TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Normal-subgroup quotients of a fixed free group give a canonical solution set for the underlying-set functor on groups

Statement

Fix a set $S$ and a chosen free group $(F (S), i_{S})$ . For every normal subgroup $N ⊴ F (S)$ , let $η_{N} : S \overset{i_{S}}{\to} U (F (S)) \overset{U (q_{N})}{\to} U (F (S) / N) .$ The family $(η_{N})$ , indexed by the set of normal subgroups of the fixed group $F (S)$ , is a solution set at $S$ for the underlying-set functor $U : G r p \to S e t$ .

Facts & Assumptions

Given: A set $S$ and the chosen free group on $S$ .

[L1]

Every function $f : S \to U (G)$ extends uniquely to a homomorphism $\hat{f} : F (S) \to G$ (The free-group functor is left adjoint to the underlying-set functor).

[L2]

If a homomorphism kills a normal subgroup $N$ , it factors uniquely through the quotient by $N$ (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L3]

A subgroup $N \leq G$ is normal when it is invariant under conjugation by every element of $G$ ; in particular a normal subgroup is a subset of its ambient group (Normal subgroup: invariance under conjugation).

[L4]

A solution set at $S$ is a supplied set of arrows through one of which every arrow $S \to U (G)$ factors (The solution-set condition for a functor, stated object by object).

[L5]

For a group homomorphism, the image is a subgroup of the codomain and the kernel is a normal subgroup of the domain (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

Proof

technique · constructive

1.1L3L4construct

The normal subgroups of $F (S)$ form a set because they are among the subsets of the fixed underlying set. This includes the empty- $S$ case, where $F (S)$ is trivial. Hence the displayed quotient arrows form a supplied set-indexed family.

2.1step 1.1L1L2L5

Given $f : S \to U (G)$ , extend it by [L1] to $\hat{f} : F (S) \to G$ and put $N = \ker \hat{f}$ . By [L5], $N$ is a normal subgroup of $F (S)$ , so $\hat{f}$ kills $N$ and [L2] gives a unique $\overset{ˉ}{f} : F (S) / N \to G$ with $\hat{f} = \overset{ˉ}{f} q_{N}$ . Therefore $f = U (\overset{ˉ}{f}) η_{N}$ .

3.1step 1.1step 2.1L4discharge-construct∎

The factorisation in step 2.1 is exactly the clause of [L4]. The index $N$ is computed as a kernel rather than chosen from isomorphism representatives, so the family is canonical once the free group is chosen.

Depends on

Used by

Dependency tree · two levels

20 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

E. Riehl, Category Theory in Context, example 4.7.6 (standard reference, not scraped)
T. Leinster, Basic Category Theory, example 6.3.11 (standard reference, not scraped)