TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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

1.1F2algebra

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 $[G, G] \subseteq \ker f$ .

1.2F1F2algebra

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.

2.1step 1.1F1F3

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

3.1step 1.2step 2.1construct

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 $q_{G}$ .

4.1step 3.1F3

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.

5.1step 2.1step 4.1L1∎

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

Depends on

Used by

Dependency tree · two levels

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