小科普之epimorphism in Grp is surjective

This proof is due to Arturo Magidin.

In category theory, as we don't consider the 'internal structure' of objects, many properties of maps are defined in an abstract way using interaction with other morphisms. 'Epimorphism' is such an analog of surjection. A f:A\rightarrow B is called epimorphism if for all C and g_1:B\rightarrow C, g_2:B\rightarrow C with g_1\circ f=g_2\circ f we have g_1=g_2. It's easy to demonstrate that in the category \mathrm{Set} epimorphisms are precisely surjections. Yet that's not always the case: in category \mathrm{Ring}, i:\mathbb Z\hookrightarrow \mathbb Q, the inclusion map from \mathbb Z into \mathbb Q, is an epimorphism, since any ring homomorphism f:\mathbb Q\rightarrow R is determined uniquely by f\restriction \mathbb Z. In category \mathrm{Grp}, however, it's possible, though not trivially, to demonstrate that any epimorphism is surjective, using our favorite toy of group actions.

The idea is pretty primitive: for any epimorphism f:G\rightarrow H, f(G) is a subgroup of H, and then H can be considered as a permutation group of the set of left cosets of f(G), denoted as X, via Hacting on X by left translation. If you have learnt any group theory with Sylow theorems, you should be quite familiar with this trick. Hence there's already one g_1:H\rightarrow \mathrm{Sym}(Y) for some Y\supset X, and g_1\circ f(x) keeps f(G) fixed for all x\in G. If we can have another g_2 with g_2\circ f=g_1\circ f, we can conclude g_1=g_2. When finding another homomorphism with one existing, one natural thought is to conjugate the image, namely we try if g_2(y)=\sigma^{-1}g_1(y)\sigma for some \sigma\in\mathrm{Sym}(Y) can work. Luckily it's easy to make g_2\circ f=g_1\circ f with this simple thought: as g_1(f(x)) fixes f(G), as long as \sigma keeps other cosets fixed the equation holds. Now we return to the property of permutations. When \sigma is a transpose, if g_1(y)=\sigma^{-1}g_1(y)\sigma, every g_1(y) should keep the support of \sigma fixed or contain \sigma as a disjoint cycle. The latter case can be excluded when \sigma permutes some element outside X. The construction is then obvious: \sigma=(f(G)\ \mathrm{zyq}) can work for some zyq not in X!