CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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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.

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A limit in a full subcategory need not be the ambient limit

Statement refuted

The inclusion of every full subcategory preserves all limits that exist in the subcategory and the ambient category.

Facts & Assumptions

Given: The poset $P = {0, q, m, a, b}$ with $0 < q < m < a$ , $0 < q < m < b$ , and $a, b$ incomparable, and its full subposet $Q = {0, q, a, b}$ .

[F1]

Products represent common lower bounds universally (Products and coproducts as limits and colimits of discrete diagrams, including their existence-and-uniqueness equations).

[F2]

A full functor is surjective on every hom-set map (Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors).

[F3]

Posets form categories with an arrow exactly for an order relation (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

[L1]

Preservation is detected by the canonical comparison to the ambient chosen limit (A functor preserves a chosen limit exactly when its canonical comparison to the chosen target limit is an isomorphism, and dually for colimits).

Counterexample

technique · finite posets

1.1

The inclusion $i : Q ↪ P$ is full: for retained objects, an arrow exists in $Q$ exactly when it exists in $P$ .

F2F3

1.2

In $P$ , the greatest common lower bound of $a, b$ is $m$ , so [F1] makes $m = a \times b$ . In $Q$ , the object $m$ is absent and the greatest common lower bound of $a, b$ is $q$ , so $q$ is their product there.

F1F3

2.1

The image under $i$ of the product cone in $Q$ has apex $q$ , whereas the ambient product has apex $m$ . Its canonical comparison is $q \to m$ , not an isomorphism because $m ≰ q$ . By [L1], the full inclusion does not preserve this product, refuting the statement.

L1step 1.1step 1.2∎

Depends on

Used by

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Dependency tree · two levels

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