CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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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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Two formally distinct words can become the same object

Counterexample

Let $M$ be the one-object category with object $*$ and endomorphism monoid ${1, u}$ , where $u \neq 1$ and $u^{2} = u$ . Define the tensor on objects by $* \otimes * = *$ and on morphisms by the same monoid multiplication. Then $M$ is a strict monoidal category in which the formal words $x$ and $(x \otimes 1)$ evaluate to the same object $*$ .

Facts & Assumptions

Given: The one-object strict monoidal category $M$ just described.

[L1]

In a strict monoidal category, the associator and unitors are identities, so $x$ and $(x \otimes 1)$ evaluate to the same object on the nose (Strict monoidal category).

[L2]

The slogan "every diagram commutes" fails because formally different vertices can coincide and then a noncanonical endomorphism may be inserted (Why 'every diagram commutes' is false as stated).

Verification

technique · counterexample

1.1L1

By [L1], the two formal vertices $x$ and $(x \otimes 1)$ both evaluate to the single object $*$ of $M$ .

1.2given

Consider the square all of whose vertices are $*$ , whose top edge is $u$ , and whose other three edges are $1$ . The clockwise composite is $u$ , while the counterclockwise composite is $1$ , so the square does not commute because $u \neq 1$ .

2.1L2step 1.1step 1.2∎

This realizes the mechanism described in [L2]: formally distinct words can collapse to one object, and a noncanonical endomorphism then breaks the blanket slogan. Hence the slogan is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

S. Mac Lane, Categories for the Working Mathematician, Chapter VII.2 (standard reference, not scraped)