A morphism in a category is said to be an **epimorphism** or **epic** if the relation , for any two morphisms and an arbitrary object , implies . It is not too difficult to see that the epimorphisms of the category of sets are exactly the surjective functions.

Informally, we said that a category is **set-based**, if its objects are sets with additional structures (e.g., semigroups, groups, vector spaces, rings, topological spaces or topological groups), and its morphisms are functions preserving structure.

Proposition 1.In any set-based category, surjective morphisms are epimorphisms.

*Proof:* Let be a surjective morphism in a set-based category . Now consider morphisms in with . Let . Since is surjective, there is an with . Thus . Hence .

In general, the converse of Proposition 1 is false. It is certainly true in the category of sets or in the category of topological spaces, but it is not in the category of semigroups; indeed, the inclusion is an epic. We also find non-surjective epimorphisms in the category of Hausdorff spaces and continuous maps.

In this post, we deal with the category of groups, and provide two proofs of the following result.

Theorem 2.In the category of groups, every epimorphism is surjective.

** First Proof. **

Let be an epimorphism in the category of groups. We recall that, in each category, if the composition is epic, so is . In view of the Theorem of canonical decomposition of group morphisms, there is thus no loss in generality if we suppose that is a subgroup of , and that is the inclusion morphism. First, we assume that the index of in is . Then is a normal subgroup of . Let be the quotient and constant morphisms, respectively. Then . Since is an epimorphism, it follows that . Thus the factor group is trivial. This contradicts . The assumption is so false. Now we suppose that there are at least three cosets of in . Then we find two elements such that the cosets , and are pairwise disjoint. Since is a bijection, the assignments , for , and for , yield a member of the group of all permutations of the set . For , let denote the right translation . Then and are group morphisms. For and , we compute

Thus

and

Thus . However

for , since . That is, . This contradicts the fact that is an epimorphism. Consequently, , that is, . The surjectivity of is so proved.

** Second Proof. **

As stressed in the preceding proof, it is sufficient to show that for a group and an epimorphically embedded subgroup , the equality holds. We consider a -module and form the semidirect product with multiplication . Let denote a function. If the set is a subgroup of , then, for all , we have

that is, . Conversely, if for all , then , and

Thus is a subgroup of if and only if the equality holds for all . Such functions are called **-cocycles**.

Assume that is a proper subgroup of . If there exists a non-constant -cocycle with , then the morphisms given by and agrees on , but are different. This contradicts the fact that is an epimorphically embedded subgroup. To conclude the proof, it is sufficient to find such a cocycle.

One source of cocycles are the so-called **coboundaries** defined by for some . Indeed,

and therefore

We are thus done if we find an element such that and for some . We take an arbitrary abelian group which contains at least two elements (for instance, for some prime number ) and set . Then is a -module via the action

given by . If , then there is an element . Consider a function such that . Then the composition of the quotient map after yields a function such that for all and . Thus for all and , but . Hence and .

Todd TrimbleMarch 8, 2015 / 10:39 pmHi; this is a nice post. I think however there is a little detail in the second proof that needs fixing. Namely, the left action of on isn’t actually a left action; I think you probably want instead .

Christian Nguembou TagneApril 4, 2018 / 11:24 pmThank you for your comment. But, I think it is really a left action. Indeed, for and , we compute

.

Hence .

Michael WeissJuly 21, 2017 / 10:28 pmThere’s another permutation-based proof, which I prefer. I think it was first published by Carl Linderholm (AMM 77(2) Feb. 1970 176–177). As already noted, it’s enough to show that if H is a proper subgroup of G, then there are two distinct homomorphisms from G to some other group that agree on the subgroup H. Elements of G act on the set of left cosets G/H by left translation; this gives a homomorphism from G to Perm(G/H), say g–>L_g. Now add a new element to G/H, call it A; let S be the new set, i.e., S is G/H union {A}. Extend L_g to S by declaring that L_g(A)=A. So L is a homomorphism G–>Perm(S).

Now let s be the permutation switching A and H (both elements of S). Let M_g be L_g conjugated by s, i.e., M_g(X)=s(L_g(s^{-1}(X))). (Since s^2=1, I don’t really have to write s^{-1}, but it really doesn’t matter.) Obviously M_g is also a homomorphism G–>Perm(S).

For h in H, L_h and s commute— obvious, since L_h leaves A and H fixed,

while s leaves everything but A and H fixed. So M_h and L_h are the same permutation. On the other hand, if g is not in H (so gH =/= H) then it’s immediate that L_g and s don’t commute: just compute sL_g(A) and compare it with L_gs(A) (or do the same with H instead of A). So L_g and M_g differ when g is not in H. QED

One reason I prefer this: you don’t have to treat G:H=2 as a special case.

Christian Nguembou TagneApril 4, 2018 / 11:12 pmThank you for your comment. I’m going to look this proof up.

Cyrille Nganteu(from cameroon)August 30, 2018 / 2:57 pmAlways a good mathematician. Short results but very usefull. Bravo.

evelynNovember 10, 2018 / 3:57 amIn the second proof, why use 1 and not general, {m(g_{0})\neq m(\mathbf{1})}. Or simply, why was it necessary to use 1 and not general?

Christian Nguembou TagneNovember 10, 2018 / 3:00 pmIn the last part of the second proof, we want to find a function from to , where is a prime number, such that and . Note that , as well as , is a function from to . Therefore, to show that , it suffices to establish that for some . However, by definition, . For example, if is the identity , we compute . Thus, if we find that , we can conclude that . That was our approach here. It is also not a necessity to use the identity of . One could theoretically use any element of the group . However, in this context, the identity is the optimal choice.