TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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 hom-assignment $C (-, -) : C^{o p} \times C \to S e t$ is a bifunctor

Statement

For every locally small category $C$ , the hom-assignment

$C (-, -) : C^{o p} \times C ⟶ S e t$

of The covariant and contravariant hom-assignments and the hom-bifunctor of a locally small category is a functor. Its restrictions in the two variables are the contravariant and covariant hom-functors of The assignments $C (a, -)$ and $C (-, a)$ are functors to $S e t$ .

Facts & Assumptions

Given: A locally small category $C$ ; morphisms $h : a^{'} \to a$ , $k : a^{''} \to a^{'}$ , $u : b \to b^{'}$ , and $v : b^{'} \to b^{''}$ ; and the category axioms of $C$ .

[L1]

The one-variable assignments $C (a, -)$ and $C (-, a)$ are functors to $S e t$ (The assignments $C (a, -)$ and $C (-, a)$ are functors to $S e t$ ).

[F1]

A morphism in a product category is a pair, and identities and composition are componentwise (Product category and its projection functors).

[F2]

The hom-assignment sends $(h, u)$ to $f \mapsto u \circ f \circ h$ (The covariant and contravariant hom-assignments and the hom-bifunctor of a locally small category).

Proof

technique · direct

1.1givenF2

If $f : a \to b$ , then $u \circ f \circ h : a^{'} \to b^{'}$ , so [F2] defines a function $C (a, b) \to C (a^{'}, b^{'})$ .

1.2givenF1F2

For every $f : a \to b$ , the identity pair acts by $1_{b} \circ f \circ 1_{a} = f$ .

1.3givenF1F2

Applying $(h, u)$ and then $(k, v)$ sends $f$ to $v \circ (u \circ f \circ h) \circ k = (v \circ u) \circ f \circ (h \circ k)$ , which is the action of their componentwise composite in $C^{o p} \times C$ .

2.1step 1.1step 1.2step 1.3L1∎

Steps 1.1--1.3 prove the functor laws, and fixing either variable recovers the postcomposition or precomposition action of [L1]; hence the hom-assignment is the asserted bifunctor.

The hom-assignment $C (-, -) : C^{o p} \times C \to S e t$ is a bifunctor

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The hom-assignment C(−,−):Cop×C→SetC(−,−):Cop×C→Set is a bifunctor

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The hom-assignment $C (-, -) : C^{o p} \times C \to S e t$ is a bifunctor