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