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

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.

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Endofunctor composition as a strict tensor product

Example

Let $C$ be the discrete category on the two objects $0$ and $1$ . Define endofunctors $S, T : C \to C$ by $S (0) = 1$ , $S (1) = 0$ , $T (0) = 0$ , and $T (1) = 0$ . Since $C$ is discrete, these object assignments determine the whole functors.

Facts & Assumptions

Given: The endofunctor category of a small category under composition.

[L1]

For a small category, endofunctor composition makes the endofunctor category strict monoidal (The endofunctor category of a small category is strict monoidal under composition).

Verification

technique · direct

1.1

The category $C$ is small, so [L1] applies. Its endofunctors are determined by their action on the two objects because every morphism is an identity.

2.1

The composites are easy to compute: $S \circ S = 1_{C}$ , $T \circ S = T$ , and $S \circ T$ is the constant functor at $1$ . Hence $((S \circ T) \circ S) = S \circ (T \circ S)$ as literal equalities of functors, and $1_{C} \circ S = S = S \circ 1_{C}$ .

step 1.1

3.1

This concrete pair therefore realizes composition as a strict tensor product.

step 2.1L1∎

Depends on

The endofunctor category of a small category is strict monoidal under composition

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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