epimorphisms of groups are surjective

- group, ∞-group
- group object, group object in an (∞,1)-category
- abelian group, spectrum
- group action, ∞-action
- representation, ∞-representation
- progroup
- homogeneous space

Every epimorphism in the category of groups is surjective (a regular epimorphism). Equivalently, the category of groups is balanced (every monic epimorphism is an isomorphism).

There are probably many proofs of this statement. Some (for example, I think the one in Mac Lane's *Categories for the Working Mathematician*) involve some mild case analysis. The following (originally written up here) is more uniform (and constructive).

Suppose $i: H \hookrightarrow G$ is a monic epi. Let $A$ be a nontrivial abelian group, say $\mathbb{Z}/(2)$. There is a left action of $G$ on $A^G = \prod_{g \in G} A$, the group of *functions* $G \to A$, where $(g \cdot \phi)(g') \coloneqq \phi(g'g^{-1})$; this gives rise to a left $G$-module. It suffices to prove that any element $m \in A^G$ of the form

$G \stackrel{proj}{\to} G/H \to A$

is a constant function, i.e., is fixed under the action of $G$, or in other words that the map $\phi_m: G \to A^G$ mapping $g \in G$ to $g \cdot m - m$ is identically $0$. As $\phi_m$ is an $A^G$-valued derivation, it yields a splitting of the exact sequence of groups

$0 \to A^G \to G \ltimes A^G \to G \to 1$

given by $k: G \to G \ltimes A^G$, with $k(g) \coloneqq (g, \phi_m(g))$. Of course we also have the trivial splitting $j(g) \coloneqq (g, 0)$. But note that the restrictions along $i: H \hookrightarrow G$ coincide: $j \circ i = k \circ i$ since $\phi_m(h) = 0$ for all $h \in H$ (as $H$ acts trivially on the right on cosets $g H$). Since $i$ is epic, we conclude $j = k$, or that $\phi_m \equiv 0$, as was to be shown.

Last revised on November 6, 2020 at 13:09:20. See the history of this page for a list of all contributions to it.