ExampleConstruction: AI-generatedVerification: AI-generatedprecheck 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.

A tagged disjoint union represents $X \mapsto S e t (A, X) \times S e t (B, X)$

Example

For sets $A$ and $B$ , put

$A ⊔ B = (A \times {0}) \cup (B \times {1})$

and define $i_{A} (a) = (a, 0)$ and $i_{B} (b) = (b, 1)$ . The covariant functor

$G (X) = S e t (A, X) \times S e t (B, X)$

is represented by $A ⊔ B$ . The representing isomorphism sends $q : A ⊔ B \to X$ to $(q \circ i_{A}, q \circ i_{B})$ .

Facts & Assumptions

Given: Sets $A, B$ , the category $S e t$ , and an arbitrary set $X$ .

[F1]

A covariant set-valued functor is represented by $R$ when it is naturally isomorphic to $S e t (R, -)$ (Presheaves, covariantly and contravariantly representable functors, and representations).

[F2]

Sets and functions form a category with ordinary composition (Sets and functions form the large locally small category $S e t$ ).

[F3]

Cartesian products contain exactly the ordered pairs with entries in the two factors, and membership in a binary union is membership in at least one of its two members (The Cartesian product $A \times B : = {z \in P (P (A \cup B)) : \exists a \in A \exists b \in B z = (a, b)}$ , The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ , $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[F4]

Ordered pairs satisfy $(a, s) = (b, t)$ if and only if $a = b$ and $s = t$ ; the natural numbers $0 = \emptyset$ and $1 = {0}$ are distinct, so the two tagged parts are disjoint (The Kuratowski ordered pair $(a, b) : = {{a}, {a, b}}$ , $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ , The natural numbers $N$ (von Neumann)).

[F5]

Two functions with the same domain and codomain are equal exactly when their values agree everywhere (Functions $f$ and $g$ are equal if and only if $dom f = dom g$ and $f (x) = g (x)$ for every $x$ in that common domain).

Verification

technique · constructive

1.1

Define $Φ_{X} (q) = (q \circ i_{A}, q \circ i_{B})$ for $q : A ⊔ B \to X$ .

F2construct

1.2

For $(g, h) \in G (X)$ define $[g, h] : A ⊔ B \to X$ by $[g, h] (a, 0) = g (a)$ and $[g, h] (b, 1) = h (b)$ . Every element has one of these forms by [F3], and the forms cannot overlap by [F4], so this is a function.

F3F4construct

2.1

Restricting $[g, h]$ along $i_{A}$ and $i_{B}$ gives $g$ and $h$ , so $Φ_{X} ([g, h]) = (g, h)$ .

step 1.1step 1.2

2.2

Every $z \in A ⊔ B$ is in exactly one tagged part; there $[q \circ i_{A}, q \circ i_{B}] (z) = q (z)$ . Thus $[q \circ i_{A}, q \circ i_{B}] = q$ by [F5].

step 1.1step 1.2F3F4F5

2.3

If $k : X \to Y$ , then $Φ_{Y} (k \circ q) = (k \circ q \circ i_{A}, k \circ q \circ i_{B})$ , obtained by applying $G (k)$ to $Φ_{X} (q)$ . Hence $Φ$ is natural.

step 1.1F2

3.1

Steps 2.1--2.3 give a natural isomorphism $S e t (A ⊔ B, -) ≅ G$ , so [F1] proves the claim, including $A = \emptyset$ or $B = \emptyset$ .

step 2.1step 2.2step 2.3F1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 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

Tom Leinster, Basic Category Theory, Section 3.1 (sums of sets) (standard reference, not scraped)

A tagged disjoint union represents X↦Set(A,X)×Set(B,X)X↦Set(A,X)×Set(B,X)