Epimorphisms of groups

In my previous post, we saw that injections (surjections) in concrete categories are always monomorphisms (epimorphisms), and in some cases the converse holds.

We now wish to classify all epimorphisms of groups. To show that all epimorphisms are surjective, for any strict subgroup H \subseteq G we want to construct maps f_1, f_2 \colon G \to G' to some group G' that differ on G but agree on H. In the case of abelian groups this is relatively easy, because we can take G' to be the cokernel, f_1 the quotient map, and f_2 the zero map. But in general the cokernel only exists if the image is normal, so a different argument is needed.

Lemma. Let f \colon H \to G be a group homomorphism. Then f is an epimorphism if and only if f is surjective.

Proof. We already saw that surjections are epimorphisms. Conversely, let f \colon H \to G be an epimorphism of groups. We may replace H by its image in G, since the map \im(f) \to G is still an epimorphism. Let X = G/H be the coset space, viewed as a pointed set with distinguished element * = H. Let Y = X \amalg_{X\setminus *} X be the set “X with the distinguished point doubled”, and write *_1 and *_2 for these distinguished points.

Let S(Y) be the symmetric group on Y, and define homomorphisms f_i \colon G \to S(Y) by letting G act naturally on the i^{\text{th}} copy of X in Y (for i \in \{1,2\}). Since the action of H on X = G/H fixes the trivial coset *, we see that the maps f_i|_H agree. Since f is an epimorphism, this forces f_1 = f_2. But then

    \[H = \Stab_{f_1}(*_1) = \Stab_{f_2}(*_1) = G,\]

showing that f is surjective (and a fortiori X = \{*\}). \qedsymbol

Note however that the result is not true in every algebraic category. For example, the map \mathbf Z \to \mathbf Q is an epimorphism of (commutative) rings that is not surjective. More generally, every localisation R \to R[S^{-1}] is an epimorphism, by the universal property of localisation; these maps are rarely surjective.

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