Y ⨿ {⊥}$ is a partial function from $X$ to $Y$ , with $⊥$ marking undefined values, so [L1] identifies the Kleisli category with sets and partial functions.

L1construct

1.2

A maybe-monad algebra chooses a point $y_{0}$ as the value of $⊥$ and fixes every $y \in Y$ by the unit law; the multiplication law adds no further data. Algebra homomorphisms are precisely basepoint-preserving maps, so the Eilenberg–Moore category is the category of pointed sets.

givenconstruct

2.1

By [L6] the Kleisli adjunction induces the maybe monad on the nose, so [L5] applies to it and its comparison is the canonical one of [L7]. That comparison sends a Kleisli object $X$ to the free algebra $(T (X), μ_{X})$ , which step 1.2 identifies with the free pointed set $X ⨿ {⊥}$ based at $⊥$ , and sends a Kleisli arrow to its underlying function. A quasi-inverse sends a pointed set $(Y, y_{0})$ to $Y ∖ {y_{0}}$ ; adjoining and deleting the basepoint give the natural isomorphisms required by [L2], including the empty Kleisli object and the one-point algebra.

step 1.1step 1.2L2L5L6L7

3.1

Thus the comparison is an equivalence, and by [L6] the Kleisli right adjoint induces the maybe monad, so it is monadic by [L4].

step 2.1L2L4L6

4.1

By [L7] the strict image of the comparison is the full subcategory of free algebras. Under the fixed tagged-coproduct encoding, the pointed singleton $({0}, 0)$ is isomorphic to but not literally equal to the free pointed set on the empty set, whose point is the distinguished tag $⊥$ , so it is not in that image. The comparison is therefore not surjective on objects and is not an isomorphism by [L3], so strict monadicity fails by [L4].

step 2.1L3L4L7∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 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

E. Riehl, Category Theory in Context, 2nd ed., Examples 5.1.4(i), 5.2.11(i), and 5.3.2 (standard reference, not scraped)