CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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.

Free products are unique up to a unique factor-compatible isomorphism

Statement

Any two free products of the same family are connected by a unique isomorphism commuting with every canonical factor map.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

For a family $(G_{i})_{i \in I}$ , a free product is a group $F$ with homomorphisms $ι_{i} : G_{i} \to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_{i} : G_{i} \to H$ , there is a unique homomorphism $f : F \to H$ satisfying $f \circ ι_{i} = f_{i}$ for all $i$ . It is denoted $*_{i \in I} G_{i}$ . Injectivity of the maps $ι_{i}$ is not part of this definition. (The free product of an arbitrary family of groups).

[L2]

Group isomorphisms, automorphisms and the set $Aut (G)$ . An isomorphism $f : G \to H$ is a bijective group homomorphism (def-group-homomorphism, def-injection-surjection-bijection). When $G = H$ , it is an automorphism of $G$ . Write $Aut (G) : = {f : G \to G : f is an automorphism} .$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1

The universal properties give unique factor-compatible homomorphisms $u : F \to F^{'}$ and $v : F^{'} \to F$ .

givenL1L2

2.1

Both $v \circ u$ and ${i d}_{F}$ agree with every factor map, so uniqueness gives $v \circ u = {i d}_{F}$ ; similarly $u \circ v = {i d}_{F^{'}}$ .

step 1.1

3.1

Hence $u$ is the unique compatible isomorphism. For the empty family both free products are trivial.

step 2.1∎

Depends on

Used by

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

George D. Torres, Combinatorial Group Theory, §2 (standard reference, not scraped)
B. H. Neumann, Lectures on Topics in the Theory of Infinite Groups, Ch. 9 (standard reference, not scraped)