TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Abelianisation is left adjoint to the inclusion of abelian groups

Statement

Abelianisation defines a functor $(-)^{a b} : G r p \to A b$ , and it is left adjoint to the inclusion $I : A b ↪ G r p$ . Explicitly, every homomorphism $f : G \to I (A)$ with $A$ abelian factors uniquely through the quotient $q_{G} : G \to G^{a b}$ .

Facts & Assumptions

Given: A group $G$ , an abelian group $A$ , and a homomorphism $f : G \to A$ .

[F1]

The abelianisation of $G$ is $G^{a b} : = G / [G, G]$ (The abelianisation $G^{a b} : = G / [G, G]$ and its canonical map).

[F2]

For $g, h \in G$ , their commutator is $[g, h] : = g h g^{- 1} h^{- 1}$ , and $[G, G]$ is generated by all commutators (Commutators $[g, h] = g h g^{- 1} h^{- 1}$ and the commutator subgroup $[G, G]$ ).

[F3]

A homomorphism that kills a normal subgroup factors uniquely through the quotient group (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L1]

Chosen objectwise universal arrows assemble uniquely into a left adjoint (Chosen objectwise universal arrows assemble uniquely into a left adjoint).

Proof

technique · direct

1.1

Since $A$ is abelian, $f ([g, h]) = f (g) f (h) f (g)^{- 1} f (h)^{- 1} = 1_{A}$ for all $g, h \in G$ , so $[G, G] \subseteq \ker f$ .

F2algebra

1.2

The group $G^{a b}$ is abelian: $q_{G}$ has kernel $[G, G]$ by [F1], so for $g, h \in G$ the commutator $q_{G} (g) q_{G} (h) q_{G} (g)^{- 1} q_{G} (h)^{- 1} = q_{G} ([g, h])$ is trivial by [F2], and $q_{G}$ is surjective.

F1F2algebra

2.1

By [F3], there is a unique homomorphism $\overset{ˉ}{f} : G^{a b} \to A$ with $\overset{ˉ}{f} q_{G} = f$ .

step 1.1F1F3

3.1

For a homomorphism $a : G \to H$ , the target $H^{a b}$ is abelian by step 1.2, so step 2.1 applies to $q_{H} a : G \to H^{a b}$ ; define $a^{a b}$ as its unique factor through $q_{G}$ .

step 1.2step 2.1construct

4.1

Uniqueness in [F3] gives $(1_{G})^{a b} = 1$ and $(b a)^{a b} = b^{a b} a^{a b}$ , so abelianisation is a functor and $q$ is natural.

step 3.1F3

5.1

Step 2.1 is the universal-arrow property of $(G^{a b}, q_{G})$ from $G$ to the inclusion $I$ ; [L1] therefore gives $(-)^{a b} ⊣ I$ .

step 2.1step 4.1L1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Tom Leinster, Basic Category Theory, Example 2.1.3 (standard reference, not scraped)
Emily Riehl, Category Theory in Context, 2nd ed., Example 4.1.10 (standard reference, not scraped)