Exercise 5.2.26


In , the inclusion is monic and epic but not an isomorphism. If we slightly strengthen the requirements, however, we obtain:

Theorem: A map is an isomorphism if and only if it is monic and \underline{regular} epic.

Proof: Assume is monic and regular epic. Then it is the coequalizer of some parallel pair . This means that , but since is monic, . Since this is the case, equalizes and , so we obtain a unique map such that . Then, . Since is epic (regular epic epic), we have . So, is an isomorphism.

Suppose is an isomorphism. It’s trivially monic, since given , we can left-compose by to obtain , and similarly for epic. To show it is regular epic, we must exhibit it as the coequalizer of a parallel pair. Take the parallel pair . Clearly coequalizes the pair. Now suppose coequalizes the pair (any map does, trivially). Define as . Clearly this makes the appropriate triangle commute, since . Suppose there were another map with . Right compose with to get . So is the unique such map and is a coequalizer.