ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The adjoint functor theorem for ordered sets

Example

Let $A$ and $B$ be complete partially ordered sets, and let $g : B \to A$ preserve arbitrary meets, including the empty meet. Then $g$ has a left adjoint $f : A \to B$ , given by $f (a) = ⋀ {b \in B : a \leq g (b)} .$

Facts & Assumptions

Given: Complete posets $A, B$ and a meet-preserving monotone map $g : B \to A$ .

[L1]

Preorders are thin categories and monotone maps are exactly the functors between them (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

[L2]

A Galois connection $f ⊣ g$ is characterised by $f (a) \leq b$ if and only if $a \leq g (b)$ (Galois connection between preorders).

[L3]

Completeness means that all small limits exist (Finite, small, and large limits and colimits; complete and cocomplete categories).

Verification

technique · constructive

1.1

For each $a \in A$ , let $S_{a} = {b \in B : a \leq g (b)}$ . It is nonempty: because $g$ preserves the empty meet, $g (⊤_{B}) = ⊤_{A}$ , so $a \leq g (⊤_{B})$ . Completeness [L3] therefore supplies $f (a) = ⋀ S_{a}$ .

L3construct

2.1

If $a \leq a^{'}$ , then $S_{a^{'}} \subseteq S_{a}$ , so $⋀ S_{a} \leq ⋀ S_{a^{'}}$ . Hence $f$ is monotone and therefore a functor under [L1].

step 1.1L1

2.2

If $a \leq g (b)$ , then $b \in S_{a}$ , so $f (a) = ⋀ S_{a} \leq b$ .

step 1.1

2.3

Conversely, since $g$ preserves the meet of $S_{a}$ , one has $g (f (a)) = ⋀_{c \in S_{a}} g (c)$ . Every $g (c)$ on the right lies above $a$ , hence $a \leq g (f (a))$ . Thus $f (a) \leq b$ implies $a \leq g (f (a)) \leq g (b)$ by monotonicity.

step 1.1

3.1

Steps 2.2 and 2.3 give the equivalence in [L2] for every $a, b$ , so $f ⊣ g$ . The empty-meet case in step 1.1 is what prevents the defining set from being empty.

step 2.2step 2.3L2discharge-construct∎

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 5 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

T. Leinster, Basic Category Theory, example 6.3.14 (standard reference, not scraped)