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.

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Each factor is a retract of a free product when all other factors are sent trivially

Statement

For every $i \in I$ , the factor $G_{i}$ is a retract of $*_{j \in I} G_{j}$ : there is $r_{i} : *_{j} G_{j} \to G_{i}$ with $r_{i} \circ ι_{i} = {i d}_{G_{i}}$ .

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]

Every canonical factor homomorphism $ι_{i} : G_{i} \to *_{j} G_{j}$ is injective. (Every canonical factor map into a free product is injective).

[L3]

Let $(M, \cdot, e)$ and $(M^{'}, \cdot^{'}, e^{'})$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M^{'}$ is a function $f : M \to M^{'}$ such that - (H1) $f (x \cdot y) = f (x) \cdot^{'} f (y)$ for all $x, y \in M$ ; - (H2) $f (e) = e^{'}$ . Let $G$ and $G^{'}$ be groups (def-group). A group homomorphism from $G$ to $G^{'}$ is a function $f : G \to G^{'}$ satisfying (H1) alone: $f (x y) = f (x) f (y) for all x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (x^{- 1}) = f (x)^{- 1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f) (x y) = g (f (x) f (y)) = g (f (x)) g (f (y))$ and $(g \circ f) (e) = g (e^{'}) = e^{''}$ ; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

Proof

technique · direct

1.1

Use the identity homomorphism on $G_{i}$ and the trivial homomorphism $G_{j} \to G_{i}$ for every $j \neq i$ .

givenL1L2L3

2.1

Free-product universality gives a unique $r_{i}$ extending this family, and its defining equation is $r_{i} \circ ι_{i} = {i d}_{G_{i}}$ .

step 1.1

3.1

Thus $ι_{i}$ is a section and $G_{i}$ is a retract. The assertion is made only for an index $i \in I$ .

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)