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epimorphisms of groups are surjective

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Statement

Proposition

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).

Proof

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

GprojG/HA G \stackrel{proj}{\to} G/H \to A

is a constant function, i.e., is fixed under the action of GG, or in other words that the map ϕ m:GA G\phi_m: G \to A^G mapping gGg \in G to gmmg \cdot m - m is identically 00. As ϕ m\phi_m is an A GA^G-valued derivation, it yields a splitting of the exact sequence of groups

0A GGA GG10 \to A^G \to G \ltimes A^G \to G \to 1

given by k:GGA Gk: G \to G \ltimes A^G, with k(g)(g,ϕ m(g))k(g) \coloneqq (g, \phi_m(g)). Of course we also have the trivial splitting j(g)(g,0)j(g) \coloneqq (g, 0). But note that the restrictions along i:HGi: H \hookrightarrow G coincide: ji=kij \circ i = k \circ i since ϕ m(h)=0\phi_m(h) = 0 for all hHh \in H (as HH acts trivially on the right on cosets gHg H). Since ii is epic, we conclude j=kj = k, or that ϕ m0\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.